The Imomi archipelago consists of $n \geqslant 2$ islands. Between each pair of distinct islands is a unique ferry line that runs in both directions, and each ferry line is operated by one of $k$ companies. It is known that if any one of the $k$ companies closes all its ferry lines, then it becomes impossible for a traveller, no matter where the traveller starts at, to visit all the islands exactly once (in particular, not returning to the island the traveller started at).  Determine the maximal possible value of $k$ in terms of $n$. (Ukraine)

Solution: We reformulate the problem using graph theory. We have a complete graph $K_{n}$ on $n$ nodes (corresponding to islands), and we want to colour the edges (corresponding to ferry lines) with $k$ colours (corresponding to companies), so that every Hamiltonian path contains all $k$ different colours. For a fixed set of $k$ colours, we say that an edge colouring of $K_{n}$ is good if every Hamiltonian path contains an edge of each one of these $k$ colours.  We first construct a good colouring of $K_{n}$ using $k=\left\lfloor\log _{2} n\right\rfloor$ colours. Claim 1. Take $k=\left\lfloor\log _{2} n\right\rfloor$. Consider the complete graph $K_{n}$ in which the nodes are labelled by $1,2, \ldots, n$. Colour node $i$ with colour $\min \left(\left\lfloor\log _{2} i\right\rfloor+1, k\right)$ (so the colours of the first nodes are $1,2,2,3,3,3,3,4, \ldots$ and the last $n-2^{k-1}+1$ nodes have colour $k$ ), and for $1 \leqslant i<j \leqslant n$, colour the edge $i j$ with the colour of the node $i$. Then the resulting edge colouring of $K_{n}$ is good. Proof. We need to check that every Hamiltonian path contains edges of every single colour. We first observe that the number of nodes assigned colour $k$ is $n-2^{k-1}+1$. Since $n \geqslant 2^{k}$, we have  $$ n-2^{k-1}+1 \geqslant \frac{n}{2}+1 $$  This implies that in any Hamiltonian path, there exists an edge between two nodes with colour $k$. Then that edge must have colour $k$.  We next show that for each $1 \leqslant i<k$, every Hamiltonian path contains an edge of colour $i$. Suppose the contrary, that some Hamiltonian path does not contain an edge of colour $i$. Then nodes with colour $i$ can only be adjacent to nodes with colour less than $i$ inside the Hamiltonian path. Since there are $2^{i-1}$ nodes with colour $i$ and $2^{i-1}-1$ nodes with colour less than $i$, the Hamiltonian path must take the form  $$ (i) \leftrightarrow(<i) \leftrightarrow(i) \leftrightarrow(<i) \leftrightarrow \cdots \leftrightarrow(<i) \leftrightarrow(i), $$  where $(i)$ denotes a node with colour $i,(<i)$ denotes a node with colour less than $i$, and $\leftrightarrow$ denotes an edge. But this is impossible, as the Hamiltonian path would not have any nodes with colours greater than $i$.  Fix a set of $k$ colours, we now prove that if there exists a good colouring of $K_{n}$, then $k \leqslant\left\lfloor\log _{2} n\right\rfloor$. For $n=2$, this is trivial, so we assume $n \geqslant 3$. For any node $v$ of $K_{n}$ and $1 \leqslant i \leqslant k$, we denote by $d_{i}(v)$ the number of edges with colour $i$ incident with the node $v$. Lemma 1. Consider a good colouring of $K_{n}$, and let $A B$ be an arbitrary edge with colour $i$. If $d_{i}(A)+d_{i}(B) \leqslant n-1$, then the colouring will remain good after recolouring edge $A B$ with any other colour. Proof. Suppose there exists a good colouring together with an edge $A B$ of colour $i$, such that if $A B$ is recoloured with another colour, the colouring will no longer be good. The failure of the new colouring being good will come from colour $i$, and thus there exists a Hamiltonian path containing edge $A B$ such that initially (i.e. before recolouring) $A B$ is the only edge of colour $i$ in this path. Writing $A=A_{0}$ and $B=B_{0}$, denote this Hamiltonian path by  $$ A_{s} \leftrightarrow A_{s-1} \leftrightarrow \cdots \leftrightarrow A_{1} \leftrightarrow A_{0} \leftrightarrow B_{0} \leftrightarrow B_{1} \leftrightarrow \cdots \leftrightarrow B_{t-1} \leftrightarrow B_{t}, $$  where $s, t \geqslant 0$ and $s+t+2=n$. In the initial colouring, we observe the following.  - The edge $B_{0} A_{s}$ must have colour $i$, since otherwise the path  $$ A_{0} \leftrightarrow A_{1} \leftrightarrow \cdots \leftrightarrow A_{s-1} \leftrightarrow A_{s} \leftrightarrow B_{0} \leftrightarrow B_{1} \leftrightarrow \cdots \leftrightarrow B_{t-1} \leftrightarrow B_{t} $$  has no edges of colour $i$.  - Similarly, the edge $A_{0} B_{t}$ must have colour $i$. - For each $0 \leqslant p<s$, at least one of the edges $B_{0} A_{p}$ and $A_{0} A_{p+1}$ must have colour $i$, since otherwise the path  $$ A_{s} \leftrightarrow \cdots \leftrightarrow A_{p+2} \leftrightarrow A_{p+1} \leftrightarrow A_{0} \leftrightarrow A_{1} \leftrightarrow \cdots \leftrightarrow A_{p-1} \leftrightarrow A_{p} \leftrightarrow B_{0} \leftrightarrow B_{1} \leftrightarrow \cdots \leftrightarrow B_{t} $$  has no edges of colour $i$.  - Similarly, for each $0 \leqslant q<t$, at least one of the edges $A_{0} B_{q}$ and $B_{0} B_{q+1}$ must have colour $i$.  In the above list, each edge $A_{0} X$ appears exactly once and also each edge $B_{0} X$ appears exactly once (where $A_{0} B_{0}$ and $B_{0} A_{0}$ are counted separately). Adding up the contributions to $d_{i}(A)+$ $d_{i}(B)$, we obtain  $$ d_{i}(A)+d_{i}(B) \geqslant(s+1)+(t+1)=n . $$  This contradicts our assumption that $d_{i}(A)+d_{i}(B) \leqslant n-1$. Our strategy now is to repeatedly recolour the edges using Lemma 1 until the colouring has a simple structure. For a node $v$, we define $m(v)$ to be the largest value of $d_{i}(v)$ over all colours $i$. Lemma 2. Assume we have a good colouring of $K_{n}$. Let $A, B$ be two distinct nodes, and let $j$ be the colour of edge $A B$ where $1 \leqslant j \leqslant k$. If  - $m(A) \geqslant m(B)$ and - $m(A)=d_{i}(A)$ for some $i \neq j$, then after recolouring edge $A B$ with colour $i$, the colouring remains good. Proof. Note that  $$ d_{j}(A)+d_{j}(B) \leqslant(n-1-m(A))+m(B) \leqslant n-1, $$  and so we may apply Lemma 1. Lemma 3. Assume we have a good colouring of $K_{n}$. Let $S$ be a nonempty set of nodes. Let $A \in S$ be a node such that $m(A) \geqslant m(B)$ for all $B \in S$, and choose $1 \leqslant i \leqslant k$ for which $d_{i}(A)=m(A)$. Then after recolouring the edge $A B$ with colour $i$ for all $B \in S$ distinct from $A$, the colouring remains good. Proof. We repeatedly perform the following operation until all edges $A B$ with $B \in S$ have colour $i$ : choose an edge $A B$ with $B \in S$ that does not have colour $i$, and recolour it with colour $i$. By Lemma 2, the colouring remains good after one operation. Moreover, $m(A)$ increase by 1 during an operation, and all other $m(B)$ may increase by at most 1 . This shows that $m(A)$ will remain maximal amongst $m(B)$ for $B \in S$. We will also have $d_{i}(A)=m(A)$ after the operation, since both sides increase by 1 . Therefore the operation can be performed repeatedly, and the colouring remains good.  We first apply Lemma 3 to the set of all $n$ nodes in $K_{n}$. After recolouring, there exists a node $A_{1}$ such that every edge incident with $A_{1}$ has colour $c_{1}$. We then apply Lemma 3 to the set of nodes excluding $A_{1}$, and we obtain a colouring where  - every edge incident with $A_{1}$ has colour $c_{1}$, - every edge incident with $A_{2}$ except for $A_{1} A_{2}$ has colour $c_{2}$.  Repeating this process, we arrive at the following configuration:  - the $n$ nodes of $K_{n}$ are labelled $A_{1}, A_{2}, \ldots, A_{n}$, - the node $A_{i}$ has a corresponding colour $c_{i}$ (as a convention, we also colour $A_{i}$ with $c_{i}$ ), - for all $1 \leqslant u<v \leqslant n$, the edge between $A_{u}$ and $A_{v}$ has colour $c_{u}$, - this colouring is good.  Claim 2. For every colour $i$, there exists a $1 \leqslant p \leqslant n$ such that the number of nodes of colour $i$ amongst $A_{1}, \ldots, A_{p}$ is greater than $p / 2$. Proof. Suppose the contrary, that for every $1 \leqslant p \leqslant n$, there are at most $\lfloor p / 2\rfloor$ nodes of colour $i$. We then construct a Hamiltonian path not containing any edge of colour $i$. Let $A_{x_{1}}, \ldots, A_{x_{t}}$ be the nodes with colour $i$, where $x_{1}<x_{2}<\cdots<x_{t}$, and let $A_{y_{1}}, A_{y_{2}}, \ldots, A_{y_{s}}$ be the nodes with colour different from $i$, where $y_{1}<y_{2}<\cdots<y_{s}$. We have $s+t=n$ and $t \leqslant\lfloor n / 2\rfloor$, so $t \leqslant s$. We also see that $y_{j}<x_{j}$ for all $1 \leqslant j \leqslant t$, because otherwise, $A_{1}, A_{2}, \ldots, A_{x_{j}}$ will have $j$ nodes of colour $i$ and less than $j$ nodes of colour different from $i$. Then we can construct a Hamiltonian path  $$ A_{x_{1}} \leftrightarrow A_{y_{1}} \leftrightarrow A_{x_{2}} \leftrightarrow A_{y_{2}} \leftrightarrow A_{x_{3}} \leftrightarrow \cdots \leftrightarrow A_{x_{t}} \leftrightarrow A_{y_{t}} \leftrightarrow A_{y_{t+1}} \leftrightarrow \cdots \leftrightarrow A_{y_{s}} $$  that does not contain an edge with colour $i$. This contradicts that the colouring is good. So for every colour $i$, there has to be an integer $p_{i}$ with $1 \leqslant p_{i} \leqslant n$ such that there are more than $p_{i} / 2$ nodes assigned colour $i$ amongst $A_{1}, \ldots, A_{p_{i}}$. Choose the smallest such $p_{i}$ for every $i$, and without loss of generality assume  $$ p_{1}<p_{2}<\cdots<p_{k} $$  Note that the inequalities are strict by the definition of $p_{i}$. Then amongst the nodes $A_{1}, \ldots, A_{p_{i}}$, there are at least $\left\lceil\left(p_{j}+1\right) / 2\right\rceil$ nodes of colour $j$ for all $1 \leqslant j \leqslant i$. Then  $$ p_{i} \geqslant\left\lceil\frac{p_{1}+1}{2}\right\rceil+\left\lceil\frac{p_{2}+1}{2}\right\rceil+\cdots+\left\lceil\frac{p_{i}+1}{2}\right\rceil . $$  This inductively shows that  $$ p_{i} \geqslant 2^{i}-1 $$  for all $1 \leqslant i \leqslant k$, and this already proves $n \geqslant 2^{k}-1$. It remains to show that $n=2^{k}-1$ is not possible. If $n=2^{k}-1$, then all inequalities have to be equalities, so $p_{i}=2^{i}-1$ and there must be exactly $2^{i-1}$ nodes of colour $i$. Moreover, there cannot be a node of colour $i$ amongst $A_{1}, A_{2}, \ldots, A_{p_{i-1}}$, and so the set of nodes of colour $i$ must precisely be  $$ A_{2^{i-1}}, A_{2^{i-1}+1}, \ldots, A_{2^{i}-1} $$  Then we can form a Hamiltonian path  $$ A_{2^{k-1}} \leftrightarrow A_{1} \leftrightarrow A_{2^{k-1}+1} \leftrightarrow A_{2} \leftrightarrow A_{2^{k-1}+2} \leftrightarrow A_{3} \leftrightarrow \ldots \leftrightarrow A_{n} $$  which does not contain an edge of colour $k$. This is a contradiction, and therefore $n \geqslant 2^{k}$.


Let $A B C D E$ be a convex pentagon such that $\angle A B C=\angle A E D=90^{\circ}$. Suppose that the midpoint of $C D$ is the circumcentre of triangle $A B E$. Let $O$ be the circumcentre of triangle $A C D$.  Prove that line $A O$ passes through the midpoint of segment $B E$. (Slovakia)


Let $A B C$ be a triangle with $A C>B C$. Let $\omega$ be the circumcircle of triangle $A B C$ and let $r$ be the radius of $\omega$. Point $P$ lies on segment $A C$ such that $B C=C P$ and point $S$ is the foot of the perpendicular from $P$ to line $A B$. Let ray $B P$ intersect $\omega$ again at $D$ and let $Q$ lie on line $S P$ such that $P Q=r$ and $S, P, Q$ lie on the line in that order. Finally, let the line perpendicular to $C Q$ from $A$ intersect the line perpendicular to $D Q$ from $B$ at $E$.  Prove that $E$ lies on $\omega$.


Let $A B C D$ be a cyclic quadrilateral with $\angle B A D<\angle A D C$. Let $M$ be the midpoint of the arc $C D$ not containing $A$. Suppose there is a point $P$ inside $A B C D$ such that $\angle A D B=$ $\angle C P D$ and $\angle A D P=\angle P C B$.  Prove that lines $A D, P M, B C$ are concurrent.

Solution: Let $X$ and $Y$ be the intersection points of $A M$ and $B M$ with $P D$ and $P C$ respectively. Since $A B C M D$ is cyclic and $C M=M D$, we have  $$ \angle X A D=\angle M A D=\angle C B M=\angle C B Y $$  Combining this with $\angle A D X=\angle Y C B$, we get $\angle D X A=\angle B Y C$, and so $\angle P X M=\angle M Y P$. Moreover, $\angle Y P X=\angle C P D=\angle A D B=\angle A M B$. The quadrilateral $M X P Y$ therefore has equal opposite angles and so is a parallelogram. ![](https://cdn.mathpix.com/cropped/2025_01_09_9538693ed865d47d2de5g-63.jpg?height=1298&width=1126&top_left_y=1007&top_left_x=471)  Let $R$ and $S$ be the intersection points of $A M$ and $B M$ with $B C$ and $A D$ respectively. Due to $A M \| P C$ and $B M \| P D$, we have $\angle A S B=\angle A D P=\angle P C B=\angle A R B$ and so the quadrilateral $A B R S$ is cyclic. We then have $\angle S R B=180^{\circ}-\angle B A S=\angle D C B$ and so $S R \| C D$. In triangles $P C D$ and $M R S$, the corresponding sides are parallel so they are homothetic meaning lines $D S, P M, C R$ concur at the centre of this homothety.


Solution: Let $A D$ and $B C$ meet at $T$. Denote by $p_{a}, p_{b}, m_{a}$ and $m_{b}$ the distances between line $T A$ and $P, T B$ and $P, T A$ and $M$ and between $T B$ and $M$ respectively. Our goal is to prove $p_{a}: p_{b}=m_{a}: m_{b}$ which is equivalent to the collinearity of $T, P$ and $M$. ![](https://cdn.mathpix.com/cropped/2025_01_09_9538693ed865d47d2de5g-64.jpg?height=986&width=1198&top_left_y=432&top_left_x=429)  Let $\angle B A C=\angle B D C=\alpha, \angle D B A=\angle D C A=\beta, \angle A D B=\angle A M B=\angle A C B=$ $\angle C P D=\mu, \angle A D P=\angle P C B=\nu$ and $\angle M A D=\angle C A M=\angle M B D=\angle C B M=\chi$.  From $\angle A D P=\angle P C B=\nu$ and $\angle M A D=\angle C B M=\chi$ we get  $$ \frac{p_{a}}{p_{b}}=\frac{P D \cdot \sin \nu}{P C \cdot \sin \nu}=\frac{P D}{P C} \quad \text { and } \quad \frac{m_{a}}{m_{b}}=\frac{M A \cdot \sin \chi}{M B \cdot \sin \chi}=\frac{M A}{M B} . $$  Hence $p_{a}: p_{b}=m_{a}: m_{b}$ is equivalent to $P D: P C=M A: M B$, and since $\angle C P D=\angle A M B=$ $\mu$, this means we have to show that triangles $P D C$ and $M A B$ are similar.  In triangle $P D C$ we have  $$ \begin{aligned} & \angle P D C+\angle D C P=180^{\circ}-\angle C P D=180^{\circ}-\mu \\ & \angle P D C-\angle D C P=(\alpha+\mu-\nu)-(\beta+\mu-\nu)=\alpha-\beta . \end{aligned} $$  Similarly, in triangle $M A B$ we have  $$ \begin{aligned} & \angle B A M+\angle M B A=180^{\circ}-\angle A M B=180^{\circ}-\mu, \\ & \angle B A M-\angle M B A=(\alpha+\chi)-(\beta+\chi)=\alpha-\beta . \end{aligned} $$  Therefore, $(\angle B A M, \angle M B A)$ and $(\angle P D C, \angle D C P)$ satisfy the same system of linear equations. The common solution is  $$ \angle B A M=\angle P D C=\frac{180^{\circ}-\mu+\alpha-\beta}{2} \text { and } \angle M B A=\angle D C P=\frac{180^{\circ}-\mu-\alpha+\beta}{2} . $$  Hence triangles $P D C$ and $M A B$ have equal angles and so are similar. This completes the proof.


