Solution: For any point $X$, denote by $t(X)$ the line tangent to $g(X)$ at $X$; notice that $f(t(X))=X$, so $f$ is surjective.  Step 1: We find a point $P$ for which there are at least two different lines $p_{1}$ and $p_{2}$ such that $f\left(p_{i}\right)=P$.  Choose any point $X$. If $X$ does not have this property, take any $Y \in g(X) \backslash\{X\}$; then $f(X Y)=Y$. If $Y$ does not have the property, $t(Y)=X Y$, and the circles $g(X)$ and $g(Y)$ meet again at some point $Z$. Then $f(X Z)=Z=f(Y Z)$, so $Z$ has the required property.  We will show that $P$ is the desired point. From now on, we fix two different lines $p_{1}$ and $p_{2}$ with $f\left(p_{1}\right)=f\left(p_{2}\right)=P$. Assume for contradiction that $f(\ell)=Q \neq P$ for some line $\ell$ through $P$. We fix $\ell$, and note that $Q \in g(P)$.  Step 2: We prove that $P \in g(Q)$. Take an arbitrary point $X \in \ell \backslash\{P, Q\}$. Two cases are possible for the position of $t(X)$ in relation to the $p_{i}$; we will show that each case (and subcase) occurs for only finitely many positions of $X$, yielding a contradiction.  Case 2.1: $t(X)$ is parallel to one of the $p_{i}$; say, to $p_{1}$. Let $t(X) \operatorname{cross} p_{2}$ at $R$. Then $g(R)$ is the circle $(P R X)$, as $f(R P)=P$ and $f(R X)=X$. Let $R Q$ cross $g(R)$ again at $S$. Then $f(R Q) \in\{R, S\} \cap g(Q)$, so $g(Q)$ contains one of the points $R$ and $S$.  If $R \in g(Q)$, then $R$ is one of finitely many points in the intersection $g(Q) \cap p_{2}$, and each of them corresponds to a unique position of $X$, since $R X$ is parallel to $p_{1}$.  If $S \in g(Q)$, then $\angle(Q S, S P)=\angle(R S, S P)=\angle(R X, X P)=\angle\left(p_{1}, \ell\right)$, so $\angle(Q S, S P)$ is constant for all such points $X$, and all points $S$ obtained in such a way lie on one circle $\gamma$ passing through $P$ and $Q$. Since $g(Q)$ does not contain $P$, it is different from $\gamma$, so there are only finitely many points $S$. Each of them uniquely determines $R$ and thus $X$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-083.jpg?height=843&width=909&top_left_y=207&top_left_x=568)  So, Case 2.1 can occur for only finitely many points $X$. Case 2.2: $t(X)$ crosses $p_{1}$ and $p_{2}$ at $R_{1}$ and $R_{2}$, respectively. Clearly, $R_{1} \neq R_{2}$, as $t(X)$ is the tangent to $g(X)$ at $X$, and $g(X)$ meets $\ell$ only at $X$ and $Q$. Notice that $g\left(R_{i}\right)$ is the circle $\left(P X R_{i}\right)$. Let $R_{i} Q$ meet $g\left(R_{i}\right)$ again at $S_{i}$; then $S_{i} \neq Q$, as $g\left(R_{i}\right)$ meets $\ell$ only at $P$ and $X$. Then $f\left(R_{i} Q\right) \in\left\{R_{i}, S_{i}\right\}$, and we distinguish several subcases. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-083.jpg?height=824&width=866&top_left_y=1501&top_left_x=595)  Subcase 2.2.1: $f\left(R_{1} Q\right)=S_{1}, f\left(R_{2} Q\right)=S_{2}$; so $S_{1}, S_{2} \in g(Q)$. In this case we have $0=\angle\left(R_{1} X, X P\right)+\angle\left(X P, R_{2} X\right)=\angle\left(R_{1} S_{1}, S_{1} P\right)+\angle\left(S_{2} P, S_{2} R_{2}\right)=$ $\angle\left(Q S_{1}, S_{1} P\right)+\angle\left(S_{2} P, S_{2} Q\right)$, which shows $P \in g(Q)$.  Subcase 2.2.2: $f\left(R_{1} Q\right)=R_{1}, f\left(R_{2} Q\right)=R_{2}$; so $R_{1}, R_{2} \in g(Q)$. This can happen for at most four positions of $X$ - namely, at the intersections of $\ell$ with a line of the form $K_{1} K_{2}$, where $K_{i} \in g(Q) \cap p_{i}$.  Subcase 2.2.3: $f\left(R_{1} Q\right)=S_{1}, f\left(R_{2} Q\right)=R_{2}$ (the case $f\left(R_{1} Q\right)=R_{1}, f\left(R_{2} Q\right)=S_{2}$ is similar). In this case, there are at most two possible positions for $R_{2}$ - namely, the meeting points of $g(Q)$ with $p_{2}$. Consider one of them. Let $X$ vary on $\ell$. Then $R_{1}$ is the projection of $X$ to $p_{1}$ via $R_{2}, S_{1}$ is the projection of $R_{1}$ to $g(Q)$ via $Q$. Finally, $\angle\left(Q S_{1}, S_{1} X\right)=\angle\left(R_{1} S_{1}, S_{1} X\right)=$ $\angle\left(R_{1} P, P X\right)=\angle\left(p_{1}, \ell\right) \neq 0$, so $X$ is obtained by a fixed projective transform $g(Q) \rightarrow \ell$ from $S_{1}$. So, if there were three points $X$ satisfying the conditions of this subcase, the composition of the three projective transforms would be the identity. But, if we apply it to $X=Q$, we successively get some point $R_{1}^{\prime}$, then $R_{2}$, and then some point different from $Q$, a contradiction.  Thus Case 2.2 also occurs for only finitely many points $X$, as desired. Step 3: We show that $f(P Q)=P$, as desired. The argument is similar to that in Step 2, with the roles of $Q$ and $X$ swapped. Again, we show that there are only finitely many possible positions for a point $X \in \ell \backslash\{P, Q\}$, which is absurd. Case 3.1: $t(Q)$ is parallel to one of the $p_{i}$; say, to $p_{1}$. Let $t(Q)$ cross $p_{2}$ at $R$; then $g(R)$ is the circle $(P R Q)$. Let $R X \operatorname{cross} g(R)$ again at $S$. Then $f(R X) \in\{R, S\} \cap g(X)$, so $g(X)$ contains one of the points $R$ and $S$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-084.jpg?height=729&width=909&top_left_y=1126&top_left_x=568)  Subcase 3.1.1: $S=f(R X) \in g(X)$. We have $\angle(t(X), Q X)=\angle(S X, S Q)=\angle(S R, S Q)=\angle(P R, P Q)=\angle\left(p_{2}, \ell\right)$. Hence $t(X) \| p_{2}$. Now we recall Case 2.1: we let $t(X)$ cross $p_{1}$ at $R^{\prime}$, so $g\left(R^{\prime}\right)=\left(P R^{\prime} X\right)$, and let $R^{\prime} Q$ meet $g\left(R^{\prime}\right)$ again at $S^{\prime}$; notice that $S^{\prime} \neq Q$. Excluding one position of $X$, we may assume that $R^{\prime} \notin g(Q)$, so $R^{\prime} \neq f\left(R^{\prime} Q\right)$. Therefore, $S^{\prime}=f\left(R^{\prime} Q\right) \in g(Q)$. But then, as in Case 2.1, we get $\angle(t(Q), P Q)=\angle\left(Q S^{\prime}, S^{\prime} P\right)=\angle\left(R^{\prime} X, X P\right)=\angle\left(p_{2}, \ell\right)$. This means that $t(Q)$ is parallel to $p_{2}$, which is impossible. Subcase 3.1.2: $R=f(R X) \in g(X)$. In this case, we have $\angle(t(X), \ell)=\angle(R X, R Q)=\angle\left(R X, p_{1}\right)$. Again, let $R^{\prime}=t(X) \cap p_{1}$; this point exists for all but at most one position of $X$. Then $g\left(R^{\prime}\right)=\left(R^{\prime} X P\right)$; let $R^{\prime} Q$ meet $g\left(R^{\prime}\right)$ again at $S^{\prime}$. Due to $\angle\left(R^{\prime} X, X R\right)=\angle(Q X, Q R)=\angle\left(\ell, p_{1}\right), R^{\prime}$ determines $X$ in at most two ways, so for all but finitely many positions of $X$ we have $R^{\prime} \notin g(Q)$. Therefore, for those positions we have $S^{\prime}=f\left(R^{\prime} Q\right) \in g(Q)$. But then $\angle\left(R X, p_{1}\right)=\angle\left(R^{\prime} X, X P\right)=\angle\left(R^{\prime} S^{\prime}, S^{\prime} P\right)=$ $\angle\left(Q S^{\prime}, S^{\prime} P\right)=\angle(t(Q), Q P)$ is fixed, so this case can hold only for one specific position of $X$ as well.  Thus, in Case 3.1, there are only finitely many possible positions of $X$, yielding a contradiction.  Case 3.2: $t(Q)$ crosses $p_{1}$ and $p_{2}$ at $R_{1}$ and $R_{2}$, respectively. By Step $2, R_{1} \neq R_{2}$. Notice that $g\left(R_{i}\right)$ is the circle $\left(P Q R_{i}\right)$. Let $R_{i} X$ meet $g\left(R_{i}\right)$ at $S_{i}$; then $S_{i} \neq X$. Then $f\left(R_{i} X\right) \in\left\{R_{i}, S_{i}\right\}$, and we distinguish several subcases. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-085.jpg?height=854&width=866&top_left_y=424&top_left_x=589)  Subcase 3.2.1: $f\left(R_{1} X\right)=S_{1}$ and $f\left(R_{2} X\right)=S_{2}$, so $S_{1}, S_{2} \in g(X)$. As in Subcase 2.2.1, we have $0=\angle\left(R_{1} Q, Q P\right)+\angle\left(Q P, R_{2} Q\right)=\angle\left(X S_{1}, S_{1} P\right)+\angle\left(S_{2} P, S_{2} X\right)$, which shows $P \in g(X)$. But $X, Q \in g(X)$ as well, so $g(X)$ meets $\ell$ at three distinct points, which is absurd. Subcase 3.2.2: $f\left(R_{1} X\right)=R_{1}, f\left(R_{2} X\right)=R_{2}$, so $R_{1}, R_{2} \in g(X)$. Now three distinct collinear points $R_{1}, R_{2}$, and $Q$ belong to $g(X)$, which is impossible. Subcase 3.2.3: $f\left(R_{1} X\right)=S_{1}, f\left(R_{2} X\right)=R_{2}$ (the case $f\left(R_{1} X\right)=R_{1}, f\left(R_{2} X\right)=S_{2}$ is similar). We have $\angle\left(X R_{2}, R_{2} Q\right)=\angle\left(X S_{1}, S_{1} Q\right)=\angle\left(R_{1} S_{1}, S_{1} Q\right)=\angle\left(R_{1} P, P Q\right)=\angle\left(p_{1}, \ell\right)$, so this case can occur for a unique position of $X$.  Thus, in Case 3.2 , there is only a unique position of $X$, again yielding the required contradiction.