Let $A B C$ be an acute-angled triangle with $A B<A C$. Denote its circumcircle by $\Omega$ and denote the midpoint of arc $C A B$ by $S$. Let the perpendicular from $A$ to $B C$ meet $B S$ and $\Omega$ at $D$ and $E \neq A$ respectively. Let the line through $D$ parallel to $B C$ meet line $B E$ at $L$ and denote the circumcircle of triangle $B D L$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$.  Prove that the line tangent to $\omega$ at $P$, and line $B S$ intersect on the internal bisector of $\angle B A C$. (Portugal)


Let $A B C$ be an acute-angled triangle with circumcircle $\omega$ and circumcentre $O$. Points $D \neq B$ and $E \neq C$ lie on $\omega$ such that $B D \perp A C$ and $C E \perp A B$. Let $C O$ meet $A B$ at $X$, and $B O$ meet $A C$ at $Y$.  Prove that the circumcircles of triangles $B X D$ and $C Y E$ have an intersection on line $A O$. (Malaysia)


Let $A B C$ be an acute-angled triangle with circumcircle $\omega$. A circle $\Gamma$ is internally tangent to $\omega$ at $A$ and also tangent to $B C$ at $D$. Let $A B$ and $A C$ intersect $\Gamma$ at $P$ and $Q$ respectively. Let $M$ and $N$ be points on line $B C$ such that $B$ is the midpoint of $D M$ and $C$ is the midpoint of $D N$. Lines $M P$ and $N Q$ meet at $K$ and intersect $\Gamma$ again at $I$ and $J$ respectively. The ray $K A$ meets the circumcircle of triangle $I J K$ at $X \neq K$.  $$ \text { Prove that } \angle B X P=\angle C X Q \text {. } $$


Let $A B C$ be an acute, scalene triangle with orthocentre $H$. Let $\ell_{a}$ be the line through the reflection of $B$ with respect to $C H$ and the reflection of $C$ with respect to $B H$. Lines $\ell_{b}$ and $\ell_{c}$ are defined similarly. Suppose lines $\ell_{a}, \ell_{b}$, and $\ell_{c}$ determine a triangle $\mathcal{T}$.  Prove that the orthocentre of $\mathcal{T}$, the circumcentre of $\mathcal{T}$ and $H$ are collinear.

Solution: ![](https://cdn.mathpix.com/cropped/2025_01_09_9538693ed865d47d2de5g-76.jpg?height=1314&width=1326&top_left_y=762&top_left_x=391)  We write $\triangle P_{1} P_{2} P_{3} \stackrel{ \pm}{\sim} \triangle Q_{1} Q_{2} Q_{3}$ (resp. $\triangle P_{1} P_{2} P_{3} \approx \triangle Q_{1} Q_{2} Q_{3}$ ) to indicate that two triangles are directly (resp. oppositely) similar. We use directed angles throughout denoted with $\Varangle$.  Denote by $A_{b}, A_{c}$ the reflections of $A$ in $B H$ and $C H$ respectively. $B_{c}, B_{a}$ and $C_{a}, C_{b}$ are defined similarly. By definition, $\ell_{a}=B_{c} C_{b}, \ell_{b}=C_{a} A_{c}, \ell_{c}=A_{b} B_{a}$. Let $A_{1}=\ell_{b} \cap \ell_{c}$, $B_{1}=\ell_{c} \cap \ell_{a}, C_{1}=\ell_{a} \cap \ell_{b}$ and let $O_{1}, H_{1}$ be the orthocentre and circumcentre of $\mathcal{T} \equiv \triangle A_{1} B_{1} C_{1}$ respectively. Claim 1. $\triangle A A_{b} A_{c} \approx \triangle A B C$. Proof. Let $P=B H \cap A C, Q=C H \cap A B$, then it is well known that $\triangle A P Q \approx \triangle A B C$. By the dilation with factor 2 centred at $A, \triangle A P Q$ is sent to $\triangle A A_{b} A_{c}$, so we have $\triangle A A_{b} A_{c} \sim \triangle A B C$.  Claim 2. $\triangle A A_{b} A_{c} \stackrel{\downarrow}{\sim} \triangle A B_{a} C_{a}$ and $A_{1}$ lies on the circumcircle of $\triangle A A_{b} A_{c}$ which is centred at $H$. Proof. Since $B_{a}, C_{a}$ are reflections of $B, C$ in $A H$, we have $\triangle A B_{a} C_{a} \approx \triangle A B C$. Combining this with Claim 1, we have $\triangle A A_{b} A_{c} \stackrel{\downarrow}{\sim} \triangle A B_{a} C_{a}$, where $A$ is the centre of this similarity. Therefore, $\Varangle A_{c} A_{1} A_{b}=\Varangle A_{c} A A_{b}$ meaning $A_{1}$ lies on $\odot A A_{b} A_{c}$. By symmetry, $H A_{b}=H A=H A_{c}$, so $H$ is centre of this circle.  Claim 3. $\triangle A_{1} B_{1} C_{1} \approx \triangle A B C$. Proof. From Claim 2 we have  $$ \Varangle C_{1} A_{1} B_{1}=\Varangle A_{c} A_{1} A_{b} \stackrel{\text { Claim } 2}{=} \npreceq A_{c} A A_{b}=-\Varangle C A B $$  and similarly $\Varangle A_{1} B_{1} C_{1}=-\Varangle A B C, \Varangle B_{1} C_{1} A_{1}=-\Varangle B C A$, which imply $\triangle A_{1} B_{1} C_{1} \approx \triangle A B C$.  Denote the ratio of similitude of $\triangle A_{1} B_{1} C_{1}$ and $\triangle A B C$ by $\lambda\left(=\frac{B_{1} C_{1}}{B C}\right)$, then  $$ \lambda=\frac{H_{1} A_{1}}{H A}=\frac{H_{1} B_{1}}{H B}=\frac{H_{1} C_{1}}{H C} . $$  Since $H A=H A_{1}$ and similarly $H B=H B_{1}, H C=H C_{1}$ from Claim 2, we get  $$ \lambda=\frac{H_{1} A_{1}}{H A_{1}}=\frac{H_{1} B_{1}}{H B_{1}}=\frac{H_{1} C_{1}}{H C_{1}} . $$  Therefore, the circle $A_{1} B_{1} C_{1}$ is the Apollonian circle of the segment $H H_{1}$ with ratio $\lambda$ so the line $H H_{1}$ passes through $O_{1}$.


Solution: We use the same notation $A_{b}, A_{c}, B_{c}, B_{a}, C_{a}, C_{b}$ and $A_{1}, B_{1}, C_{1}, O_{1}, H_{1}$ as in   Let $O$ be the circumcentre of $\triangle A B C$ and $A_{2}$ be the reflection of $A_{1}$ in $A H$. As $\odot A A_{b} A_{c}$ is centred at $H, A_{2}$ also lies on this circle.  By Claim 2, $\Varangle B_{a} A_{1} C_{a}=\Varangle A_{b} A A_{c}=\Varangle B_{a} A C_{a}$, so $A_{1}$ lies on $\odot A B_{a} C_{a}$. Reflecting this in $A H$ gives that $A_{2}$ lies on $\odot A B C$. We now have circles centred at $O$ and $H$ passing through $A$ and $A_{2}$ so these points are symmetric with respect to $O H$. Define $B_{2}$ and $C_{2}$ similarly then $\triangle A B C$ and $\triangle A_{2} B_{2} C_{2}$ are symmetric with respect to $O H$ and also $\odot A B C=\odot A_{2} B_{2} C_{2}$. ![](https://cdn.mathpix.com/cropped/2025_01_09_9538693ed865d47d2de5g-78.jpg?height=1512&width=1475&top_left_y=723&top_left_x=296)  Claim 4. $A_{1} A_{2}, B_{1} B_{2}$ and $C_{1} C_{2}$ have an intersection on $\odot A B C$ which we denote by $T$. Proof. Let $T=A_{1} A_{2} \cap B_{1} B_{2}$. Since $A_{1} A_{2} \| B C$ and $B_{1} B_{2} \| A C$, we have  $$ \Varangle A_{2} T B_{2}=\Varangle B C A=-\Varangle B_{2} C_{2} A_{2}=\Varangle A_{2} C_{2} B_{2} . $$  So $T$ lies on $\odot A_{2} B_{2} C_{2}=\odot A B C$. Similarly the intersection of $A_{1} A_{2}$ and $C_{1} C_{2}$ lies on $\odot A B C$, so $C_{1} C_{2}$ also passes through $T$.  Claim 5. $T$ also lies on $\odot A_{1} B_{1} C_{1}$ and $T$ corresponds to $T$ itself under the similarity $\triangle A_{1} B_{1} C_{1} \approx$ $\triangle A B C$.  Proof. We know $\triangle A_{1} B_{1} C_{1} \sim \triangle A B C$ by Claim 3. We also have  $$ \Varangle B_{1} T C_{1}=\Varangle B_{2} T C_{2}=\Varangle B_{2} A_{2} C_{2}=-\Varangle B A C \stackrel{\text { Claim } 3}{=} \not B_{1} A_{1} C_{1}, $$  so $T$ lies on $\odot A_{1} B_{1} C_{1}$. The remaining part is concluded by the following angle chase:  $$ \Varangle A_{1} B_{1} T=\Varangle A_{1} B_{1} B_{2} \stackrel{B_{1} B_{2} \| A C}{\underline{\|}} \npreceq A_{1} A_{b} A=\Varangle A_{1} A_{2} A=-\Varangle A A_{2} T=-\Varangle A B T . $$  Claim 6. The circumradius of $\triangle A_{1} B_{1} C_{1}$ is equal to $H O$. Proof. Two circles centred at $H$ intersect $\ell_{c}$ at $A_{1}, A_{b}$ and $B_{1}, B_{a}$, so $A_{1} A_{b}$ and $B_{1} B_{a}$ have the same midpoint and thus $A_{1} B_{1}=A_{b} B_{a}$. Consider the spiral symmetry $\triangle A A_{b} A_{c} \stackrel{\downarrow}{\sim} \triangle A B_{a} C_{a}$. This takes $H$, the circumcentre of $\triangle A A_{b} A_{c}$, to the circumcentre of $\triangle A B_{a} C_{a}$, denoted by $O_{a}$, which is symmetric to $O$ in $A H$. Hence $\triangle A A_{b} B_{a} \stackrel{ \pm}{\sim} \triangle A H O_{a}$, so  $$ \frac{A A_{b}}{B_{a} A_{b}}=\frac{A H}{H O_{a}}=\frac{A H}{H O} \Longrightarrow \frac{A H}{A A_{b}}=\frac{H O}{A_{b} B_{a}} . $$  Also since $\triangle A_{1} O_{1} B_{1} \stackrel{\downarrow}{\sim} \triangle A H A_{b}$ (both of them are $\approx \triangle A O B$ ), we have  $$ \frac{A_{1} O_{1}}{A_{1} B_{1}}=\frac{A H}{A A_{b}}=\frac{H O}{A_{b} B_{a}}=\frac{H O}{A_{1} B_{1}} \Longrightarrow A_{1} O_{1}=H O $$  as desired.


Let $A B C$ be an equilateral triangle. Points $A_{1}, B_{1}, C_{1}$ lie inside triangle $A B C$ such that triangle $A_{1} B_{1} C_{1}$ is scalene, $B A_{1}=A_{1} C, C B_{1}=B_{1} A, A C_{1}=C_{1} B$ and  $$ \angle B A_{1} C+\angle C B_{1} A+\angle A C_{1} B=480^{\circ} . $$  Lines $B C_{1}$ and $C B_{1}$ intersect at $A_{2}$; lines $C A_{1}$ and $A C_{1}$ intersect at $B_{2}$; and lines $A B_{1}$ and $B A_{1}$ intersect at $C_{2}$.  Prove that the circumcircles of triangles $A A_{1} A_{2}, B B_{1} B_{2}, C C_{1} C_{2}$ have two common points. (U.S.A.)

Solution: Let $\delta_{A}, \delta_{B}, \delta_{C}$ be the circumcircles of $\triangle A A_{1} A_{2}, \triangle B B_{1} B_{2}, \triangle C C_{1} C_{2}$. The general strategy of the solution is to find two different points having equal power with respect to $\delta_{A}, \delta_{B}, \delta_{C}$. Claim. $A_{1}$ is the circumcentre of $A_{2} B C$ and cyclic variations. Proof. Since $A_{1}$ lies on the perpendicular bisector of $B C$ and inside $\triangle B A_{2} C$, it suffices to prove $\angle B A_{1} C=2 \angle B A_{2} C$. This follows from  $$ \begin{aligned} \angle B A_{2} C & =\angle A_{2} B A+\angle B A C+\angle A C A_{2} \\ & =\frac{1}{2}\left(\left(180^{\circ}-\angle A C_{1} B\right)+\left(180^{\circ}-\angle C B_{1} A\right)\right)+60^{\circ} \\ & =240^{\circ}-\frac{1}{2}\left(480^{\circ}-\angle B A_{1} C\right) \\ & =\frac{1}{2} \angle B A_{1} C \end{aligned} $$  ![](https://cdn.mathpix.com/cropped/2025_01_09_9538693ed865d47d2de5g-80.jpg?height=849&width=1009&top_left_y=1546&top_left_x=529)  The circumcentres above give  $$ \angle B_{1} B_{2} C_{1}=\angle B_{1} B_{2} A=\angle B_{2} A B_{1}=\angle C_{1} A C_{2}=\angle A C_{2} C_{1}=\angle B_{1} C_{2} C_{1} $$  and so $B_{1} C_{1} B_{2} C_{2}$ is cyclic. Likewise $C_{1} A_{1} C_{2} A_{2}$ and $A_{1} B_{1} A_{2} B_{2}$ are cyclic. Note that hexagon $A_{1} B_{2} C_{1} A_{2} B_{1} C_{2}$ is not cyclic since  $$ \angle C_{2} A_{1} B_{2}+\angle B_{2} C_{1} A_{2}+\angle A_{2} B_{1} C_{2}=480^{\circ} \neq 360^{\circ} . $$  Thus we can apply radical axis theorem to the three circles to show that $A_{1} A_{2}, B_{1} B_{2}, C_{1}, C_{2}$ concur at a point $X$ and this point has equal power with respect to $\delta_{A}, \delta_{B}, \delta_{C}$.  Let the circumcircle of $\triangle A_{2} B C$ meet $\delta_{A}$ at $A_{3} \neq A_{2}$. Define $B_{3}$ and $C_{3}$ similarly. Claim. $\mathrm{BCB}_{3} C_{3}$ cyclic. Proof. Using directed angles  $$ \begin{aligned} \Varangle B C_{3} C & =\Varangle B C_{3} C_{2}+\Varangle C_{2} C_{3} C \\ & =\Varangle B A C_{2}+\Varangle C_{2} C_{1} C \\ & =90^{\circ}+\Varangle\left(C_{1} C, A C_{2}\right)+\Varangle C_{2} C_{1} C \\ & =90^{\circ}+\Varangle C_{1} C_{2} B_{1} . \end{aligned} $$  Similarly $\Varangle C B_{3} B=90^{\circ}+\Varangle B_{1} B_{2} C_{1}$. Hence, using $B_{1} C_{1} B_{2} C_{2}$ cyclic  $$ \Varangle B B_{3} C=90^{\circ}+\Varangle C_{1} B_{2} B_{1}=90^{\circ}+\Varangle C_{1} C_{2} B_{1}=\Varangle B C_{3} C $$  as required. ![](https://cdn.mathpix.com/cropped/2025_01_09_9538693ed865d47d2de5g-81.jpg?height=1041&width=1477&top_left_y=1093&top_left_x=292)  Similarly $C A C_{3} A_{3}$ and $A B A_{3} B_{3}$ are cyclic. $A C_{3} B A_{3} C B_{3}$ is not cyclic because then $A B_{2} C B_{3}$ cyclic would mean $B_{2}$ lies on $\odot A B C$ which is impossible since $B_{2}$ lies inside $\triangle A B C$. Thus we can apply radical axis theorem to the three circles to get $A A_{3}, B B_{3}, C C_{3}$ concur at a point $Y$ which has equal power with respect to $\delta_{A}, \delta_{B}, \delta_{C}$.  We now make some technical observations before finishing.  - Let $O$ be the centre of $\triangle A B C$. We have that  $$ \angle B A_{1} C=480^{\circ}-\angle C B_{1} A-\angle A C_{1} B>480^{\circ}-180^{\circ}-180^{\circ}=120^{\circ} . $$  so $A_{1}$ lies inside $\triangle B O C$. We have similar results for $B_{1}, C_{1}$ and thus $\triangle B A_{1} C, \triangle C B_{1} A$, $\triangle A C_{1} B$ have disjoint interiors. It follows that $A_{1} B_{2} C_{1} A_{2} B_{1} C_{2}$ is a convex hexagon thus $X$ lies on segment $A_{1} A_{2}$ and therefore is inside $\delta_{A}$.  - Since $A_{1}$ is the centre of $A_{2} B C$ we have that $A_{1} A_{2}=A_{1} A_{3}$ so, from cyclic quadrilateral $A A_{2} A_{1} A_{3}$ we get that lines $A A_{2}$ and $A A_{3} \equiv A Y$ are reflections in line $A A_{1}$. As $X$ lies on segment $A_{1} A_{2}$, the only way $X \equiv Y$ is if $A_{1}$ and $A_{2}$ both lie on the perpendicular bisector of $B C$. But this forces $B_{1}$ and $C_{1}$ to also be reflections in this line meaning $A_{1} B_{1}=A_{1} C_{1}$ contradicting the scalene condition.  Summarising, we have distinct points $X, Y$ with equal power with respect to $\delta_{A}, \delta_{B}, \delta_{C}$ thus these circles have a common radical axis. As $X$ lies inside $\delta_{A}$ (and similarly $\delta_{B}, \delta_{C}$ ), this radical axis intersects the circles at two points and so $\delta_{A}, \delta_{B}, \delta_{C}$ have two points in common.  Comment. An alternative construction for $Y$ comes by observing that  $$ \frac{\sin \angle B A A_{2}}{\sin \angle A_{2} A C}=\frac{\frac{A_{2} B}{A A_{2}} \sin \angle A_{2} B A}{\frac{A_{2} C}{A A_{2}} \sin \angle A C A_{2}}=\frac{A_{2} B}{A_{2} C} \cdot \frac{\sin \angle C_{1} B A}{\sin \angle A C B_{1}}=\frac{\sin \angle B_{1} C B}{\sin \angle C B C_{1}} \cdot \frac{\sin \angle C_{1} B A}{\sin \angle A C B_{1}} $$  and hence  $$ \frac{\sin \angle B A A_{2}}{\sin \angle A_{2} A C} \cdot \frac{\sin \angle C B B_{2}}{\sin \angle B_{2} B A} \cdot \frac{\sin \angle A C C_{2}}{\sin \angle C_{2} C B}=1 $$  so by Ceva's theorem, $A A_{2}, B B_{2}, C C_{2}$ concur and thus we can construct the isogonal conjugate of this point of concurrency which turns out to be $Y$.


Determine all positive, composite integers $n$ that satisfy the following property: if the positive divisors of $n$ are $1=d_{1}<d_{2}<\cdots<d_{k}=n$, then $d_{i}$ divides $d_{i+1}+d_{i+2}$ for every $1 \leqslant i \leqslant k-2$. (Colombia)

Solution: It is easy to see that such an $n=p^{r}$ with $r \geqslant 2$ satisfies the condition as $d_{i}=p^{i-1}$ with $1 \geqslant i \geqslant k=r+1$ and clearly  $$ p^{i-1} \mid p^{i}+p^{i+1} $$  Now, let us suppose that there is a positive integer $n$ that satisfies the divisibility condition of the problem and that has two different prime divisors $p$ and $q$. Without lost of generality, we assume $p<q$ and that they are the two smallest prime divisors of $n$. Then there is a positive integer $j$ such that  $$ d_{1}=1, d_{2}=p, \ldots, d_{j}=p^{j-1}, d_{j+1}=p^{j}, d_{j+2}=q, $$  and it follows that  $$ d_{k-j-1}=\frac{n}{q}, d_{k-j}=\frac{n}{p^{j}}, d_{k-j+1}=\frac{n}{p^{j-1}}, \ldots, d_{k-1}=\frac{n}{p}, d_{k}=n . $$  Thus  $$ d_{k-j-1}=\frac{n}{q} \left\lvert\, d_{k-j}+d_{k-j+1}=\frac{n}{p^{j}}+\frac{n}{p^{j-1}}=\frac{n}{p^{j}}(p+1) .\right. $$  This gives $p^{j} \mid q(p+1)$, which is a contradiction since $\operatorname{gcd}(p, p+1)=1$ and $p \neq q$.


Solution: Since $d_{i} d_{k+1-i}=n$, we have the equivalence:  $$ d_{k-i-1}\left|d_{k-i}+d_{k-i+1} \Longleftrightarrow \frac{n}{d_{i+2}}\right| \frac{n}{d_{i+1}}+\frac{n}{d_{i}} . $$  We multiply both sides by $d_{i} d_{i+1} d_{i+2}$ and cancel the $n$ 's to get  $$ d_{i} d_{i+1} \mid d_{i} d_{i+2}+d_{i+1} d_{i+2} . $$  Hence,  $$ d_{i} \mid d_{i+1} d_{i+2} $$  Moreover, by the condition of the problem,  $$ d_{i} \mid d_{i+1}\left(d_{i+1}+d_{i+2}\right)=d_{i+1}^{2}+d_{i+1} d_{i+2} . $$  Combining this with (2) we get that $d_{i} \mid d_{i+1}^{2}$ for all $1 \leqslant i \leqslant k-2$. Let $d_{2}=p$ be the smallest prime divisor of $n$. By induction on $i$ we prove that $p \mid d_{i}$ for all $2 \leqslant i \leqslant k-1$. The base case $d_{2}=p$ is obvious. Let us suppose that $p \mid d_{j}$ for some $2 \leqslant j \leqslant k-2$. Then we have that  $$ p\left|d_{j}\right| d_{j+1}^{2} \Longrightarrow p \mid d_{j+1} $$  as $p$ is prime, which completes the induction. This implies that $n$ has to be a prime power, as otherwise there would be another prime $q$ that divides $n$ and we would get that $p \mid q$ which is obviously false.  We finally check that the powers of $p$ satisfy the condition in the statement of the problem as in Solution 1.


Solution: We start by proving the following claim: Claim. $d_{i} \mid d_{i+1}$ for every $1 \leqslant i \leqslant k-1$. Proof. We prove the Claim by induction on $i$; it is trivial for $i=1$ because $d_{1}=1$. Suppose that $2 \leqslant i \leqslant k-1$ and the Claim is true for $i-1$, i.e. $d_{i-1} \mid d_{i}$. By the induction hypothesis and the problem condition, $d_{i-1} \mid d_{i}$ and $d_{i-1} \mid d_{i}+d_{i+1}$, so $d_{i-1} \mid d_{i+1}$.  Now consider the divisors $d_{k-i}=\frac{n}{d_{i+1}}, d_{k-i+1}=\frac{n}{d_{i}}, d_{k-i+2}=\frac{n}{d_{i-1}}$. By the problem condition,  $$ \frac{d_{k-i+1}+d_{k-i+2}}{d_{k-i}}=\frac{\frac{n}{d_{i}}+\frac{n}{d_{i-1}}}{\frac{n}{d_{i+1}}}=\frac{d_{i+1}}{d_{i}}+\frac{d_{i+1}}{d_{i-1}} $$  is an integer. We conclude that $\frac{d_{i+1}}{d_{i}}$ is an integer, so $d_{i} \mid d_{i+1}$. By the Claim, $n$ cannot have two different prime divisors because the smallest one would divide the other one. Hence, $n$ must be a power of a prime, and powers of primes satisfy the condition of the problem as we saw in Solution 1.


Solution: We present here a more technical way of finishing   $$ v_{p}\left(\frac{n}{p^{j}}(p+1)\right)=v_{p}(n)-j $$  as $\operatorname{gcd}(p, p+1)=1$. But (1) implies  $$ v_{p}(n)=v_{p}(n / q) \leqslant v_{p}\left(\frac{n}{p^{j}}(p+1)\right)=v_{p}(n)-j $$  which is a contradiction. Thus $n$ has only one prime divisor as desired.


Determine all pairs $(a, p)$ of positive integers with $p$ prime such that $p^{a}+a^{4}$ is a perfect square. (Bangladesh)

Solution: Let $p^{a}+a^{4}=b^{2}$ for some positive integer $b$. Then we have  $$ p^{a}=b^{2}-a^{4}=\left(b+a^{2}\right)\left(b-a^{2}\right) . $$  Hence both $b+a^{2}$ and $b-a^{2}$ are powers of $p$. Let $b-a^{2}=p^{x}$ for some integer $x$. Then $b+a^{2}=p^{a-x}$ and $a-x>x$. Therefore, we have  $$ 2 a^{2}=\left(b+a^{2}\right)-\left(b-a^{2}\right)=p^{a-x}-p^{x}=p^{x}\left(p^{a-2 x}-1\right) . $$  We shall consider two cases according to whether $p=2$ or $p \neq 2$. We let $v_{p}(m)$ denote the $p$-adic valuation of $m$.  Case $1(p=2)$ : In this case,  $$ a^{2}=2^{x-1}\left(2^{a-2 x}-1\right)=2^{2 v_{2}(a)}\left(2^{a-2 x}-1\right) $$  where the first equality comes from (1) and the second one from $\operatorname{gcd}\left(2,2^{a-2 x}-1\right)=1$. So, $2^{a-2 x}-1$ is a square.  If $v_{2}(a)>0$, then $2^{a-2 x}$ is also a square. So, $2^{a-2 x}-1=0$, and $a=0$ which is a contradiction. If $v_{2}(a)=0$, then $x=1$, and $a^{2}=2^{a-2}-1$. If $a \geqslant 4$, the right hand side is congruent to 3 modulo 4 , thus cannot be a square. It is easy to see that $a=1,2,3$ do not satisfy this condition.  Therefore, we do not get any solutions in this case. Case $2(p \neq 2)$ : In this case, we have $2 v_{p}(a)=x$. Let $m=v_{p}(a)$. Then we have $a^{2}=p^{2 m} \cdot n^{2}$ for some integer $n \geqslant 1$. So, $2 n^{2}=p^{a-2 x}-1=p^{p^{m} \cdot n-4 m}-1$.  We consider two subcases. Subcase 2-1 $(p \geqslant 5)$ : By induction, one can easily prove that $p^{m} \geqslant 5^{m}>4 m$ for all $m$. Then we have  $$ 2 n^{2}+1=p^{p^{m} \cdot n-4 m}>p^{p^{m} \cdot n-p^{m}} \geqslant 5^{5^{m} \cdot(n-1)} \geqslant 5^{n-1} . $$  But, by induction, one can easily prove that $5^{n-1}>2 n^{2}+1$ for all $n \geqslant 3$. Therefore, we conclude that $n=1$ or 2 . If $n=1$ or 2 , then $p=3$, which is a contradiction. So there are no solutions in this subcase.  Subcase 2-2 $(p=3)$ : Then we have $2 n^{2}+1=3^{3^{m} \cdot n-4 m}$. If $m \geqslant 2$, one can easily prove by induction that $3^{m}>4 m$. Then we have  $$ 2 n^{2}+1=3^{3^{m} \cdot n-4 m}>3^{3^{m} \cdot n-3^{m}}=3^{3^{m} \cdot(n-1)} \geqslant 3^{9(n-1)} . $$  Again, by induction, one can easily prove that $3^{9(n-1)}>2 n^{2}+1$ for all $n \geqslant 2$. Therefore, we conclude that $n=1$. Then we have $2 \cdot 1^{2}+1=3^{3^{m}-4 m}$ hence $3=3^{3^{m}-4 m}$. Consequently, we have $3^{m}-4 m=1$. The only solution of this equation is $m=2$ in which case we have $a=3^{m} \cdot n=3^{2} \cdot 1=9$. If $m \leqslant 1$, then there are two possible cases: $m=0$ or $m=1$.  - If $m=1$, then we have $2 n^{2}+1=3^{3 n-4}$. Again, by induction, one can easily prove that $3^{3 n-4}>2 n^{2}+1$ for all $n \geqslant 3$. By checking $n=1,2$, we only get $n=2$ as a solution. This gives $a=3^{m} \cdot n=3^{1} \cdot 2=6$. - If $m=0$, then we have $2 n^{2}+1=3^{n}$. By induction, one can easily prove that $3^{n}>2 n^{2}+1$ for all $n \geqslant 3$. By checking $n=1,2$, we find the solutions $a=3^{0} \cdot 1=1$ and $a=3^{0} \cdot 2=2$.  Therefore, $(a, p)=(1,3),(2,3),(6,3),(9,3)$ are all the possible solutions.


For positive integers $n$ and $k \geqslant 2$ define $E_{k}(n)$ as the greatest exponent $r$ such that $k^{r}$ divides $n$ !. Prove that there are infinitely many $n$ such that $E_{10}(n)>E_{9}(n)$ and infinitely many $m$ such that $E_{10}(m)<E_{9}(m)$. (Brazil) $\mathbf{N 4 .}$ Let $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ be $2 n$ positive integers such that the $n+1$ products  $$ \begin{gathered} a_{1} a_{2} a_{3} \cdots a_{n}, \\ b_{1} a_{2} a_{3} \cdots a_{n} \\ b_{1} b_{2} a_{3} \cdots a_{n}, \\ \vdots \\ b_{1} b_{2} b_{3} \cdots b_{n} \end{gathered} $$  form a strictly increasing arithmetic progression in that order. Determine the smallest positive integer that could be the common difference of such an arithmetic progression. (Canada)

Solution: We let $v_{p}(m)$ denote the $p$-adic valuation of $m$. By Legendre's Formula we know, for $p$ prime, that $v_{p}(n!)=\lfloor n / p\rfloor+\left\lfloor n / p^{2}\right\rfloor+\cdots$. We can see that $E_{9}(n)=\left\lfloor\frac{v_{3}(n!)}{2}\right\rfloor$. Since $v_{5}(n!) \leqslant v_{2}(n!)$ and $E_{10}(n)=\min \left(v_{5}(n!), v_{2}(n!)\right)$, we have $E_{10}(n)=v_{5}(n!)$.  Let $l$ be a positive integer. Set $n=5^{2 l-1}$. Then we have  $$ E_{10}(n)=v_{5}(n!)=5^{2 l-2}+5^{2 l-3}+\cdots+5+1=\frac{5^{2 l-1}-1}{4}=\frac{n-1}{4} $$  Since $n=5^{2 l-1} \equiv 2(\bmod 3)$, we have $\left\lfloor\frac{n}{3}\right\rfloor=\frac{n-2}{3}$ and it implying  $$ v_{3}(n!)=\left\lfloor\frac{n}{3}\right\rfloor+\left\lfloor\frac{n}{3^{2}}\right\rfloor+\left\lfloor\frac{n}{3^{3}}\right\rfloor+\cdots<\frac{n-2}{3}+\frac{n}{3^{2}}+\frac{n}{3^{3}}+\cdots=\frac{n}{2}-\frac{2}{3} . $$  From this we obtain  $$ E_{9}(n)=\left\lfloor\frac{v_{3}(n!)}{2}\right\rfloor \leqslant \frac{v_{3}(n!)}{2} \leqslant \frac{n}{4}-\frac{1}{3}<\frac{n}{4}-\frac{1}{4}=E_{10}(n) . $$  In a similar way, we set now $m=3^{4 l-2}$. Then we have  $$ v_{3}(m!)=3^{4 l-3}+3^{4 l-4}+\cdots+3+1=\frac{3^{4 l-2}-1}{2}=\frac{m-1}{2} . $$  Note that $m=3^{4 l-2} \equiv 1(\bmod 4)$ and hence $E_{9}(m)=\left\lfloor\frac{v_{3}(m!)}{2}\right\rfloor=\left\lfloor\frac{m-1}{4}\right\rfloor=\frac{m-1}{4}$. We also have $m=3^{4 l-2} \equiv 4(\bmod 5)$ implying $\left\lfloor\frac{m}{5}\right\rfloor=\frac{m-4}{5}$. Therefore we obtain  $$ E_{10}(m)=v_{5}(m!)=\left\lfloor\frac{m}{5}\right\rfloor+\left\lfloor\frac{m}{5^{2}}\right\rfloor+\cdots<\frac{m-4}{5}+\frac{m}{5^{2}}+\cdots=\frac{m}{4}-\frac{4}{5}<\frac{m}{4}-\frac{1}{4}=E_{9}(m) $$  We can take infinitely many $n=5^{2 l-1}$ and $m=3^{4 l-2}$ completing the proof.