Let $\mathcal{L}$ be the set of all lines in the plane and let $f$ be a function that assigns to each  line $\ell \in \mathcal{L}$ a point $f(\ell)$ on $\ell$. Suppose that for any point $X$, and for any three lines $\ell_{1}, \ell_{2}, \ell_{3}$ passing through $X$, the points $f\left(\ell_{1}\right), f\left(\ell_{2}\right), f\left(\ell_{3}\right)$ and $X$ lie on a circle.  Prove that there is a unique point $P$ such that $f(\ell)=P$ for any line $\ell$ passing through $P$. (Australia) Common remarks. The condition on $f$ is equivalent to the following: There is some function $g$ that assigns to each point $X$ a circle $g(X)$ passing through $X$ such that for any line $\ell$ passing through $X$, the point $f(\ell)$ lies on $g(X)$. (The function $g$ may not be uniquely defined for all points, if some points $X$ have at most one value of $f(\ell)$ other than $X$; for such points, an arbitrary choice is made.)  If there were two points $P$ and $Q$ with the given property, $f(P Q)$ would have to be both $P$ and $Q$, so there is at most one such point, and it will suffice to show that such a point exists.

Solution: We will get an upper bound on $n$ from the speed at which $v_{2}\left(L_{n}\right)$ grows. From  $$ L_{n}=\left(2^{n}-1\right)\left(2^{n}-2\right) \cdots\left(2^{n}-2^{n-1}\right)=2^{1+2+\cdots+(n-1)}\left(2^{n}-1\right)\left(2^{n-1}-1\right) \cdots\left(2^{1}-1\right) $$  we read  $$ v_{2}\left(L_{n}\right)=1+2+\cdots+(n-1)=\frac{n(n-1)}{2} . $$  On the other hand, $v_{2}(m!)$ is expressed by the Legendre formula as  $$ v_{2}(m!)=\sum_{i=1}^{\infty}\left\lfloor\frac{m}{2^{i}}\right\rfloor . $$  As usual, by omitting the floor functions,  $$ v_{2}(m!)<\sum_{i=1}^{\infty} \frac{m}{2^{i}}=m $$  Thus, $L_{n}=m$ ! implies the inequality  $$ \frac{n(n-1)}{2}<m . $$  In order to obtain an opposite estimate, observe that  $$ L_{n}=\left(2^{n}-1\right)\left(2^{n}-2\right) \cdots\left(2^{n}-2^{n-1}\right)<\left(2^{n}\right)^{n}=2^{n^{2}} . $$  We claim that  $$ 2^{n^{2}}<\left(\frac{n(n-1)}{2}\right)!\text { for } n \geqslant 6 $$  For $n=6$ the estimate (3) is true because $2^{6^{2}}<6.9 \cdot 10^{10}$ and $\left(\frac{n(n-1)}{2}\right)!=15!>1.3 \cdot 10^{12}$. For $n \geqslant 7$ we prove (3) by the following inequalities:  $$ \begin{aligned} \left(\frac{n(n-1)}{2}\right)! & =15!\cdot 16 \cdot 17 \cdots \frac{n(n-1)}{2}>2^{36} \cdot 16^{\frac{n(n-1)}{2}-15} \\ & =2^{2 n(n-1)-24}=2^{n^{2}} \cdot 2^{n(n-2)-24}>2^{n^{2}} . \end{aligned} $$  Putting together (2) and (3), for $n \geqslant 6$ we get a contradiction, since  $$ L_{n}<2^{n^{2}}<\left(\frac{n(n-1)}{2}\right)!<m!=L_{n} $$  Hence $n \geqslant 6$ is not possible. Checking manually the cases $n \leqslant 5$ we find  $$ \begin{gathered} L_{1}=1=1!, \quad L_{2}=6=3!, \quad 5!<L_{3}=168<6!, \\ 7!<L_{4}=20160<8!\quad \text { and } \quad 10!<L_{5}=9999360<11! \end{gathered} $$  So, there are two solutions:  $$ (m, n) \in\{(1,1),(3,2)\} $$


Solution: Like in the previous solution, the cases $n=1,2,3,4$ are checked manually. We will exclude $n \geqslant 5$ by considering the exponents of 3 and 31 in (1).  For odd primes $p$ and distinct integers $a, b$, coprime to $p$, with $p \mid a-b$, the Lifting The Exponent lemma asserts that  $$ v_{p}\left(a^{k}-b^{k}\right)=v_{p}(a-b)+v_{p}(k) . $$  Notice that 3 divides $2^{k}-1$ if only if $k$ is even; moreover, by the Lifting The Exponent lemma we have  $$ v_{3}\left(2^{2 k}-1\right)=v_{3}\left(4^{k}-1\right)=1+v_{3}(k)=v_{3}(3 k) $$  Hence,  $$ v_{3}\left(L_{n}\right)=\sum_{2 k \leqslant n} v_{3}\left(4^{k}-1\right)=\sum_{k \leqslant\left\lfloor\frac{n}{2}\right\rfloor} v_{3}(3 k) $$  Notice that the last expression is precisely the exponent of 3 in the prime factorisation of (3 $\left.\left.\frac{n}{2}\right\rfloor\right)$ !. Therefore  $$ \begin{gathered} v_{3}(m!)=v_{3}\left(L_{n}\right)=v_{3}\left(\left(3\left\lfloor\frac{n}{2}\right\rfloor\right)!\right) \\ 3\left\lfloor\frac{n}{2}\right\rfloor \leqslant m \leqslant 3\left\lfloor\frac{n}{2}\right\rfloor+2 . \end{gathered} $$  Suppose that $n \geqslant 5$. Note that every fifth factor in $L_{n}$ is divisible by $31=2^{5}-1$, and hence we have $v_{31}\left(L_{n}\right) \geqslant\left\lfloor\frac{n}{5}\right\rfloor$. Then  $$ \frac{n}{10} \leqslant\left\lfloor\frac{n}{5}\right\rfloor \leqslant v_{31}\left(L_{n}\right)=v_{31}(m!)=\sum_{k=1}^{\infty}\left\lfloor\frac{m}{31^{k}}\right\rfloor<\sum_{k=1}^{\infty} \frac{m}{31^{k}}=\frac{m}{30} $$  By combining (4) and (5),  $$ 3 n<m \leqslant \frac{3 n}{2}+2 $$  so $n<\frac{4}{3}$ which is inconsistent with the inequality $n \geqslant 5$. Comment 1. There are many combinations of the ideas above; for example combining (2) and (4) also provides $n<5$. Obviously, considering the exponents of any two primes in (1), or considering one prime and the magnitude orders lead to an upper bound on $n$ and $m$.  Comment 2. This problem has a connection to group theory. Indeed, the left-hand side is the order of the group $G L_{n}\left(\mathbb{F}_{2}\right)$ of invertible $n$-by- $n$ matrices with entries modulo 2 , while the right-hand side is the order of the symmetric group $S_{m}$ on $m$ elements. The result thus shows that the only possible isomorphisms between these groups are $G L_{1}\left(\mathbb{F}_{2}\right) \cong S_{1}$ and $G L_{2}\left(\mathbb{F}_{2}\right) \cong S_{3}$, and there are in fact isomorphisms in both cases. In general, $G L_{n}\left(\mathbb{F}_{2}\right)$ is a simple group for $n \geqslant 3$, as it is isomorphic to $\operatorname{PS} L_{n}\left(\mathbb{F}_{2}\right)$.  There is also a near-solution of interest: the left-hand side for $n=4$ is half of the right-hand side when $m=8$; this turns out to correspond to an isomorphism $G L_{4}\left(\mathbb{F}_{2}\right) \cong A_{8}$ with the alternating group on eight elements.  However, while this indicates that the problem is a useful one, knowing group theory is of no use in solving it!


Find all triples $(a, b, c)$ of positive integers such that $a^{3}+b^{3}+c^{3}=(a b c)^{2}$. (Nigeria) Answer: The solutions are $(1,2,3)$ and its permutations. Common remarks. Note that the equation is symmetric. In all solutions, we will assume without loss of generality that $a \geqslant b \geqslant c$, and prove that the only solution is $(a, b, c)=(3,2,1)$.  The first two solutions all start by proving that $c=1$.

Solution: We will start by proving that $c=1$. Note that  $$ 3 a^{3} \geqslant a^{3}+b^{3}+c^{3}>a^{3} . $$  So $3 a^{3} \geqslant(a b c)^{2}>a^{3}$ and hence $3 a \geqslant b^{2} c^{2}>a$. Now $b^{3}+c^{3}=a^{2}\left(b^{2} c^{2}-a\right) \geqslant a^{2}$, and so  $$ 18 b^{3} \geqslant 9\left(b^{3}+c^{3}\right) \geqslant 9 a^{2} \geqslant b^{4} c^{4} \geqslant b^{3} c^{5} $$  so $18 \geqslant c^{5}$ which yields $c=1$. Now, note that we must have $a>b$, as otherwise we would have $2 b^{3}+1=b^{4}$ which has no positive integer solutions. So  $$ a^{3}-b^{3} \geqslant(b+1)^{3}-b^{3}>1 $$  and  $$ 2 a^{3}>1+a^{3}+b^{3}>a^{3} $$  which implies $2 a^{3}>a^{2} b^{2}>a^{3}$ and so $2 a>b^{2}>a$. Therefore  $$ 4\left(1+b^{3}\right)=4 a^{2}\left(b^{2}-a\right) \geqslant 4 a^{2}>b^{4} $$  so $4>b^{3}(b-4)$; that is, $b \leqslant 4$. Now, for each possible value of $b$ with $2 \leqslant b \leqslant 4$ we obtain a cubic equation for $a$ with constant coefficients. These are as follows:  $$ \begin{array}{rlrl} b=2: & & & a^{3}-4 a^{2}+9=0 \\ b=3: & & a^{3}-9 a^{2}+28=0 \\ b=4: & & a^{3}-16 a^{2}+65=0 . \end{array} $$  The only case with an integer solution for $a$ with $b \leqslant a$ is $b=2$, leading to $(a, b, c)=(3,2,1)$. Comment 1.1. Instead of writing down each cubic equation explicitly, we could have just observed that $a^{2} \mid b^{3}+1$, and for each choice of $b$ checked each square factor of $b^{3}+1$ for $a^{2}$.  We could also have observed that, with $c=1$, the relation $18 b^{3} \geqslant b^{4} c^{4}$ becomes $b \leqslant 18$, and we can simply check all possibilities for $b$ (instead of working to prove that $b \leqslant 4$ ). This check becomes easier after using the factorisation $b^{3}+1=(b+1)\left(b^{2}-b+1\right)$ and observing that no prime besides 3 can divide both of the factors.  Comment 1.2. Another approach to finish the problem after establishing that $c \leqslant 1$ is to set $k=b^{2} c^{2}-a$, which is clearly an integer and must be positive as it is equal to $\left(b^{3}+c^{3}\right) / a^{2}$. Then we divide into cases based on whether $k=1$ or $k \geqslant 2$; in the first case, we have $b^{3}+1=a^{2}=\left(b^{2}-1\right)^{2}$ whose only positive root is $b=2$, and in the second case we have $b^{2} \leqslant 3 a$, and so  $$ b^{4} \leqslant(3 a)^{2} \leqslant \frac{9}{2}\left(k a^{2}\right)=\frac{9}{2}\left(b^{3}+1\right), $$  which implies that $b \leqslant 4$.