Solution: In the setting of  First, we take $n=5^{3^{b-1}}$ with $b \geqslant 2$. Because 5 is not a square modulo 3 and $\varphi\left(3^{b}\right)=2 \cdot 3^{b-1}$, we have $n \equiv-1\left(\bmod 3^{b}\right)$. Hence,  $$ v_{3}(n!)=\left\lfloor\frac{n}{3}\right\rfloor+\left\lfloor\frac{n}{3^{2}}\right\rfloor+\cdots<\frac{n-2}{3}+\frac{n-8}{9}+\cdots+\frac{n-\left(3^{b}-1\right)}{3^{b}}+\frac{n}{3^{b+1}}+\cdots<\frac{n}{2}-b+\frac{1}{2}, $$  and $E_{10}(n)=\frac{n-1}{4}>\frac{n+1-2 b}{4}>E_{9}(n)$. In the same way, for $m=3^{2 \cdot 5^{b-1}} \equiv-1\left(\bmod 5^{b}\right)$ with $b \geqslant 2$, $E_{10}(m)=\left\lfloor\frac{m}{5}\right\rfloor+\left\lfloor\frac{m}{5^{2}}\right\rfloor+\cdots<\frac{m-4}{5}+\frac{m-24}{25}+\cdots+\frac{m-\left(5^{b}-1\right)}{5^{b}}+\frac{m}{5^{b+1}}+\cdots<\frac{m}{4}-b+\frac{1}{4}$, and $E_{9}(m)=\frac{m-1}{4}>E_{10}(m)$ holds. Comment. From Solution 2 we can see that for any positive real $B$, there exist infinitely many positive integers $m$ and $n$ such that $E_{10}(n)-E_{9}(n)>B$ and $E_{10}(m)-E_{9}(m)<-B$.


Let $a_{1}<a_{2}<a_{3}<\cdots$ be positive integers such that $a_{k+1}$ divides $2\left(a_{1}+a_{2}+\cdots+a_{k}\right)$ for every $k \geqslant 1$. Suppose that for infinitely many primes $p$, there exists $k$ such that $p$ divides $a_{k}$. Prove that for every positive integer $n$, there exists $k$ such that $n$ divides $a_{k}$. (Netherlands)

Solution: For every $k \geqslant 2$ define the quotient $b_{k}=2\left(a_{1}+\cdots+a_{k-1}\right) / a_{k}$, which must be a positive integer. We first prove the following properties of the sequence $\left(b_{k}\right)$ : Claim 1. We have $b_{k+1} \leqslant b_{k}+1$ for all $k \geqslant 2$. Proof. By subtracting $b_{k} a_{k}=2\left(a_{1}+\cdots+a_{k-1}\right)$ from $b_{k+1} a_{k+1}=2\left(a_{1}+\cdots+a_{k}\right)$, we find that $b_{k+1} a_{k+1}=b_{k} a_{k}+2 a_{k}=\left(b_{k}+2\right) a_{k}$. From $a_{k}<a_{k+1}$ it follows that $b_{k}+2>b_{k+1}$. Claim 2. The sequence $\left(b_{k}\right)$ is unbounded. Proof. We start by rewriting $b_{k+1} a_{k+1}=\left(b_{k}+2\right) a_{k}$ as  $$ \left.a_{k+1}=a_{k} \cdot \frac{b_{k}+2}{b_{k+1}} \Longrightarrow a_{k+1} \right\rvert\, a_{k}\left(b_{k}+2\right) . $$  If the sequence $\left(b_{k}\right)$ were bounded, say by some positive integer $B$, then the prime factors of the terms of the sequence $\left(a_{k}\right)$ could only be primes less than or equal to $B+2$ or those dividing $a_{1}$ or $a_{2}$, which contradicts the property in the statement of the problem.  Consider now an arbitrary positive integer $n$. We assume $n>b_{2}$, otherwise we replace $n$ by an arbitrary multiple of $n$ that is bigger than $b_{2}$. By Claim 2 , there exists $k$ such that $b_{k+1} \geqslant n$. Consider the smallest such $k$. From Claim 1, it follows that we must have $b_{k}=n-1$ and $b_{k+1}=n$ (we assumed $n>b_{2}$ to ensure that $k \geqslant 2$ ). We now find that  $$ a_{k+1}=a_{k} \cdot \frac{b_{k}+2}{b_{k+1}}=a_{k} \cdot \frac{n+1}{n} . $$  Because $n$ and $n+1$ are coprime, this immediately implies that $a_{k}$ is divisible by $n$. Comment. For $c$ a positive integer, the sequence $a_{k}=c k$ satisfies the conditions of the problem. Another example is  $$ a_{1}=1, \quad a_{2}=2, \quad a_{k}=3(k-1) \text { for } k \geqslant 3 . $$


A sequence of integers $a_{0}, a_{1}, a_{2}, \ldots$ is called kawaii, if $a_{0}=0, a_{1}=1$, and, for any positive integer $n$, we have  $$ \left(a_{n+1}-3 a_{n}+2 a_{n-1}\right)\left(a_{n+1}-4 a_{n}+3 a_{n-1}\right)=0 . $$  An integer is called kawaii if it belongs to a kawaii sequence. Suppose that two consecutive positive integers $m$ and $m+1$ are both kawaii (not necessarily belonging to the same kawaii sequence). Prove that 3 divides $m$, and that $m / 3$ is kawaii. (China)

Solution: We start by rewriting the condition in the problem as:  $$ a_{n+1}=3 a_{n}-2 a_{n-1}, \text { or } a_{n+1}=4 a_{n}-3 a_{n-1} . $$  We have $a_{n+1} \equiv a_{n}$ or $a_{n-1}(\bmod 2)$ and $a_{n+1} \equiv a_{n-1}$ or $a_{n}(\bmod 3)$ for all $n \geqslant 1$. Now, since $a_{0}=0$ and $a_{1}=1$, we have that $a_{n} \equiv 0,1 \bmod 3$ for all $n \geqslant 0$. Since $m$ and $m+1$ are kawaii integers, then necessarily $m \equiv 0 \bmod 3$.  We also observe that $a_{2}=3$ or $a_{2}=4$. Moreover, (1) If $a_{2}=3$, then $a_{n} \equiv 1(\bmod 2)$ for all $n \geqslant 1$ since $a_{1} \equiv a_{2} \equiv 1(\bmod 2)$. (2) If $a_{2}=4$, then $a_{n} \equiv 1(\bmod 3)$ for all $n \geqslant 1$ since $a_{1} \equiv a_{2} \equiv 1(\bmod 3)$.  Since $m \equiv 0(\bmod 3)$, any kawaii sequence containing $m$ does not satisfy $(2)$, so it must satisfy (1). Hence, $m$ is odd and $m+1$ is even.  Take a kawaii sequence $\left(a_{n}\right)$ containing $m+1$. Let $t \geqslant 2$ be such that $a_{t}=m+1$. As $\left(a_{n}\right)$ does not satisfy (1), it must satisfy (2). Then $a_{n} \equiv 1(\bmod 3)$ for all $n \geqslant 1$. We define the sequence $a_{n}^{\prime}=\left(a_{n+1}-1\right) / 3$. This is a kawaii sequence: $a_{0}^{\prime}=0, a_{1}^{\prime}=1$ and for all $n \geqslant 1$,  $$ \left(a_{n+1}^{\prime}-3 a_{n}^{\prime}+2 a_{n-1}^{\prime}\right)\left(a_{n+1}^{\prime}-4 a_{n}^{\prime}+3 a_{n-1}^{\prime}\right)=\left(a_{n+2}-3 a_{n+1}+2 a_{n}\right)\left(a_{n+2}-4 a_{n+1}+3 a_{n}\right) / 9=0 . $$  Finally, we notice that the term $a_{t-1}^{\prime}=m / 3$ which implies that $m / 3$ is kawaii.


Solution: We start by proving the following: Claim 1. We have $a_{n} \equiv 0,1 \bmod 3$ for all $n \geqslant 0$. Proof. We have $a_{n+1}=3 a_{n}-2 a_{n-1}=3\left(a_{n}-a_{n-1}\right)+a_{n-1}$ or $a_{n+1}=4 a_{n}-3 a_{n-1}=3\left(a_{n}-\right.$ $\left.a_{n-1}\right)+a_{n}$, so $a_{n+1} \equiv a_{n}$ or $a_{n-1} \bmod 3$, and since $a_{0}=0$ and $a_{1}=1$ the result follows.  Hence if $m$ and $m+1$ are kawaii, then necessarily $m \equiv 0 \bmod 3$. Claim 2. An integer $\geqslant 2$ is kawaii if and only if it can be written as $1+b_{2}+\cdots+b_{n}$ for some $n \geqslant 2$ with $b_{i}=2^{r_{i}} 3^{s_{i}}$ satisfying $r_{i}+s_{i}=i-1$ for $i=2, \ldots, n$ and $b_{i} \mid b_{i+1}$ for all $i=2, \ldots, n-1$. Proof. For a kawaii sequence $\left(a_{n}\right)$, we can write $a_{n+1}=3 a_{n}-2 a_{n-1}=a_{n}+2\left(a_{n}-a_{n-1}\right)$ or $a_{n+1}=4 a_{n}-3 a_{n-1}=a_{n}+3\left(a_{n}-a_{n-1}\right)$, so $a_{n+1}-a_{n}=2\left(a_{n}-a_{n-1}\right)$ or $3\left(a_{n}-a_{n-1}\right)$. Hence, $a_{n}=1+b_{2}+\cdots+b_{n}$ where $b_{2}=2$ or 3 and $b_{i+1}=2 b_{i}$ or $3 b_{i}$.  Conversely, given a number that can be written in that way, we consider any sequence given by $a_{0}=0, a_{1}=1$ and $a_{i}=1+b_{2}+\cdots+b_{i}$ for $2 \leqslant i \leqslant n$ and $a_{i}$ given by the kawaii condition for $i \geqslant n+1$. This defines a kawaii sequence containing the given number as $a_{n}$.  Let us suppose that $m$ and $m+1$ are kawaii, then they belong to some kawaii sequences and we can write them as in Claim 2 as $m=1+2+\cdots+2^{\ell}+2^{\ell} \cdot 3 \cdot A$ and $m+1=1+2+\cdots+2^{\ell^{\prime}}+2^{\ell^{\prime}} \cdot 3 \cdot A^{\prime}$ where $\ell$ is odd and $\ell^{\prime}$ is even because of modulo 3 reasons. Since $m+1 \equiv m\left(\bmod 2^{\min \left(\ell, \ell^{\prime}\right)}\right)$, we have $\min \left(\ell, \ell^{\prime}\right)=0$, so $\ell^{\prime}=0$.  Then $m+1=1+b_{2}+\cdots+b_{j}$ for some $b_{i}$ 's as in Claim 2 with $b_{2}=3$ and $b_{i} \mid b_{i+1}$ : so with $3 \mid b_{i}$ for all $i=2, \ldots, j$. Then $\frac{m}{3}=1+b_{1}^{\prime}+\cdots+b_{j-1}^{\prime}$ with $b_{i}^{\prime}=\frac{b_{i+1}}{3}$ as in Claim 2 and $\frac{m}{3}$ is a kawaii integer.


Solution: (This solution is just a different combination of the ideas in   For $n \geqslant 1$, put $b_{n}=a_{n}-a_{n-1}$. We have  $$ a_{t}=a_{0}+\sum_{k=1}^{t}\left(a_{k}-a_{k-1}\right)=\sum_{k=1}^{t} b_{k} . $$  Note that  $$ a_{n+1}-3 a_{n}+2 a_{n-1}=b_{n+1}-2 b_{n} \quad \text { and } \quad a_{n+1}-4 a_{n}+3 a_{n-1}=b_{n+1}-3 b_{n} . $$  The conditions on the $b_{i}$ 's for defining a kawaii sequence are  $$ b_{1}=a_{1}-a_{0}=1, \quad \text { and } \quad \frac{b_{n+1}}{b_{n}} \in\{2,3\} \quad \text { for } \quad n \geqslant 1 $$  1. If we have $\frac{b_{n+1}}{b_{n}}=2$ for any $n$ with $1 \leqslant n \leqslant t-1$, then (*) implies that  $$ a_{t}=\sum_{k=1}^{t} 2^{k-1}=2^{t}-1 \equiv 0,1(\bmod 3) . $$  2. If there exists some integer $s$ with $2 \leqslant s \leqslant t-1$ such that  $$ \frac{b_{2}}{b_{1}}=\frac{b_{3}}{b_{2}}=\cdots=\frac{b_{s}}{b_{s-1}}=2, \quad \frac{b_{s+1}}{b_{s}}=3 $$  it implies that $3 \mid b_{n}$ for any $n \geqslant s+1$. Similarly to the argument in (1), we obtain  $$ a_{t} \equiv \sum_{k=1}^{s} b_{k} \equiv 0,1(\bmod 3) $$  3. If $\frac{b_{2}}{b_{1}}=b_{2}=3$, we have $3 \mid b_{n}$ for any $n \geqslant 2$, and hence $a_{t} \equiv 1(\bmod 3)$.  Combining these, we have proved that $a_{t} \equiv 0,1(\bmod 3)$. We next prove that no positive kawaii integer is divisible by both 2 and 3 . If $b_{2}=2$ for some kawaii sequence, then $2 \mid b_{n}$ and $a_{n} \equiv 1(\bmod 2)$ for all $n \geqslant 2$ in it. If $b_{2}=3$ in some kawaii sequence, then $3 \mid b_{n}$ and $a_{n} \equiv 1(\bmod 3)$ for all $n \geqslant 2$ in it.  Now, consider the original problem. Since $m$ and $m+1$ are both kawaii integer, it means  $$ m \equiv 0,1(\bmod 3), \quad \text { and } \quad m+1 \equiv 0,1(\bmod 3) $$  and hence we easily obtain $3 \mid m$. Since a kawaii integer $m$ is divisible by $3, m$ must be odd, and hence $m+1$ is even. Take a kawaii sequence $a_{0}, a_{1}, a_{2}, \ldots$ containing $m+1$ as $a_{t}$. The fact that $m+1$ is even implies that $b_{2}=3$ and so $3 \mid b_{n}$ for all $n \geqslant 2$ in this sequence. Set $b_{n}^{\prime}=\frac{b_{n+1}}{3}$ for $n \geqslant 1$. Thus $b_{1}^{\prime}=\frac{b_{2}}{3}=1$, and $\frac{b_{n+1}^{\prime}}{b_{n}^{\prime}}=\frac{b_{n+2}}{b_{n+1}} \in\{2,3\}$ for all $n \geqslant 1$. Define $a_{0}^{\prime}=0$ and $a_{n}^{\prime}=\sum_{k=1}^{n} b_{k}^{\prime}$ for $n \geqslant 1$, then $a_{0}^{\prime}, a_{1}^{\prime}, a_{2}^{\prime}, \ldots$ is a kawaii sequence. Now,  $$ a_{t-1}^{\prime}=\sum_{k=1}^{t-1} b_{k}^{\prime}=\frac{1}{3} \sum_{k=2}^{t} b_{k}=\frac{1}{3}\left(-b_{1}+\sum_{k=1}^{t} b_{k}\right)=\frac{1}{3}\left(-1+a_{t}\right)=\frac{m}{3} . $$  This means that $\frac{m}{3}$ is a kawaii integer. Comment. There are infinitely many positive integers $m$ such that $m, m+1, m / 3$ are kawaii. To see this, let $k \geqslant 1$ be a kawaii integer. Then $2 k+1$ and $3 k+1$ are kawaii by Claim 2 in Solution 2, and $3(2 k+1)+1=6 k+4$ and $2(3 k+1)+1=6 k+3$ are also kawaii.


Let $a, b, c, d$ be positive integers satisfying  $$ \frac{a b}{a+b}+\frac{c d}{c+d}=\frac{(a+b)(c+d)}{a+b+c+d} $$  Determine all possible values of $a+b+c+d$. (Netherlands)

Solution: First, note that if we take $a=\ell, b=k \ell, c=k \ell, d=k^{2} \ell$ for some positive integers $k$ and $\ell$, then we have  $$ \frac{a b}{a+b}+\frac{c d}{c+d}=\frac{k \ell^{2}}{\ell+k \ell}+\frac{k^{3} \ell^{2}}{k \ell+k^{2} \ell}=\frac{k \ell}{k+1}+\frac{k^{2} \ell}{k+1}=k \ell $$  and  $$ \frac{(a+b)(c+d)}{a+b+c+d}=\frac{(\ell+k \ell)\left(k \ell+k^{2} \ell\right)}{\ell+k \ell+k \ell+k^{2} \ell}=\frac{k(k+1)^{2} \ell^{2}}{\ell(k+1)^{2}}=k \ell $$  so that  $$ \frac{a b}{a+b}+\frac{c d}{c+d}=k \ell=\frac{(a+b)(c+d)}{a+b+c+d} $$  This means that $a+b+c+d=\ell\left(1+2 k+k^{2}\right)=\ell(k+1)^{2}$ can be attained. We conclude that all non-square-free positive integers can be attained.  Now, we will show that if  $$ \frac{a b}{a+b}+\frac{c d}{c+d}=\frac{(a+b)(c+d)}{a+b+c+d} $$  then $a+b+c+d$ is not square-free. We argue by contradiction. Suppose that $a+b+c+d$ is square-free, and note that after multiplying by $(a+b)(c+d)(a+b+c+d)$, we obtain  $$ (a b(c+d)+c d(a+b))(a+b+c+d)=(a+b)^{2}(c+d)^{2} . $$  A prime factor of $a+b+c+d$ must divide $a+b$ or $c+d$, and therefore divides both $a+b$ and $c+d$. Because $a+b+c+d$ is square-free, the fact that every prime factor of $a+b+c+d$ divides $a+b$ implies that $a+b+c+d$ itself divides $a+b$. Because $a+b<a+b+c+d$, this is impossible. So $a+b+c+d$ cannot be square-free.  Comment 1. Another way to conclude after obtaining (1) is by observing that  $$ (a+b)^{2}(c+d)^{2} \equiv(a+b)^{4} \quad(\bmod a+b+c+d) $$  Hence $a+b+c+d \mid(a+b)^{4}$. But if $a+b+c+d$ is square-free, this forces $a+b+c+d \mid a+b$, which is clearly a contradiction.  Comment 2. It seems difficult to characterise all quadruples ( $a, b, c, d$ ) that satisfy the equality in the problem. Many of them, including those used in the solution, are of the general form $(a, b, c, d)=$ $\left(x y^{2}, x y z, x y z, x z^{2}\right)$ for some positive integers $x, y$, and $z$, but there are more solutions than that, such as $(a, b, c, d)=(2,7,8,10)$ or $(a, b, c, d)=(13,14,16,38)$.


Let $\mathbb{Z}_{>0}$ be the set of positive integers. Determine all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that  $$ f^{b f(a)}(a+1)=(a+1) f(b) $$  holds for all $a, b \in \mathbb{Z}_{>0}$, where $f^{k}(n)=f(f(\cdots f(n) \cdots))$ denotes the composition of $f$ with itself $k$ times.

Solution: We divide the solution into 5 steps. Step 1. ( $f$ is injective) Claim 1. For any $a \geqslant 2$, the set $\left\{f^{n}(a) \mid n \in \mathbb{Z}_{>0}\right\}$ is infinite. Proof. First, we have $f^{f(a)}(a+1) \stackrel{P(a, 1)}{=}(a+1) f(1)$. Varying $a$, we see that $f\left(\mathbb{Z}_{>0}\right)$ is infinite. Next, we have $f^{b f(a-1)}(a) \stackrel{P(a-1, b)}{=} a f(b)$. So, varying $b, f^{b f(a-1)}(a)$ takes infinitely many values. Claim 2. For any $a \geqslant 2$ and $n \in \mathbb{Z}_{>0}$, we have $f^{n}(a) \neq a$. Proof. Otherwise we would get a contradiction with Claim 1. Assume $f(b)=f(c)$ for some $b<c$. Then we have  $$ \begin{aligned} (a+1) f(c) & \stackrel{P(a, c)}{=} f^{c f(a)}(a+1) \\ & =f^{(c-b) f(a)}\left(f^{b f(a)}(a+1)\right) \\ & \stackrel{P(a, b)}{=} f^{(c-b) f(a)}((a+1) f(b)) \\ & =f^{(c-b) f(a)}((a+1) f(c)), \end{aligned} $$  which contradicts Claim 2. So, $f$ is injective. Step 2. $\left(f\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{\geqslant 2}\right)$ Claim 3. 1 is not in the range of $f$. Proof. If $f(b)=1$, then $f^{f(a)}(a+1)=a+1$ by $P(a, 1)$, which contradicts Claim 2. We say that $a$ is a descendant of $b$ if $f^{n}(b)=a$ for some $n \in \mathbb{Z}_{>0}$. Claim 4. For any $a, b \geqslant 1$, both of the following cannot happen at the same time:  - $a$ is a descendant of $b$; - $b$ is a descendant of $a$.  Proof. If both of the above hold, then $a=f^{m}(b)$ and $b=f^{n}(a)$ for some $m, n \in \mathbb{Z}_{>0}$. Then $a=f^{m+n}(a)$, which contradicts Claim 2. Claim 5. For any $a, b \geqslant 2$, exactly one of the following holds:  - $a$ is a descendant of $b$; - $b$ is a descendant of $a$; - $a=b$.  Proof. For any $c \geqslant 2$, taking $m=f^{c f(a-1)-1}(a)$ and $n=f^{c f(b-1)-1}(b)$, we have  $$ f(m)=f^{c f(a-1)}(a) \stackrel{P(a-1, c)}{=} a f(c) \text { and } f(n)=f^{c f(b-1)}(b) \stackrel{P(b-1, c)}{=} b f(c) . $$  Hence  $$ f^{n f(a-1)}(a) \stackrel{P(a-1, n)}{=} a f(n)=a b f(c)=b f(m) \stackrel{P(a-1, m)}{=} f^{m f(b-1)}(b) . $$  The assertion then follows from the injectivity of $f$ and Claim 2. Now, we show that any $a \geqslant 2$ is in the range of $f$. Let $b=f(1)$. If $a=b$, then $a$ is in the range of $f$. If $a \neq b$, either $a$ is a descendant of $b$, or $b$ is a descendant of $a$ by Claim 5. If $b$ is a descendant of $a$, then $b=f^{n}(a)$ for some $n \in \mathbb{Z}_{>0}$, so $1=f^{n-1}(a)$. Then, by Claim 3, we have $n=1$, so $1=a$, which is absurd. So, $a$ is a descendant of $b$. In particular, $a$ is in the range of $f$. Thus, $f\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{\geqslant 2}$.  Step 3. $(f(1)=2)$ Claim 6. Let $a, n \geqslant 2$, then $n a$ is a descendant of $a$. Proof. We write $n=f(m)$ by Step 2. We have $n a=f(m) a \stackrel{P(a-1, m)}{=} f^{m f(a-1)}(a)$, which shows $n a$ is a descendant of $a$.  By Claim 6, all even integers $\geqslant 4$ are descendants of 2 . Hence $2=f(2 k+1)$ for some $k \geqslant 0$. Next, we show $f(2 k+1) \geqslant f(1)$, which implies $f(1)=2$. It trivially holds if $k=0$. If $k \geqslant 1$, let $n$ be the integer such that $f^{n}(2)=2 k+2$. For any $b>n / f(1)$, we have  $$ f^{b f(1)-n}(2 k+2)=f^{b f(1)}(2) \stackrel{P(1, b)}{=} 2 f(b) \text { and } f^{b f(2 k+1)}(2 k+2) \stackrel{P(2 k+1, b)}{=}(2 k+2) f(b) . $$  By Claim 6, $(2 k+2) f(b)$ is a descendant of $2 f(b)$. By Claim 2, we have $b f(2 k+1)>b f(1)-n$. By taking $b$ large enough, we conclude $f(2 k+1) \geqslant f(1)$.  Step 4. $(f(2)=3$ and $f(3)=4)$ From $f(1)=2$ and $P(1, b)$, we have $f^{2 b}(2)=2 f(b)$. So taking $b=1$, we obtain $f^{2}(2)=2 f(1)=4$; and taking $b=f(2)$, we have $f^{2 f(2)}(2)=2 f^{2}(2)=8$. Hence, $f^{2 f(2)-2}(4)=f^{2 f(2)}(2)=8$ and $f^{f(3)}(4) \stackrel{P(3,1)}{=} 8$ give $f(3)=2 f(2)-2$.  Claim 7. For any $m, n \in \mathbb{Z}_{>0}$, if $f(m)$ divides $f(n)$, then $m \leqslant n$. Proof. If $f(m)=f(n)$, the assertion follows from the injectivity of $f$. If $f(m)<f(n)$, by $P(a, m), P(a, n)$ and Claim 6, we have that $f^{n f(a)}(a+1)$ is a descendant of $f^{m f(a)}(a+1)$ for any $a \in \mathbb{Z}_{>0}$. So $m f(a)<n f(a)$, and $m<n$.  By Claim 7, every possible divisor of $f(2)$ is in $\{1, f(1)=2, f(2)\}$. Thus $f(2)$ is an odd prime or $f(2)=4$. Since $f^{2}(2)=4$, we have $f(2) \neq 4$, and hence $f(2)$ is an odd prime. We set $p=f(2)$.  Now, $f(3)=2 f(2)-2=2(p-1)$. Since $p-1$ divides $f(3)$, we have $p-1 \in\{1, f(1)=$ $2, f(2)=p\}$ by Claim 7, so $p-1=2$. Thus, $f(2)=p=3$ and $f(3)=2(p-1)=4$.  Step 5. $(f(n)=n+1)$ Claim 8. For any $b \geqslant 1, f(2 f(b)-1)=2 b+2$. Proof. Since $f^{2}(2)=4$, we have $f^{2 b-2}(4)=f^{2 b}(2)=2 f(b)$, so  $$ f^{f(2 f(b)-1)+2 b-2}(4)=f^{f(2 f(b)-1)}(2 f(b))^{P(2 f(b)-1,1)} 4 f(b) \stackrel{P(3, b)}{=} f^{4 b}(4), $$  which gives us $f(2 f(b)-1)=2 b+2$.  Finally, we prove $f(n)=n+1$ by induction on $n$. Suppose $f(n)=n+1$ for all $1 \leqslant n \leqslant 2 b+1$. Replace $b$ by $b+1$ in $f(2 f(b)-1)=2 b+2$ to get  $$ f(2 b+3)=f(2 f(b+1)-1)=2(b+1)+2=2 b+4 $$  By induction hypothesis, we have $f^{b}(b+2)=2 b+2$. Hence  $$ f(f(2 b+2))=f^{b+2}(b+2)=f^{f(b+1)}(b+2) \stackrel{P(b+1,1)}{=} 2(b+2)=f(2 b+3) . $$  By injectivity, $f(2 b+2)=2 b+3$. Then $f(n)=n+1$ for all $n \in \mathbb{Z}_{>0}$, which is indeed a solution.


Solution: In the same way as Steps 1-2 of   We first note that Claim 2 in Solution 1 is also true for $a=1$. Claim 2'. For any $a, n \in \mathbb{Z}_{>0}$, we have $f^{n}(a) \neq a$. Proof. If $a \geqslant 2$, the assertion was proved in Claim 2 in Solution 1. If $a=1$, we have that 1 is not in the range of $f$ by Claim 3 in Solution 1 . So, $f^{n}(1) \neq 1$ for every $n \in \mathbb{Z}_{>0}$.  For any $a, b \in \mathbb{Z}_{>0}$, we have  $$ f^{b f(f(a)-1)+1}(a)=f^{b f(f(a)-1)}(f(a)) \stackrel{P(f(a)-1, b)}{=} f(a) f(b) $$  Since the right-hand side is symmetric in $a, b$, we have  $$ f^{b f(f(a)-1)+1}(a)=f(a) f(b)=f^{a f(f(b)-1)+1}(b) $$  Since $f$ is injective, we have $f^{b f(f(a)-1)}(a)=f^{a f(f(b)-1)}(b)$. We set $g(n)=f(f(n)-1)$. Then we have $f^{b g(a)}(a)=f^{a g(b)}(b)$ for any $a, b \in \mathbb{Z}_{>0}$. We set $n_{a, b}=b g(a)-a g(b)$. Then, for sufficiently large $n$, we have $f^{n+n_{a, b}}(a)=f^{n}(b)$. For any $a, b, c \in \mathbb{Z}_{>0}$ and sufficiently large $n$, we have  $$ f^{n+n_{a, b}+n_{b, c}+n_{c, a}}(a)=f^{n}(a) . $$  By Claim 2' above, we have $n_{a, b}+n_{b, c}+n_{c, a}=0$, so  $$ (a-b) g(c)+(b-c) g(a)+(c-a) g(b)=0 . $$  Taking $(a, b, c)=(n, n+1, n+2)$, we have $g(n+1)-g(n)=g(n+2)-g(n+1)$. So, $\{g(n)\}_{n \geqslant 1}$ is an arithmetic progression.  There exist $C, D \in \mathbb{Z}$ such that $g(n)=f(f(n)-1)=C n+D$ for all $n \in \mathbb{Z}_{>0}$. By Step 2 of Solution 1, we have $f\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{\geqslant 2}$, so $C=1$. Since $2=\min _{n \in \mathbb{Z}_{>0}}\{f(f(n)-1)\}$, we have $D=1$.  Thus, $g(n)=f(f(n)-1)=n+1$ for all $n \geqslant 1$. For any $a, b \in \mathbb{Z}_{>0}$, we have $f^{b(a+1)}(a)=$ $f^{a(b+1)}(b)$. By the injectivity of $f$, we have $f^{b}(a)=f^{a}(b)$. For any $n \in \mathbb{Z}_{>0}$, taking $(a, b)=(1, n)$, we have $f^{n}(1)=f(n)$, so $f^{n-1}(1)=n$ again by the injectivity of $f$. For any $n \geqslant 1$, we have $f(n)=f\left(f^{n-1}(1)\right)=f^{n}(1)=n+1$.


Solution: The following is another way of finishing   $$ \begin{aligned} f^{b g(a)}(f(a)) & =f^{b f(f(a)-1)}(f(a)) \stackrel{P(f(a)-1, b)}{=} f(a) f(b) \stackrel{P(f(b)-1, a)}{=} f^{a f(f(b)-1)}(f(b)) \\ & =f^{a g(b)+k}(f(a)) \end{aligned} $$  By Claim 2' in Solution 2, we have $b g(a)=a g(b)+k$, so $f^{k}(a) \cdot g(a)=a \cdot g\left(f^{k}(a)\right)+k$. Therefore, for any $n \geqslant 0$, we put $a_{n}=f^{n}(1)$. We have  $$ \left\{\begin{array}{l} a_{n+1} \cdot g\left(a_{n}\right)=a_{n} \cdot g\left(a_{n+1}\right)+1 \\ a_{n+2} \cdot g\left(a_{n+1}\right)=a_{n+1} \cdot g\left(a_{n+2}\right)+1 \\ a_{n+2} \cdot g\left(a_{n}\right)=a_{n} \cdot g\left(a_{n+2}\right)+2 \end{array}\right. $$  Then we have  $$ \begin{aligned} a_{n} \cdot a_{n+1} \cdot g\left(a_{n+2}\right)+2 a_{n+1} & =a_{n+1} \cdot a_{n+2} \cdot g\left(a_{n}\right) \\ & =a_{n+2}\left(a_{n} \cdot g\left(a_{n+1}\right)+1\right) \\ & =a_{n} \cdot a_{n+2} \cdot g\left(a_{n+1}\right)+a_{n+2} \\ & =a_{n}\left(a_{n+1} \cdot g\left(a_{n+2}\right)+1\right)+a_{n+2} \\ & =a_{n} \cdot a_{n+1} \cdot g\left(a_{n+2}\right)+a_{n}+a_{n+2} . \end{aligned} $$  From these, we have $2 a_{n+1}=a_{n}+a_{n+2}$. Thus $\left(a_{n}\right)$ is an arithmetic progression, and we have $a_{n}=f^{n}(1)=C n+D$ for some $C, D \in \mathbb{Z}$.  By Step 2 in Solution 1, any $a \in \mathbb{Z}_{>0}$ is a descendant of 1 , and $f\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{\geqslant 2}$. Hence $D=1$ and $C=1$, and so $f^{n}(1)=n+1$. For any $n \geqslant 1$, we have $f^{n-1}(1)=n$, so $f(n)=f\left(f^{n-1}(1)\right)=$ $f^{n}(1)=n+1$.