Solution: Again, we will start by proving that $c=1$. Suppose otherwise that $c \geqslant 2$. We have $a^{3}+b^{3}+c^{3} \leqslant 3 a^{3}$, so $b^{2} c^{2} \leqslant 3 a$. Since $c \geqslant 2$, this tells us that $b \leqslant \sqrt{3 a / 4}$. As the right-hand side of the original equation is a multiple of $a^{2}$, we have $a^{2} \leqslant 2 b^{3} \leqslant 2(3 a / 4)^{3 / 2}$. In other words, $a \leqslant \frac{27}{16}<2$, which contradicts the assertion that $a \geqslant c \geqslant 2$. So there are no solutions in this case, and so we must have $c=1$.  Now, the original equation becomes $a^{3}+b^{3}+1=a^{2} b^{2}$. Observe that $a \geqslant 2$, since otherwise $a=b=1$ as $a \geqslant b$.  The right-hand side is a multiple of $a^{2}$, so the left-hand side must be as well. Thus, $b^{3}+1 \geqslant$ $a^{2}$. Since $a \geqslant b$, we also have  $$ b^{2}=a+\frac{b^{3}+1}{a^{2}} \leqslant 2 a+\frac{1}{a^{2}} $$  and so $b^{2} \leqslant 2 a$ since $b^{2}$ is an integer. Thus $(2 a)^{3 / 2}+1 \geqslant b^{3}+1 \geqslant a^{2}$, from which we deduce $a \leqslant 8$.  Now, for each possible value of $a$ with $2 \leqslant a \leqslant 8$ we obtain a cubic equation for $b$ with constant coefficients. These are as follows:  $$ \begin{array}{ll} a=2: & \\ b^{3}-4 b^{2}+9=0 \\ a=3: & \\ b^{3}-9 b^{2}+28=0 \\ a=4: & \\ a=5: 16 b^{2}+65=0 \\ a=6: & \\ b^{3}-25 b^{2}+126=0 \\ a=7: & \\ b^{3}-36 b^{2}+217=0 \\ a=8: & \\ b^{3}-49 b^{2}+344=0 \\ b^{3}-64 b^{2}+513=0 . \end{array} $$  The only case with an integer solution for $b$ with $a \geqslant b$ is $a=3$, leading to $(a, b, c)=(3,2,1)$. Comment 2.1. As in Solution 1, instead of writing down each cubic equation explicitly, we could have just observed that $b^{2} \mid a^{3}+1$, and for each choice of $a$ checked each square factor of $a^{3}+1$ for $b^{2}$.  Comment 2.2. This solution does not require initially proving that $c=1$, in which case the bound would become $a \leqslant 108$. The resulting cases could, in principle, be checked by a particularly industrious student.


Solution: Set $k=\left(b^{3}+c^{3}\right) / a^{2} \leqslant 2 a$, and rewrite the original equation as $a+k=(b c)^{2}$. Since $b^{3}$ and $c^{3}$ are positive integers, we have $(b c)^{3} \geqslant b^{3}+c^{3}-1=k a^{2}-1$, so  $$ a+k \geqslant\left(k a^{2}-1\right)^{2 / 3} $$  As in Comment 1.2, $k$ is a positive integer; for each value of $k \geqslant 1$, this gives us a polynomial inequality satisfied by $a$ :  $$ k^{2} a^{4}-a^{3}-5 k a^{2}-3 k^{2} a-\left(k^{3}-1\right) \leqslant 0 . $$  We now prove that $a \leqslant 3$. Indeed,  $$ 0 \geqslant \frac{k^{2} a^{4}-a^{3}-5 k a^{2}-3 k^{2} a-\left(k^{3}-1\right)}{k^{2}} \geqslant a^{4}-a^{3}-5 a^{2}-3 a-k \geqslant a^{4}-a^{3}-5 a^{2}-5 a $$  which fails when $a \geqslant 4$. This leaves ten triples with $3 \geqslant a \geqslant b \geqslant c \geqslant 1$, which may be checked manually to give $(a, b, c)=(3,2,1)$.


Solution: Again, observe that $b^{3}+c^{3}=a^{2}\left(b^{2} c^{2}-a\right)$, so $b \leqslant a \leqslant b^{2} c^{2}-1$. We consider the function $f(x)=x^{2}\left(b^{2} c^{2}-x\right)$. It can be seen that that on the interval $\left[0, b^{2} c^{2}-1\right]$ the function $f$ is increasing if $x<\frac{2}{3} b^{2} c^{2}$ and decreasing if $x>\frac{2}{3} b^{2} c^{2}$. Consequently, it must be the case that  $$ b^{3}+c^{3}=f(a) \geqslant \min \left(f(b), f\left(b^{2} c^{2}-1\right)\right) $$  First, suppose that $b^{3}+c^{3} \geqslant f\left(b^{2} c^{2}-1\right)$. This may be written $b^{3}+c^{3} \geqslant\left(b^{2} c^{2}-1\right)^{2}$, and so  $$ 2 b^{3} \geqslant b^{3}+c^{3} \geqslant\left(b^{2} c^{2}-1\right)^{2}>b^{4} c^{4}-2 b^{2} c^{2} \geqslant b^{4} c^{4}-2 b^{3} c^{4} $$  Thus, $(b-2) c^{4}<2$, and the only solutions to this inequality have $(b, c)=(2,2)$ or $b \leqslant 3$ and $c=1$. It is easy to verify that the only case giving a solution for $a \geqslant b$ is $(a, b, c)=(3,2,1)$.  Otherwise, suppose that $b^{3}+c^{3}=f(a) \geqslant f(b)$. Then, we have  $$ 2 b^{3} \geqslant b^{3}+c^{3}=a^{2}\left(b^{2} c^{2}-a\right) \geqslant b^{2}\left(b^{2} c^{2}-b\right) . $$  Consequently $b c^{2} \leqslant 3$, with strict inequality in the case that $b \neq c$. Hence $c=1$ and $b \leqslant 2$. Both of these cases have been considered already, so we are done.  Comment 4.1. Instead of considering which of $f(b)$ and $f\left(b^{2} c^{2}-1\right)$ is less than $f(a)$, we may also proceed by explicitly dividing into cases based on whether $a \geqslant \frac{2}{3} b^{2} c^{2}$ or $a<\frac{2}{3} b^{2} c^{2}$. The first case may now be dealt with as follows. We have $b^{3} c^{3}+1 \geqslant b^{3}+c^{3}$ as $b^{3}$ and $c^{3}$ are positive integers, so we have  $$ b^{3} c^{3}+1 \geqslant b^{3}+c^{3} \geqslant a^{2} \geqslant \frac{4}{9} b^{4} c^{4} $$  This implies $b c \leqslant 2$, and hence $c=1$ and $b \leqslant 2$.


We say that a set $S$ of integers is rootiful if, for any positive integer $n$ and any $a_{0}, a_{1}, \ldots, a_{n} \in S$, all integer roots of the polynomial $a_{0}+a_{1} x+\cdots+a_{n} x^{n}$ are also in $S$. Find all rootiful sets of integers that contain all numbers of the form $2^{a}-2^{b}$ for positive integers $a$ and $b$. (Czech Republic) Answer: The set $\mathbb{Z}$ of all integers is the only such rootiful set.

Solution: The set $\mathbb{Z}$ of all integers is clearly rootiful. We shall prove that any rootiful set $S$ containing all the numbers of the form $2^{a}-2^{b}$ for $a, b \in \mathbb{Z}_{>0}$ must be all of $\mathbb{Z}$.  First, note that $0=2^{1}-2^{1} \in S$ and $2=2^{2}-2^{1} \in S$. Now, $-1 \in S$, since it is a root of $2 x+2$, and $1 \in S$, since it is a root of $2 x^{2}-x-1$. Also, if $n \in S$ then $-n$ is a root of $x+n$, so it suffices to prove that all positive integers must be in $S$.  Now, we claim that any positive integer $n$ has a multiple in $S$. Indeed, suppose that $n=2^{\alpha} \cdot t$ for $\alpha \in \mathbb{Z}_{\geqslant 0}$ and $t$ odd. Then $t \mid 2^{\phi(t)}-1$, so $n \mid 2^{\alpha+\phi(t)+1}-2^{\alpha+1}$. Moreover, $2^{\alpha+\phi(t)+1}-2^{\alpha+1} \in S$, and so $S$ contains a multiple of every positive integer $n$.  We will now prove by induction that all positive integers are in $S$. Suppose that $0,1, \ldots, n-$ $1 \in S$; furthermore, let $N$ be a multiple of $n$ in $S$. Consider the base- $n$ expansion of $N$, say $N=a_{k} n^{k}+a_{k-1} n^{k-1}+\cdots+a_{1} n+a_{0}$. Since $0 \leqslant a_{i}<n$ for each $a_{i}$, we have that all the $a_{i}$ are in $S$. Furthermore, $a_{0}=0$ since $N$ is a multiple of $n$. Therefore, $a_{k} n^{k}+a_{k-1} n^{k-1}+\cdots+a_{1} n-N=0$, so $n$ is a root of a polynomial with coefficients in $S$. This tells us that $n \in S$, completing the induction.


Solution: As in the previous solution, we can prove that 0,1 and -1 must all be in any rootiful set $S$ containing all numbers of the form $2^{a}-2^{b}$ for $a, b \in \mathbb{Z}_{>0}$.  We show that, in fact, every integer $k$ with $|k|>2$ can be expressed as a root of a polynomial whose coefficients are of the form $2^{a}-2^{b}$. Observe that it suffices to consider the case where $k$ is positive, as if $k$ is a root of $a_{n} x^{n}+\cdots+a_{1} x+a_{0}=0$, then $-k$ is a root of $(-1)^{n} a_{n} x^{n}+\cdots-$ $a_{1} x+a_{0}=0$.  Note that  $$ \left(2^{a_{n}}-2^{b_{n}}\right) k^{n}+\cdots+\left(2^{a_{0}}-2^{b_{0}}\right)=0 $$  is equivalent to  $$ 2^{a_{n}} k^{n}+\cdots+2^{a_{0}}=2^{b_{n}} k^{n}+\cdots+2^{b_{0}} $$  Hence our aim is to show that two numbers of the form $2^{a_{n}} k^{n}+\cdots+2^{a_{0}}$ are equal, for a fixed value of $n$. We consider such polynomials where every term $2^{a_{i}} k^{i}$ is at most $2 k^{n}$; in other words, where $2 \leqslant 2^{a_{i}} \leqslant 2 k^{n-i}$, or, equivalently, $1 \leqslant a_{i} \leqslant 1+(n-i) \log _{2} k$. Therefore, there must be $1+\left\lfloor(n-i) \log _{2} k\right\rfloor$ possible choices for $a_{i}$ satisfying these constraints.  The number of possible polynomials is then  $$ \prod_{i=0}^{n}\left(1+\left\lfloor(n-i) \log _{2} k\right\rfloor\right) \geqslant \prod_{i=0}^{n-1}(n-i) \log _{2} k=n!\left(\log _{2} k\right)^{n} $$  where the inequality holds as $1+\lfloor x\rfloor \geqslant x$. As there are $(n+1)$ such terms in the polynomial, each at most $2 k^{n}$, such a polynomial must have value at most $2 k^{n}(n+1)$. However, for large $n$, we have $n!\left(\log _{2} k\right)^{n}>2 k^{n}(n+1)$. Therefore there are more polynomials than possible values, so some two must be equal, as required.