Solution: We provide yet another (more technical) solution assuming Step 1 and Step 2 of Solution 1. By Claim 5 in Solution 1, every $a \geqslant 2$ is a descendant of 1 . Let $g$ and $h$ be the functions on $\mathbb{Z}_{\geqslant 2}$ such that $f^{g(a)}(1)=a$ and $h(a)=f(a-1)$. Then, $g: \mathbb{Z}_{\geqslant 2} \rightarrow \mathbb{Z}_{\geqslant 1}$ and $h: \mathbb{Z}_{\geqslant 2} \rightarrow \mathbb{Z}_{\geqslant 2}$ are bijections. The equation $P(a, b)$ can be rewritten as  $$ g(a h(b))=g(a)+(b-1) h(a) $$  Consider the set $S_{a}=g\left(a \cdot \mathbb{Z}_{>0}\right)$. Since $h$ is a bijection onto $\mathbb{Z}_{\geqslant 2}$, we have  $$ S_{a}=g(a)+h(a) \cdot \mathbb{Z}_{\geq 0} $$  Consider the intersection $S_{a} \cap S_{b}=S_{\operatorname{lcm}(a, b)}$. If we put $c=\operatorname{lcm}(a, b)$, this gives  $$ \left(g(a)+h(a) \cdot \mathbb{Z}_{\geqslant 0}\right) \cap\left(g(b)+h(b) \cdot \mathbb{Z}_{\geqslant 0}\right)=g(c)+h(c) \cdot \mathbb{Z}_{\geqslant 0} . $$  Then we have $h(c)=\operatorname{lcm}(h(a), h(b))$ since the left hand side must be of the form $m+$ $\operatorname{lcm}(h(a), h(b)) \cdot \mathbb{Z}_{\geqslant 0}$ for some $m$.  If $b$ is a multiple of $a$, then $\operatorname{lcm}(a, b)=b$, so $h(b)=\operatorname{lcm}(h(a), h(b))$, and hence $h(b)$ is a multiple of $h(a)$. Conversely, if $h(b)$ is a multiple of $h(a)$, then $h(b)=\operatorname{lcm}(h(a), h(b))$. On the other hand, we have $h(c)=\operatorname{lcm}(h(a), h(b))$. Since $h$ is injective, we have $c=b$, so $b$ is a multiple of $a$.  We apply the following claim for $H=h$. Claim. Suppose $H: \mathbb{Z}_{\geqslant 2} \rightarrow \mathbb{Z}_{\geqslant 2}$ is a bijection such that $a$ divides $b$ if and only if $H(a)$ divides $H(b)$. Then:  1. $H(p)$ is prime if and only if $p$ is prime; 2. $H\left(\prod_{i=1}^{m} p_{i}^{e_{i}}\right)=\prod_{i=1}^{m} H\left(p_{i}\right)^{e_{i}}$ i.e. $H$ is completely multiplicative; 3. $H$ preserves gcd and lcm.  Proof. We define $H(1)=1$, and consider the bijection $H: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$. By the conditions on $H$, for any $n \in \mathbb{Z}_{\geqslant 2}, n$ and $H(n)$ have the same number of divisors. Hence $H(p)$ is prime if and only if $p$ is prime.  Since the only prime dividing $H\left(p^{r}\right)$ is $H(p)$, we have $H\left(p^{r}\right)=H(p)^{s}$ for some $s \geqslant 1$. Counting the number of divisors, we have $s=r$, so $H\left(p^{r}\right)=H(p)^{r}$ for any prime $p$ and $r \geqslant 1$.  For $a, b \in \mathbb{Z}_{>0}$, recall that $\operatorname{gcd}(a, b)$ is a unique positive integer satisfying the following condition: for any $c \in \mathbb{Z}_{>0}, c$ divides $\operatorname{gcd}(a, b)$ if and only if $c$ divides both $a$ and $b$. By the condition on $H$, for any $c \in \mathbb{Z}_{>0}, H(c)$ divides $H(\operatorname{gcd}(a, b))$ if and only if $H(c)$ divides both $H(a)$ and $H(b)$. Hence we have $H(\operatorname{gcd}(a, b))=\operatorname{gcd}(H(a), H(b))$.  Similarly, we have $H(\operatorname{lcm}(a, b))=\operatorname{lcm}(H(a), H(b))$. Hence we have  $$ \begin{aligned} H\left(\prod_{i=1}^{m} p_{i}^{e_{i}}\right) & =H\left(\operatorname{lcm}\left(p_{1}^{e_{1}}, \ldots, p_{r}^{e_{r}}\right)\right)=\operatorname{lcm}\left(H\left(p_{1}^{e_{1}}\right), \ldots, H\left(p_{r}^{e_{r}}\right)\right)=\operatorname{lcm}\left(H\left(p_{1}\right)^{e_{1}}, \ldots, H\left(p_{r}\right)^{e_{r}}\right) \\ & =\prod_{i=1}^{m} H\left(p_{i}\right)^{e_{i}} \end{aligned} $$  since $H\left(p_{i}\right)$ and $H\left(p_{j}\right)$ are different primes for $i \neq j$. Take two primes $p \neq q$, and let $x, y$ be positive integers such that  $$ g(p)+(x-1) h(p)=g(q)+(y-1) h(q) . $$  This is possible as $h(p)$ and $h(q)$ are two distinct primes. For every $k \geqslant 0$, by $P(p, x+k h(q))$ and $P(q, y+k h(p))$, we have  $$ \left\{\begin{array}{l} g(p \cdot h(x+k h(q)))=g(p)+(x+k h(q)-1) h(p), \\ g(q \cdot h(y+k h(p)))=g(q)+(y+k h(p)-1) h(q), \end{array}\right. $$  where the right hand sides are equal. By the injectivity of $g$, we have  $$ p \cdot h(x+k h(q))=q \cdot h(y+k h(p)) . $$  So, $h(y+k h(p))$ is divisible by $p$ for all $k \geqslant 0$. By the above Claim, $h$ preserves gcd, so  $$ h(\operatorname{gcd}(y, h(p)))=\operatorname{gcd}(h(y), h(y+h(p))) $$  is divisible by $p$. Since $h(p)$ is a prime, $y$ must be divisible by $h(p)$. Moreover, $h(h(p))$ is also a prime, so we have $h(h(p))=p$. The function $h \circ h$ is completely multiplicative, so we have $h(h(n))=n$ for every $n \geqslant 2$.  By $P(a, h(b))$ and $P(b, h(a))$, we have  $$ \left\{\begin{array}{l} g(a b)=g(a \cdot h(h(b))) \stackrel{P(a, h(b))}{=} g(a)+(h(b)-1) h(a), \\ g(b a)=g(b \cdot h(h(a))) \stackrel{P(b, h(a))}{=} g(b)+(h(a)-1) h(b), \end{array}\right. $$  so  $$ g(a)+h(a)(h(b)-1)=g(a b)=g(b)+h(b)(h(a)-1) . $$  Hence $g(a)-h(a)=g(b)-h(b)$ for any $a, b \geqslant 2$, so $g-h$ is a constant function. By comparing the images of $g$ and $h$, the difference is -1 , i.e. $g(a)-h(a)=1$ for any $a \geqslant 2$.  So, we have $g(h(a))=h(h(a))-1=a-1$. By definition,  $$ f(a-1)=h(a)=f^{g(h(a))}(1)=f^{a-1}(1) . $$  By the injectivity of $f$, we have $f^{a-2}(1)=a-1$ for every $a \geqslant 2$. From this, we can deduce inductively that $f(a)=a+1$ for every $a \geqslant 1$. ![](https://cdn.mathpix.com/cropped/2025_01_09_9538693ed865d47d2de5g-99.jpg?height=293&width=529&top_left_y=2255&top_left_x=769)


Professor Oak is feeding his 100 Pokémon. Each Pokémon has a bowl whose capacity is a positive real number of kilograms. These capacities are known to Professor Oak. The total capacity of all the bowls is 100 kilograms. Professor Oak distributes 100 kilograms of food in such a way that each Pokémon receives a non-negative integer number of kilograms of food (which may be larger than the capacity of their bowl). The dissatisfaction level of a Pokémon who received $N$ kilograms of food and whose bowl has a capacity of $C$ kilograms is equal to $|N-C|$.  Find the smallest real number $D$ such that, regardless of the capacities of the bowls, Professor Oak can distribute the food in a way that the sum of the dissatisfaction levels over all the 100 Pokémon is at most $D$. (Ukraine) Answer: The answer is $D=50$.

Solution: First, consider the situation where 99 bowls have a capacity of 0.5 kilograms and the last bowl has a capacity of 50.5 kilograms. No matter how Professor Oak distributes the food, the dissatisfaction level of every Pokémon will be at least 0.5 . This amounts to a total dissatisfaction level of at least 50 , proving that $D \geqslant 50$.  Now we prove that no matter what the capacities of the bowls are, Professor Oak can always distribute food in a way that the total dissatisfaction level is at most 50 . We start by fixing some notation. We number the Pokémon from 1 to 100 . Let $C_{i}>0$ be the capacity of the bowl of the $i^{\text {th }}$ Pokémon. By assumption, we have $C_{1}+C_{2}+\cdots+C_{100}=100$. We write $F_{i}:=C_{i}-\left\lfloor C_{i}\right\rfloor$ for the fractional part of $C_{i}$. Without loss of generality, we may assume that $F_{1} \leqslant F_{2} \leqslant \cdots \leqslant F_{100}$.  Here is a strategy: Professor Oak starts by giving $\left\lfloor C_{i}\right\rfloor$ kilograms of food to the $i^{\text {th }}$ Pokémon. Let  $$ R:=100-\left\lfloor C_{1}\right\rfloor-\left\lfloor C_{2}\right\rfloor-\cdots-\left\lfloor C_{100}\right\rfloor=F_{1}+F_{2}+\cdots+F_{100} \geqslant 0 $$  be the amount of food left. He continues by giving an extra kilogram of food to the $R$ Pokémon numbered $100-R+1,100-R+2, \ldots, 100$, i.e. the Pokémon with the $R$ largest values of $F_{i}$. By doing so, Professor Oak distributed 100 kilograms of food. The total dissatisfaction level with this strategy is  $$ d:=F_{1}+\cdots+F_{100-R}+\left(1-F_{100-R+1}\right)+\cdots+\left(1-F_{100}\right) . $$  We can rewrite  $$ \begin{aligned} d & =2\left(F_{1}+\cdots+F_{100-R}\right)+R-\left(F_{1}+\cdots+F_{100}\right) \\ & =2\left(F_{1}+\cdots+F_{100-R}\right) . \end{aligned} $$  Now, observe that the arithmetic mean of $F_{1}, F_{2}, \ldots, F_{100-R}$ is not greater than the arithmetic mean of $F_{1}, F_{2}, \ldots, F_{100}$, because we assumed $F_{1} \leqslant F_{2} \leqslant \cdots \leqslant F_{100}$. Therefore  $$ d \leqslant 2(100-R) \cdot \frac{F_{1}+\cdots+F_{100}}{100}=2 \cdot \frac{R(100-R)}{100} $$  Finally, we use the AM-GM inequality to see that $R(100-R) \leqslant \frac{100^{2}}{2^{2}}$ which implies $d \leqslant 50$. We conclude that there is always a distribution for which the total dissatisfaction level is at most 50 , proving that $D \leqslant 50$.


Solution: We adopt the same notation as in   This solution uses the probabilistic method. We consider all distributions in which each Pokémon receives $\left\lfloor C_{i}\right\rfloor+\varepsilon_{i}$ kilograms of food, where $\varepsilon_{i} \in\{0,1\}$ and $\varepsilon_{1}+\varepsilon_{2}+\cdots+\varepsilon_{100}=R$. There are $\binom{100}{R}$ such distributions. Suppose each of them occurs in an equal probability. In other words,  $$ \varepsilon_{i}= \begin{cases}0 & \text { with probability } \frac{100-R}{100} \\ 1 & \text { with probability } \frac{R}{100}\end{cases} $$  The expected value of the dissatisfaction level of the $i^{\text {th }}$ Pokémon is  $$ \frac{100-R}{100}\left(C_{i}-\left\lfloor C_{i}\right\rfloor\right)+\frac{R}{100}\left(\left\lfloor C_{i}\right\rfloor+1-C_{i}\right)=\frac{100-R}{100} F_{i}+\frac{R}{100}\left(1-F_{i}\right) $$  Hence, the expected value of the total dissatisfaction level is  $$ \begin{aligned} \sum_{i=1}^{100}\left(\frac{100-R}{100} F_{i}+\frac{R}{100}\left(1-F_{i}\right)\right) & =\frac{100-R}{100} \sum_{i=1}^{100} F_{i}+\frac{R}{100} \sum_{i=1}^{100}\left(1-F_{i}\right) \\ & =\frac{100-R}{100} \cdot R+\frac{R}{100} \cdot(100-R) \\ & =2 \cdot \frac{R(100-R)}{100} . \end{aligned} $$  As in Solution 1, this is at most 50. We conclude that there is at least one distribution for which the total dissatisfaction level is at most 50 .


Let $\mathbb{R}$ be the set of real numbers. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that  $$ f(x+y) f(x-y) \geqslant f(x)^{2}-f(y)^{2} $$  for every $x, y \in \mathbb{R}$. Assume that the inequality is strict for some $x_{0}, y_{0} \in \mathbb{R}$. Prove that $f(x) \geqslant 0$ for every $x \in \mathbb{R}$ or $f(x) \leqslant 0$ for every $x \in \mathbb{R}$. (Malaysia) Common remarks. We will say that $f$ has constant sign, if $f$ satisfies the conclusion of the problem.

Solution: We introduce the new variables $s:=x+y$ and $t:=x-y$. Equivalently, $x=\frac{s+t}{2}$ and $y=\frac{s-t}{2}$. The inequality becomes  $$ f(s) f(t) \geqslant f\left(\frac{s+t}{2}\right)^{2}-f\left(\frac{s-t}{2}\right)^{2} $$  for every $s, t \in \mathbb{R}$. We replace $t$ by $-t$ to obtain  $$ f(s) f(-t) \geqslant f\left(\frac{s-t}{2}\right)^{2}-f\left(\frac{s+t}{2}\right)^{2} $$  Summing the previous two inequalities gives  $$ f(s)(f(t)+f(-t)) \geqslant 0 $$  for every $s, t \in \mathbb{R}$. This inequality is strict for $s=x_{0}+y_{0}$ and $t=x_{0}-y_{0}$ by assumption. In particular, there exists some $t_{0}=x_{0}-y_{0}$ for which $f\left(t_{0}\right)+f\left(-t_{0}\right) \neq 0$. Since $f(s)\left(f\left(t_{0}\right)+\right.$ $\left.f\left(-t_{0}\right)\right) \geqslant 0$ for every $s \in \mathbb{R}$, we conclude that $f(s)$ must have constant sign.


Solution: We do the same change of variables as in   $$ f(s) f(t) \geqslant f\left(\frac{s+t}{2}\right)^{2}-f\left(\frac{s-t}{2}\right)^{2} $$  In this solution, we replace $s$ by $-s($ instead of $t$ by $-t)$. This gives  $$ f(-s) f(t) \geqslant f\left(\frac{-s+t}{2}\right)^{2}-f\left(\frac{-s-t}{2}\right)^{2} $$  We now go back to the original inequality. Substituting $x=y$ gives $f(2 x) f(0) \geqslant 0$ for every $x \in \mathbb{R}$. If $f(0) \neq 0$, then we conclude that $f$ indeed has constant sign. From now on, we will assume that  $$ f(0)=0 . $$  Substituting $x=-y$ gives $f(-x)^{2} \geqslant f(x)^{2}$. By permuting $x$ and $-x$, we conclude that  $$ f(-x)^{2}=f(x)^{2} $$  for every $x \in \mathbb{R}$. Using the relation $f(x)^{2}=f(-x)^{2}$, we can rewrite (2) as  $$ f(-s) f(t) \geqslant f\left(\frac{s-t}{2}\right)^{2}-f\left(\frac{s+t}{2}\right)^{2} $$  Summing this inequality with (1), we obtain  $$ (f(s)+f(-s)) f(t) \geqslant 0 $$  for every $s, t \in \mathbb{R}$ and we can conclude as in Solution 1 .