Let $\mathbb{Z}_{>0}$ be the set of positive integers. A positive integer constant $C$ is given. Find all functions $f: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ such that, for all positive integers $a$ and $b$ satisfying $a+b>C$,  $$ a+f(b) \mid a^{2}+b f(a) $$  (Croatia) Answer: The functions satisfying (*) are exactly the functions $f(a)=k a$ for some constant $k \in \mathbb{Z}_{>0}$ (irrespective of the value of $C$ ).  Common remarks. It is easy to verify that the functions $f(a)=k a$ satisfy (*). Thus, in the proofs below, we will only focus on the converse implication: that condition (*) implies that $f=k a$.  A common minor part of these solutions is the derivation of some relatively easy bounds on the function $f$. An upper bound is easily obtained by setting $a=1$ in (*), giving the inequality  $$ f(b) \leqslant b \cdot f(1) $$  for all sufficiently large $b$. The corresponding lower bound is only marginally more difficult to obtain: substituting $b=1$ in the original equation shows that  $$ a+f(1) \mid\left(a^{2}+f(a)\right)-(a-f(1)) \cdot(a+f(1))=f(1)^{2}+f(a) $$  for all sufficiently large $a$. It follows from this that one has the lower bound  $$ f(a) \geqslant a+f(1) \cdot(1-f(1)) $$  again for all sufficiently large $a$. Each of the following proofs makes use of at least one of these bounds.

Solution: First, we show that $b \mid f(b)^{2}$ for all $b$. To do this, we choose a large positive integer $n$ so that $n b-f(b) \geqslant C$. Setting $a=n b-f(b)$ in (*) then shows that  $$ n b \mid(n b-f(b))^{2}+b f(n b-f(b)) $$  so that $b \mid f(b)^{2}$ as claimed. Now in particular we have that $p \mid f(p)$ for every prime $p$. If we write $f(p)=k(p) \cdot p$, then the bound $f(p) \leqslant f(1) \cdot p$ (valid for $p$ sufficiently large) shows that some value $k$ of $k(p)$ must be attained for infinitely many $p$. We will show that $f(a)=k a$ for all positive integers $a$. To do this, we substitute $b=p$ in (*), where $p$ is any sufficiently large prime for which $k(p)=k$, obtaining  $$ a+k p \mid\left(a^{2}+p f(a)\right)-a(a+k p)=p f(a)-p k a . $$  For suitably large $p$ we have $\operatorname{gcd}(a+k p, p)=1$, and hence we have  $$ a+k p \mid f(a)-k a $$  But the only way this can hold for arbitrarily large $p$ is if $f(a)-k a=0$. This concludes the proof.  Comment. There are other ways to obtain the divisibility $p \mid f(p)$ for primes $p$, which is all that is needed in this proof. For instance, if $f(p)$ were not divisible by $p$ then the arithmetic progression $p^{2}+b f(p)$ would attain prime values for infinitely many $b$ by Dirichlet's Theorem: hence, for these pairs p , b, we would have $p+f(b)=p^{2}+b f(p)$. Substituting $a \mapsto b$ and $b \mapsto p$ in $(*)$ then shows that $\left(f(p)^{2}-p^{2}\right)(p-1)$ is divisible by $b+f(p)$ and hence vanishes, which is impossible since $p \nmid f(p)$ by assumption.


Solution: First, we substitute $b=1$ in (*) and rearrange to find that  $$ \frac{f(a)+f(1)^{2}}{a+f(1)}=f(1)-a+\frac{a^{2}+f(a)}{a+f(1)} $$  is a positive integer for sufficiently large $a$. Since $f(a) \leqslant a f(1)$, for all sufficiently large $a$, it follows that $\frac{f(a)+f(1)^{2}}{a+f(1)} \leqslant f(1)$ also and hence there is a positive integer $k$ such that $\frac{f(a)+f(1)^{2}}{a+f(1)}=k$ for infinitely many values of $a$. In other words,  $$ f(a)=k a+f(1) \cdot(k-f(1)) $$  for infinitely many $a$. Fixing an arbitrary choice of $a$ in (*), we have that  $$ \frac{a^{2}+b f(a)}{a+k b+f(1) \cdot(k-f(1))} $$  is an integer for infinitely many $b$ (the same $b$ as above, maybe with finitely many exceptions). On the other hand, for $b$ taken sufficiently large, this quantity becomes arbitrarily close to $\frac{f(a)}{k}$; this is only possible if $\frac{f(a)}{k}$ is an integer and  $$ \frac{a^{2}+b f(a)}{a+k b+f(1) \cdot(k-f(1))}=\frac{f(a)}{k} $$  for infinitely many $b$. This rearranges to  $$ \frac{f(a)}{k} \cdot(a+f(1) \cdot(k-f(1)))=a^{2} $$  Hence $a^{2}$ is divisible by $a+f(1) \cdot(k-f(1))$, and hence so is $f(1)^{2}(k-f(1))^{2}$. The only way this can occur for all $a$ is if $k=f(1)$, in which case $(* *)$ provides that $f(a)=k a$ for all $a$, as desired.