Solution: We prove the contrapositive of the problem statement. Assume that there exist $a, b \in \mathbb{R}$ such that $f(a)<0$ and $f(b)>0$. We want to prove that the inequality is actually an equality, i.e. it is never strict. Lemma 1. The function $f$ is odd, i.e. $f(x)+f(-x)=0$ for every $x \in \mathbb{R}$. Proof. We plug in $x=\frac{a+u}{2}$ and $y=\frac{a-u}{2}$ in the original inequality, where $u$ is a free variable. We obtain  $$ f(a) f(u) \geqslant f\left(\frac{a+u}{2}\right)^{2}-f\left(\frac{a-u}{2}\right)^{2} . $$  Replacing $u$ with $-u$ and summing the two inequalities as in the previous solutions, we get  $$ f(a)(f(u)+f(-u)) \geqslant 0 $$  for every $u \in \mathbb{R}$. Since $f(a)<0$ by assumption, we conclude that $f(u)+f(-u) \leqslant 0$ for every $u \in \mathbb{R}$.  We can repeat the above argument with $b$ instead of $a$. Since $f(b)>0$ by assumption, we conclude that $f(u)+f(-u) \geqslant 0$ for every $u \in \mathbb{R}$. This implies that $f(u)+f(-u)=0$ for every $u \in \mathbb{R}$.  Now, using that $f$ is odd, we can write the following chain of inequalities  $$ \begin{aligned} f(x)^{2}-f(y)^{2} & \leqslant f(x+y) f(x-y) \\ & =-f(y+x) f(y-x) \\ & \leqslant-\left(f(y)^{2}-f(x)^{2}\right) \\ & =f(x)^{2}-f(y)^{2} . \end{aligned} $$  We conclude that every inequality above is actually an inequality, so  $$ f(x+y) f(x-y)=f(x)^{2}-f(y)^{2} $$  for every $x, y \in \mathbb{R}$.


Solution: As in   In this solution, we construct an argument by multiplying inequalities, rather than adding them as in Solutions 1-3. Lemma 2. $f(b) f(-b)<0$. Proof. Let $x_{1}:=\frac{a+b}{2}$ and $y_{1}:=\frac{a-b}{2}$ so that $a=x_{1}+y_{1}$ and $b=x_{1}-y_{1}$. Plugging in $x=x_{1}$ and $y=y_{1}$, we obtain  $$ 0>f(a) f(b)=f\left(x_{1}+y_{1}\right) f\left(x_{1}-y_{1}\right) \geqslant f\left(x_{1}\right)^{2}-f\left(y_{1}\right)^{2} $$  which implies $f\left(x_{1}\right)^{2}-f\left(y_{1}\right)^{2}<0$. Similarly, by plugging in $x=y_{1}$ and $y=x_{1}$, we get  $$ f(a) f(-b)=f\left(y_{1}+x_{1}\right) f\left(y_{1}-x_{1}\right) \geqslant f\left(y_{1}\right)^{2}-f\left(x_{1}\right)^{2} . $$  Using $f\left(x_{1}\right)^{2}-f\left(y_{1}\right)^{2}<0$, we conclude $f(a) f(-b)>0$. If we multiply the two inequalities $f(a) f(b)<0$ and $f(a) f(-b)>0$, we get $f(a)^{2} f(b) f(-b)<0$ and hence  $$ f(b) f(-b)<0 $$  Lemma 3. $f(x) f(-x) \leqslant 0$ for every $x \in \mathbb{R}$.  Proof. As in Solution 2, we prove that $f(x)^{2}=f(-x)^{2}$ for every $x \in \mathbb{R}$ and we rewrite the original inequality as  $$ f(s) f(t) \geqslant f\left(\frac{s+t}{2}\right)^{2}-f\left(\frac{s-t}{2}\right)^{2} $$  We replace $s$ by $-s$ and $t$ by $-t$, and use the relation $f(x)^{2}=f(-x)^{2}$, to get  $$ \begin{aligned} f(-s) f(-t) & \geqslant f\left(\frac{-s-t}{2}\right)^{2}-f\left(\frac{-s+t}{2}\right)^{2} \\ & =f\left(\frac{s+t}{2}\right)^{2}-f\left(\frac{s-t}{2}\right)^{2} \end{aligned} $$  Up to replacing $t$ by $-t$, we can assume that $f\left(\frac{s+t}{2}\right)^{2}-f\left(\frac{s-t}{2}\right)^{2} \geqslant 0$. Multiplying the two previous inequalities leads to  $$ f(s) f(-s) f(t) f(-t) \geqslant 0 $$  for every $s, t \in \mathbb{R}$. This shows that $f(s) f(-s)$ (as a function of $s$ ) has constant sign. Since $f(b) f(-b)<0$, we conclude that  $$ f(x) f(-x) \leqslant 0 $$  for every $x \in \mathbb{R}$. Lemma 3, combined with the relation $f(x)^{2}=f(-x)^{2}$, implies $f(x)+f(-x)=0$ for every $x \in \mathbb{R}$, i.e. $f$ is odd. We conclude with the same argument as in Solution 3.  Comment. The presence of squares on the right-hand side of the inequality is not crucial as Solution 1 illustrates very well. However, it allows non-constant functions such as $f(x)=|x|$ to satisfy the conditions of the problem statement.


Let $\mathbb{R}_{>0}$ be the set of positive real numbers. Determine all functions $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ such that  $$ x(f(x)+f(y)) \geqslant(f(f(x))+y) f(y) $$  for every $x, y \in \mathbb{R}_{>0}$. (Belgium) Answer: All functions $f(x)=\frac{c}{x}$ for some $c>0$.

Solution: Let $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ be a function that satisfies the inequality of the problem statement. We will write $f^{k}(x)=f(f(\cdots f(x) \cdots))$ for the composition of $f$ with itself $k$ times, with the convention that $f^{0}(x)=x$. Substituting $y=x$ gives  $$ x \geqslant f^{2}(x) . $$  Substituting $x=f(y)$ instead leads to $f(y)+f^{2}(y) \geqslant y+f^{3}(y)$, or equivalently  $$ f(y)-f^{3}(y) \geqslant y-f^{2}(y) $$  We can generalise this inequality. If we replace $y$ by $f^{n-1}(y)$ in the above inequality, we get  $$ f^{n}(y)-f^{n+2}(y) \geqslant f^{n-1}(y)-f^{n+1}(y) $$  for every $y \in \mathbb{R}_{>0}$ and for every integer $n \geqslant 1$. In particular, $f^{n}(y)-f^{n+2}(y) \geqslant y-f^{2}(y) \geqslant 0$ for every $n \geqslant 1$. Hereafter consider even integers $n=2 m$. Observe that  $$ y-f^{2 m}(y)=\sum_{i=0}^{m-1}\left(f^{2 i}(y)-f^{2 i+2}(y)\right) \geqslant m\left(y-f^{2}(y)\right) $$  Since $f$ takes positive values, it holds that $y-f^{2 m}(y)<y$ for every $m \geqslant 1$. So, we have proved that $y>m\left(y-f^{2}(y)\right)$ for every $y \in \mathbb{R}_{>0}$ and every $m \geqslant 1$. Since $y-f^{2}(y) \geqslant 0$, this holds if only if  $$ f^{2}(y)=y $$  for every $y \in \mathbb{R}_{>0}$. The original inequality becomes  $$ x f(x) \geqslant y f(y) $$  for every $x, y \in \mathbb{R}_{>0}$. Hence, $x f(x)$ is constant. We conclude that $f(x)=c / x$ for some $c>0$. We now check that all the functions of the form $f(x)=c / x$ are indeed solutions of the original problem. First, note that all these functions satisfy $f(f(x))=c /(c / x)=x$. So it's sufficient to check that $x f(x) \geqslant y f(y)$, which is true since $c \geqslant c$.


Solution: Let $f: \mathbb{R}_{>0} \rightarrow \mathbb{R}_{>0}$ be a function that satisfies the inequality of the problem statement. As in   $$ f^{n}(y) \geqslant f^{n+2}(y) $$  for every $y \in \mathbb{R}_{>0}$ and every $n \geqslant 0$. Since $f$ takes positive values, this implies that  $$ y f(y) \geqslant f(y) f^{2}(y) \geqslant f^{2}(y) f^{3}(y) \geqslant \cdots $$  In other words, $y f(y) \geqslant f^{n}(y) f^{n+1}(y)$ for every $y \in \mathbb{R}_{>0}$ and every $n \geqslant 1$.  We replace $x$ by $f^{n}(x)$ in the original inequality and get  $$ f^{n}(x)-f^{n+2}(x) \geqslant \frac{y f(y)-f^{n}(x) f^{n+1}(x)}{f(y)} . $$  Using that $x f(x) \geqslant f^{n}(x) f^{n+1}(x)$, we obtain  $$ f^{n}(x)-f^{n+2}(x) \geqslant \frac{y f(y)-x f(x)}{f(y)} $$  for every $n \geqslant 0$. The same trick as in Solution 1 gives  $$ x>x-f^{2 m}(x)=\sum_{i=0}^{m-1}\left(f^{2 i}(x)-f^{2 i+2}(x)\right) \geqslant m \cdot \frac{y f(y)-x f(x)}{f(y)} $$  for every $x, y \in \mathbb{R}_{>0}$ and every $m \geqslant 1$. Possibly permuting $x$ and $y$, we may assume that $y f(y)-x f(x) \geqslant 0$ then the above inequality implies $x f(x)=y f(y)$. We conclude as in Solution 1.


Let $a_{1}, a_{2}, \ldots, a_{2023}$ be positive integers such that  - $a_{1}, a_{2}, \ldots, a_{2023}$ is a permutation of $1,2, \ldots, 2023$, and - $\left|a_{1}-a_{2}\right|,\left|a_{2}-a_{3}\right|, \ldots,\left|a_{2022}-a_{2023}\right|$ is a permutation of $1,2, \ldots, 2022$.  Prove that $\max \left(a_{1}, a_{2023}\right) \geqslant 507$. (Australia)

Solution: For the sake of clarity, we consider and prove the following generalisation of the original problem (which is the case $N=1012$ ):  Let $N$ be a positive integer and $a_{1}, a_{2}, \ldots, a_{2 N-1}$ be positive integers such that  - $a_{1}, a_{2}, \ldots, a_{2 N-1}$ is a permutation of $1,2, \ldots, 2 N-1$, and - $\left|a_{1}-a_{2}\right|,\left|a_{2}-a_{3}\right|, \ldots,\left|a_{2 N-2}-a_{2 N-1}\right|$ is a permutation of $1,2, \ldots, 2 N-2$.  Then $a_{1}+a_{2 N-1} \geqslant N+1$ and hence $\max \left(a_{1}, a_{2 N-1}\right) \geqslant\left\lceil\frac{N+1}{2}\right\rceil$. Now we proceed to the proof of the generalised statement. We introduce the notion of score of a number $a \in\{1,2, \ldots, 2 N-1\}$. The score of $a$ is defined to be  $$ s(a):=|a-N| . $$  Note that, by the triangle inequality,  $$ |a-b| \leqslant|a-N|+|N-b|=s(a)+s(b) . $$  Considering the sum $\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\cdots+\left|a_{2 N-2}-a_{2 N-1}\right|$, we find that  $$ \begin{aligned} (N-1)(2 N-1) & =\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\cdots+\left|a_{2 N-2}-a_{2 N-1}\right| \\ & \leqslant 2\left(s\left(a_{1}\right)+s\left(a_{2}\right)+\cdots+s\left(a_{2 N-1}\right)\right)-\left(s\left(a_{1}\right)+s\left(a_{2 N-1}\right)\right) \\ & =2 N(N-1)-\left(s\left(a_{1}\right)+s\left(a_{2 N-1}\right)\right) . \end{aligned} $$  For the last equality we used that the numbers $s\left(a_{1}\right), s\left(a_{2}\right), \ldots, s\left(a_{2 N-1}\right)$ are a permutation of $0,1,1,2,2, \ldots, N-1, N-1$.  Hence, $s\left(a_{1}\right)+s\left(a_{2 N-1}\right) \leqslant 2 N(N-1)-(N-1)(2 N-1)=N-1$. We conclude that  $$ \left(N-a_{1}\right)+\left(N-a_{2 N-1}\right) \leqslant s\left(a_{1}\right)+s\left(a_{2 N-1}\right) \leqslant N-1, $$  which implies $a_{1}+a_{2 N-1} \geqslant N+1$. Comment 1. In the case $N=1012$, such a sequence with $\max \left(a_{1}, a_{2023}\right)=507$ indeed exists: $507,1517,508,1516, \ldots, 1011,1013,1012,2023,1,2022,2, \ldots, 1518,506$.  For a general even number $N$, a sequence with $\max \left(a_{1}, a_{2 N-1}\right)=\left\lceil\frac{N+1}{2}\right\rceil$ can be obtained similarly. If $N \geqslant 3$ is odd, the inequality is not sharp, because $\max \left(a_{1}, a_{2 N-1}\right)=\frac{N+1}{2}$ and $a_{1}+a_{2 N-1} \geqslant N+1$ together imply $a_{1}=a_{2 N-1}=\frac{N+1}{2}$, a contradiction.  Comment 2. The formulation of the author's submission was slightly different: Author's formulation. Consider a sequence of positive integers $a_{1}, a_{2}, a_{3}, \ldots$ such that the following conditions hold for all positive integers $m$ and $n$ :  - $a_{n+2023}=a_{n}+2023$, - If $\left|a_{n+1}-a_{n}\right|=\left|a_{m+1}-a_{m}\right|$, then $2023 \mid(n-m)$, and - The sequence contains every positive integer.  Prove that $a_{1} \geqslant 507$. The two formulations are equivalent up to relatively trivial arguments. Suppose $\left(a_{n}\right)$ is a sequence satisfying the author's formulation. From the first and third conditions, we see that $a_{1}, \ldots, a_{2023}$ is a permutation of $1, \ldots, 2023$. Moreover, the sequence $\left|a_{i}-a_{i+1}\right|$ for $i=1,2, \ldots, 2022$ consists of positive integers $\leqslant 2022$ and has pairwise distinct elements by the second condition. Hence, it is a permutation of $1, \ldots, 2022$. It also holds that $a_{1}>a_{2023}$, since if $a_{1}<a_{2023}$ then $\left|a_{2024}-a_{2023}\right|=$ $\left|2023+a_{1}-a_{2023}\right| \leqslant 2022$, which should be equal to $\left|a_{i}-a_{i+1}\right|$ for some $1 \leqslant i \leqslant 2022$, contradicting the second condition. This reduces the problem to the Shortlist formulation.  Conversely, if the numbers $a_{1}, \ldots, a_{2023}$ satisfy the conditions of the Shortlist formulation, then, after possibly reversing the sequence to ensure $a_{1}>a_{2023}$, the sequence can be extended to an infinite sequence satisfying the conditions of the author's formulation.


Let $k \geqslant 2$ be an integer. Determine all sequences of positive integers $a_{1}, a_{2}, \ldots$ for which there exists a monic polynomial $P$ of degree $k$ with non-negative integer coefficients such that  $$ P\left(a_{n}\right)=a_{n+1} a_{n+2} \cdots a_{n+k} $$  for every integer $n \geqslant 1$. (Malaysia) Answer: The sequence $\left(a_{n}\right)$ must be an arithmetic progression consisting of positive integers with common difference $d \geqslant 0$, and $P(x)=(x+d) \cdots(x+k d)$.  Common remarks. The following arguments and observations are implicit in the solutions given below.  Suppose the sequence $\left(a_{n}\right)$ is an arithmetic progression with common difference $d \geqslant 0$. Then it satisfies the condition with  $$ P(x)=(x+d) \cdots(x+k d) $$  This settles one direction. Now suppose $\left(a_{n}\right)$ is a sequence satisfying the condition. We will show that it is a non-decreasing arithmetic progression.  Since $P(x)$ has non-negative integer coefficients, it is strictly increasing on the positive real line. In particular, it holds that, for any positive integer $x, y$,  $$ P(x)<P(y) \quad \Longleftrightarrow \quad x<y $$  Furthermore, if the sequence $\left(a_{n}\right)$ is eventually constant, then $P(x)=x^{k}$ and the sequence $\left(a_{n}\right)$ is actually constant. Indeed, if $P(x)$ were not the polynomial $x^{k}$, then $P\left(a_{n}\right)=a_{n+1} \cdots a_{n+k}$ cannot be satisfied for $n$ such that $a_{n}=\cdots=a_{n+k}$. By a descending induction, we conclude that $\left(a_{n}\right)$ is constant. Thus we can restrict to the case $\left(a_{n}\right)$ is not eventually constant.

Solution: We assume that $\left(a_{n}\right)$ is not eventually constant. Step 1. The first goal is to show that the sequence must be increasing, i.e. $a_{n}<a_{n+1}$ for all $n \geqslant 1$.  First, by comparing the two equalities  $$ \begin{aligned} P\left(a_{n}\right) & =a_{n+1} a_{n+2} \cdots a_{n+k}, \\ P\left(a_{n+1}\right) & =a_{n+2} \cdots a_{n+k} a_{n+k+1}, \end{aligned} $$  we observe that  $$ \begin{aligned} & a_{n}<a_{n+1} \Longleftrightarrow P\left(a_{n}\right)<P\left(a_{n+1}\right) \\ & a_{n}>a_{n+1} \Longleftrightarrow \\ & a_{n}=a_{n+1} \Longleftrightarrow a_{n+1}<a_{n+k+1}, \\ & P\left(a_{n}\right)>P\left(a_{n+1}\right) \Longleftrightarrow a_{n+1}>a_{n+k+1}, \\ & \end{aligned}, P\left(a_{n+1}\right) \quad \Longleftrightarrow a_{n+1}=a_{n+k+1} . $$  Claim 1. $\quad a_{n} \leqslant a_{n+1}$ for all $n \geqslant 1$. Proof. Suppose, to the contrary, that $a_{n(0)-1}>a_{n(0)}$ for some $n(0) \geqslant 2$. We will give an infinite sequence of positive integers $n(0)<n(1)<\cdots$ satisfying  $$ a_{n(i)-1}>a_{n(i)} \text { and } a_{n(i)}>a_{n(i+1)} . $$  Then $a_{n(0)}, a_{n(1)}, a_{n(2)}, \ldots$ is an infinite decreasing sequence of positive integers, which is absurd. We construct such a sequence inductively. If we have chosen $n(i)$, then we let $n(i+1)$ be the smallest index larger than $n(i)$ such that $a_{n(i)}>a_{n(i+1)}$. Note that such an index always exists and satisfies $n(i)+1 \leqslant n(i+1) \leqslant n(i)+k$ because $a_{n(i)}>a_{n(i)+k}$ by (2). We need to check that $a_{n(i+1)-1}>a_{n(i+1)}$. This is immediate if $n(i+1)=n(i)+1$ by construction. If $n(i+1) \geqslant n(i)+2$, then $a_{n(i+1)-1} \geqslant a_{n(i)}$ by minimality of $n(i+1)$, and so $a_{n(i+1)-1} \geqslant a_{n(i)}>a_{n(i+1)}$.  We are now ready to prove that the sequence $a_{n}$ is increasing. Suppose $a_{n}=a_{n+1}$ for some $n \geqslant 1$. Then we also have $a_{n+1}=a_{n+k+1}$ by (3), and since the sequence is non-decreasing we have $a_{n}=a_{n+1}=a_{n+2}=\cdots=a_{n+k+1}$. We repeat the argument for $a_{n+k}=a_{n+k+1}$ and get that the sequence is eventually constant, which contradicts our assumption. Hence  $$ a_{n}<a_{n+1} \text { for all } n \geqslant 1 $$  Step 2. The next and final goal is to prove that the sequence $a_{n}$ is an arithmetic progression. Observe that we can make differences of terms appear as follows  $$ \begin{aligned} P\left(a_{n}\right) & =a_{n+1} a_{n+2} \cdots a_{n+k} \\ & =\left(a_{n}+\left(a_{n+1}-a_{n}\right)\right)\left(a_{n}+\left(a_{n+2}-a_{n}\right)\right) \cdots\left(a_{n}+\left(a_{n+k}-a_{n}\right)\right) . \end{aligned} $$  We will prove that, for $n$ large enough, the sum  $$ \left(a_{n+1}-a_{n}\right)+\left(a_{n+2}-a_{n}\right)+\cdots+\left(a_{n+k}-a_{n}\right) $$  is equal to the coefficient $b$ of the term $x^{k-1}$ in $P$. The argument is based on the following claim. Claim 2. There exists a bound $A$ with the following properties:  1. If $\left(c_{1}, \ldots, c_{k}\right)$ is a $k$-tuple of positive integers with $c_{1}+\cdots+c_{k}>b$, then for every $x \geqslant A$ we have $P(x)<\left(x+c_{1}\right)\left(x+c_{2}\right) \cdots\left(x+c_{k}\right)$. 2. If $\left(c_{1}, \ldots, c_{k}\right)$ is a $k$-tuple of positive integers with $c_{1}+\cdots+c_{k}<b$, then for every $x \geqslant A$ we have $P(x)>\left(x+c_{1}\right)\left(x+c_{2}\right) \cdots\left(x+c_{k}\right)$.  Proof. It suffices to show parts 1 and 2 separately, because then we can take the maximum of two bounds.  We first show part 1 . For each single $\left(c_{1}, \ldots, c_{k}\right)$ such a bound $A$ exists since  $$ P(x)-\left(x+c_{1}\right)\left(x+c_{2}\right) \cdots\left(x+c_{k}\right)=\left(b-\left(c_{1}+\cdots+c_{k}\right)\right) x^{k-1}+(\text { terms of degree } \leqslant k-2) $$  has negative leading coefficient and hence takes negative values for $x$ large enough. Suppose $A$ is a common bound for all tuples $c=\left(c_{1}, \ldots, c_{k}\right)$ satisfying $c_{1}+\cdots+c_{k}=b+1$ (note that there are only finitely many such tuples). Then, for any tuple $c^{\prime}=\left(c_{1}^{\prime}, \ldots, c_{k}^{\prime}\right)$ with $c_{1}^{\prime}+\cdots+c_{k}^{\prime}>b$, there exists a tuple $c=\left(c_{1}, \ldots, c_{k}\right)$ with $c_{1}+\cdots+c_{k}=b+1$ and $c_{i}^{\prime} \geqslant c_{i}$, and then the inequality for $c^{\prime}$ follows from the inequality for $c$.  We can show part 2 either in a similar way, or by using that there are only finitely many such tuples.  Take $A$ satisfying the assertion of Claim 2, and take $N$ such that $n \geqslant N$ implies $a_{n} \geqslant A$. Then for each $n \geqslant N$, we have  $$ \left(a_{n+1}-a_{n}\right)+\cdots+\left(a_{n+k}-a_{n}\right)=b . $$  By taking the difference of this equality and the equality for $n+1$, we obtain  $$ a_{n+k+1}-a_{n+1}=k\left(a_{n+1}-a_{n}\right) $$  for every $n \geqslant N$. We conclude using an extremal principle. Let $d=\min \left\{a_{n+1}-a_{n} \mid n \geqslant N\right\}$, and suppose it is attained at some index $n \geqslant N$. Since  $$ k d=k\left(a_{n+1}-a_{n}\right)=a_{n+k+1}-a_{n+1}=\sum_{i=1}^{k}\left(a_{n+i+1}-a_{n+i}\right) $$  and each summand is at least $d$, we conclude that $d$ is also attained at $n+1, \ldots, n+k$, and inductively at all $n^{\prime} \geqslant n$. We see that the equation $P(x)=(x+d)(x+2 d) \cdots(x+k d)$ is true for infinitely many values of $x$ (all $a_{n^{\prime}}$ for $n^{\prime} \geqslant n$ ), hence this is an equality of polynomials. Finally we use (backward) induction to show that $a_{n+1}-a_{n}=d$ for every $n \geqslant 1$.


Solution: We assume that $\left(a_{n}\right)$ is not eventually constant. In this solution, we first prove an alternative version of Claim 1. Claim 3. There exist infinitely many $n \geqslant 1$ with  $$ a_{n} \leqslant \min \left\{a_{n+1}, \ldots, a_{n+k}\right\} $$  Proof. Suppose not, then for all but finitely many $n \geqslant 1$, it holds that $a_{n}>\min \left\{a_{n+1}, \ldots, a_{n+k}\right\}$. Hence for all large enough $n$, there always exist some $1 \leqslant l \leqslant k$ such that $a_{n}>a_{n+l}$. This induces an infinite decreasing sequence $a_{n}>a_{n+l_{1}}>a_{n+l_{2}}>\cdots$ of positive integers, which is absurd.  We use Claim 3 to quickly settle the case $P(x)=x^{k}$. In that case, for every $n$ with $a_{n} \leqslant \min \left\{a_{n+1}, \ldots, a_{n+k}\right\}$, since $a_{n+1} \cdots a_{n+k}=a_{n}^{k}$, it implies $a_{n}=a_{n+1}=\cdots=a_{n+k}$. This shows that the sequence is eventually constant, which contradicts our assumption.  From now on, assume  $$ P(x)>x^{k} \text { for all } x>0 $$  Claim 4. For every $M>0$, there exists some $N>0$ such that $a_{n}>M$ for all $n>N$. Proof. Suppose there exists some $M>0$, such that $a_{n} \leqslant M$ for infinitely many $n$. For each $i$ with $a_{i} \leqslant M$, we consider the $k$-tuple $\left(a_{i+1}, \ldots, a_{i+k}\right)$. Then each of the terms in the $k$-tuple is bounded from above by $P\left(a_{i}\right)$, and hence by $P(M)$ too. Since the number of such $k$-tuples is bounded by $P(M)^{k}$, we deduce by the Pigeonhole Principle that there exist some indices $i<j$ such that $\left(a_{i+1}, \ldots, a_{i+k}\right)=\left(a_{j+1}, \ldots, a_{j+k}\right)$. Since $a_{n}$ is uniquely determined by the $k$ terms before it, we conclude that $a_{i+k+1}=a_{j+k+1}$ must hold, and similarly $a_{i+l}=a_{j+l}$ for all $l \geqslant 0$, so the sequence is eventually periodic, for some period $p=j-i$.  Take $K$ such that $a_{n}=a_{n+p}$ for every $n \geqslant K$. Then, by taking the products of the inequalities  $$ a_{n}^{k}<P\left(a_{n}\right)=a_{n+1} \cdots a_{n+k} $$  for $K \leqslant n \leqslant K+p-1$, we obtain  $$ \begin{aligned} \prod_{n=K}^{K+p-1} a_{n}^{k} & <\prod_{n=K}^{K+p-1} a_{n+1} \cdots a_{n+k} \\ & =a_{K+1} a_{K+2}^{2} \cdots a_{K+k-1}^{k-1}\left(\prod_{n=K+k}^{K+p} a_{n}\right)^{k} a_{K+p+1}^{k-1} \cdots a_{K+p+k-2}^{2} a_{K+p+k-1} \\ & =\left(\prod_{n=K}^{K+p-1} a_{n}\right)^{k} \quad(\text { by periodicity }), \end{aligned} $$  which is a contradiction. Write $P(x)=x^{k}+b x^{k-1}+Q(x)$, where $Q(x)$ is of degree at most $k-2$. Take $M$ such that $x>M$ implies $x^{k-1}>Q(x)$. Claim 5. There exist non-negative integers $b_{1}, \cdots, b_{k}$ such that $P(x)=\left(x+b_{1}\right) \cdots\left(x+b_{k}\right)$, and such that, for infinitely many $n \geqslant 1$, we have $a_{n+i}=a_{n}+b_{i}$ for every $1 \leqslant i \leqslant k$. Proof. By Claims 3 and 4, there are infinitely many $n$ such that  $$ a_{n}>M \text { and } a_{n} \leqslant \min \left\{a_{n+1}, \ldots, a_{n+k}\right\} . $$  Call such indices $n$ to be good. We claim that if $n$ is a good index then  $$ \max \left\{a_{n+1}, \ldots, a_{n+k}\right\} \leqslant a_{n}+b $$  Indeed, if $a_{n+i} \geqslant a_{n}+b+1$, then together with $a_{n} \leqslant \min \left\{a_{n+1}, \ldots, a_{n+k}\right\}$ and $a_{n}^{k-1}>Q\left(a_{n}\right)$, we have  $$ a_{n}^{k}+(b+1) a_{n}^{k-1}>a_{n}^{k}+b a_{n}^{k-1}+Q\left(a_{n}\right)=P\left(a_{n}\right) \geqslant\left(a_{n}+b+1\right) a_{n}^{k-1}, $$  a contradiction. Hence for each good index $n$, we may write $a_{n+i}=a_{n}+b_{i}$ for all $1 \leqslant i \leqslant k$ for some choices of $\left(b_{1}, \ldots, b_{k}\right)$ (which may depend on $n$ ) and $0 \leqslant b_{i} \leqslant b$. Again by Pigeonhole Principle, some $k$-tuple $\left(b_{1}, \ldots, b_{k}\right)$ must be chosen for infinitely such good indices $n$. This means that the equation $P\left(a_{n}\right)=\left(a_{n}+b_{1}\right) \cdots\left(a_{n}+b_{k}\right)$ is satisfied by infinitely many good indices $n$. By Claim 4, $a_{n}$ is unbounded among these $a_{n}$ 's, hence $P(x)=\left(x+b_{1}\right) \cdots\left(x+b_{k}\right)$ must hold identically. Claim 6. We have $b_{i}=i b_{1}$ for all $1 \leqslant i \leqslant k$. Proof. Call an index $n$ excellent if $a_{n+i}=a_{n}+b_{i}$ for every $1 \leqslant i \leqslant k$. From Claim 5 we know there are infinitely many excellent $n$.  We first show that for any pair $1 \leqslant i<j \leqslant k$ there is $1 \leqslant l \leqslant k$ such that $b_{j}=b_{i}+b_{l}$. Indeed, for such $i$ and $j$ and for excellent $n, a_{n}+b_{j}$ (which is equal to $a_{n+j}$ ) divides $P\left(a_{n+i}\right)=$ $\prod_{l=1}^{k}\left(a_{n}+b_{i}+b_{l}\right)$, and hence divides $\prod_{l=1}^{k}\left(b_{i}+b_{l}-b_{j}\right)$. Since $a_{n}+b_{j}$ is unbounded among excellent $n$, we have $\prod_{l=1}^{k}\left(b_{i}+b_{l}-b_{j}\right)=0$, hence there is $l$ such that $b_{j}=b_{i}+b_{l}$.  In particular, $b_{j}=b_{i}+b_{l} \geqslant b_{i}$, i.e. $\left(b_{1}, \ldots, b_{k}\right)$ is non-decreasing. Suppose $b_{1}=0$ and $n$ is an excellent number. In particular, it holds that $a_{n}=a_{n+1}$. Moreover, since  $$ a_{n+k+1} P\left(a_{n}\right)=a_{n+1} \cdots a_{n+k+1}=a_{n+1} P\left(a_{n+1}\right), $$  we have $a_{n}=a_{n+1}=a_{n+k+1}$, which divides $P\left(a_{n+i}\right)=\prod_{l=1}^{k}\left(a_{n}+b_{i}+b_{l}\right)$ for each $1 \leqslant i \leqslant k$. Hence $a_{n}$ divides $\prod_{l=1}^{k}\left(b_{i}+b_{l}\right)$. By the same reasoning, we have $b_{i}+b_{l}=0$ for some $l$, but since $b_{i}, b_{l} \geqslant 0$ we obtain $b_{i}=0$ for each $1 \leqslant i \leqslant k$.  Now suppose $b_{1} \geqslant 1$. Then, for each $1 \leqslant i<j \leqslant k$, we have $b_{j}-b_{i}=b_{l} \geqslant b_{1} \geqslant 1$, hence $\left(b_{1}, \ldots, b_{k}\right)$ is strictly increasing. Therefore, the $k-1$ elements $b_{2}<b_{3}<\cdots<b_{k}$ are exactly equal to $b_{1}+b_{1}<\cdots<b_{1}+b_{k-1}$, since they cannot be equal to $b_{1}+b_{k}$. This gives $b_{i}=i b_{1}$ for all $1 \leqslant i \leqslant k$ as desired.  Claim 6 implies $P(x)=(x+d)(x+2 d) \cdots(x+k d)$ for some $d \geqslant 1$, and there are infinitely many indices $n$ with $a_{n+i}=a_{n}+i d$ for $1 \leqslant i \leqslant k$. By backwards induction, $P\left(a_{n-1}\right)=$ $a_{n} \cdots a_{n+k-1}$ implies $a_{n-1}=a_{n}-d$, and so on. Thus $a_{1}, \ldots, a_{n}$ forms an arithmetic progression with common difference $d$. Since $n$ can be arbitrarily large, the whole sequence is an arithmetic progression too, as desired.  Comment 1. A typical solution would first show some kind of increasing property (assuming $a_{n}$ is not constant), and then use that property to deduce informations on the numbers $a_{n+i}-a_{n}(1 \leqslant i \leqslant k)$ and/or on the polynomial $P$.  Solution 1 shows a strict one: $a_{n}<a_{n+1}$ (arguments after Claim 1), which makes the latter part easier. Solution 2 (Claims 3 and 4) shows only weaker increasing properties, which require more complicated/tricky arguments in the latter part but still can solve the problem.  Comment 2. It would be interesting to sort out the case when $P$ can take negative integer coefficients, or $\left(a_{n}\right)$ is just an integer sequence. Then a decreasing arithmetic progression is possible too, yet that is not the only possibility. There exist bounded examples such as $1,-1,1,-1, \ldots$ with $P(x)=-x^{2}$, or $0,1,-1,0,1,-1, \ldots$ with $P(x)=x^{2}-1$. If furthermore $P$ is allowed to be non-monic, then the situation is even more unclear. For instance, the sequence $1,2,4,8, \ldots$ works for the polynomial $P(x)=8 x^{2}$.