Solution: Fix any two distinct positive integers $a$ and $b$. From (*) it follows that the two integers  $$ \left(a^{2}+c f(a)\right) \cdot(b+f(c)) \text { and }\left(b^{2}+c f(b)\right) \cdot(a+f(c)) $$  are both multiples of $(a+f(c)) \cdot(b+f(c))$ for all sufficiently large $c$. Taking an appropriate linear combination to eliminate the $c f(c)$ term, we find after expanding out that the integer  $$ \left[a^{2} f(b)-b^{2} f(a)\right] \cdot f(c)+[(b-a) f(a) f(b)] \cdot c+[a b(a f(b)-b f(a))] $$  is also a multiple of $(a+f(c)) \cdot(b+f(c))$. But as $c$ varies, $(\dagger)$ is bounded above by a positive multiple of $c$ while $(a+f(c)) \cdot(b+f(c))$ is bounded below by a positive multiple of $c^{2}$. The only way that such a divisibility can hold is if in fact  $$ \left[a^{2} f(b)-b^{2} f(a)\right] \cdot f(c)+[(b-a) f(a) f(b)] \cdot c+[a b(a f(b)-b f(a))]=0 $$  for sufficiently large $c$. Since the coefficient of $c$ in this linear relation is nonzero, it follows that there are constants $k, \ell$ such that $f(c)=k c+\ell$ for all sufficiently large $c$; the constants $k$ and $\ell$ are necessarily integers.  The value of $\ell$ satisfies  $$ \left[a^{2} f(b)-b^{2} f(a)\right] \cdot \ell+[a b(a f(b)-b f(a))]=0 $$  and hence $b \mid \ell a^{2} f(b)$ for all $a$ and $b$. Taking $b$ sufficiently large so that $f(b)=k b+\ell$, we thus have that $b \mid \ell^{2} a^{2}$ for all sufficiently large $b$; this implies that $\ell=0$. From ( $\dagger \dagger \dagger$ ) it then follows that $\frac{f(a)}{a}=\frac{f(b)}{b}$ for all $a \neq b$, so that there is a constant $k$ such that $f(a)=k a$ for all $a$ ( $k$ is equal to the constant defined earlier).


Solution: Let $\Gamma$ denote the set of all points $(a, f(a))$, so that $\Gamma$ is an infinite subset of the upper-right quadrant of the plane. For a point $A=(a, f(a))$ in $\Gamma$, we define a point $A^{\prime}=\left(-f(a),-f(a)^{2} / a\right)$ in the lower-left quadrant of the plane, and let $\Gamma^{\prime}$ denote the set of all such points $A^{\prime}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_9ee8c2ac2c2f5002ff14g-095.jpg?height=552&width=652&top_left_y=449&top_left_x=702)  Claim. For any point $A \in \Gamma$, the set $\Gamma$ is contained in finitely many lines through the point $A^{\prime}$. Proof. Let $A=(a, f(a))$. The functional equation (with $a$ and $b$ interchanged) can be rewritten as $b+f(a) \mid a f(b)-b f(a)$, so that all but finitely many points in $\Gamma$ are contained in one of the lines with equation  $$ a y-f(a) x=m(x+f(a)) $$  for $m$ an integer. Geometrically, these are the lines through $A^{\prime}=\left(-f(a),-f(a)^{2} / a\right)$ with gradient $\frac{f(a)+m}{a}$. Since $\Gamma$ is contained, with finitely many exceptions, in the region $0 \leqslant y \leqslant$ $f(1) \cdot x$ and the point $A^{\prime}$ lies strictly in the lower-left quadrant of the plane, there are only finitely many values of $m$ for which this line meets $\Gamma$. This concludes the proof of the claim.  Now consider any distinct points $A, B \in \Gamma$. It is clear that $A^{\prime}$ and $B^{\prime}$ are distinct. A line through $A^{\prime}$ and a line through $B^{\prime}$ only meet in more than one point if these two lines are equal to the line $A^{\prime} B^{\prime}$. It then follows from the above claim that the line $A^{\prime} B^{\prime}$ must contain all but finitely many points of $\Gamma$. If $C$ is another point of $\Gamma$, then the line $A^{\prime} C^{\prime}$ also passes through all but finitely many points of $\Gamma$, which is only possible if $A^{\prime} C^{\prime}=A^{\prime} B^{\prime}$.  We have thus seen that there is a line $\ell$ passing through all points of $\Gamma^{\prime}$ and through all but finitely many points of $\Gamma$. We claim that this line passes through the origin $O$ and passes through every point of $\Gamma$. To see this, note that by construction $A, O, A^{\prime}$ are collinear for every point $A \in \Gamma$. Since $\ell=A A^{\prime}$ for all but finitely many points $A \in \Gamma$, it thus follows that $O \in \ell$. Thus any $A \in \Gamma$ lies on the line $\ell=A^{\prime} O$.  Since $\Gamma$ is contained in a line through $O$, it follows that there is a real constant $k$ (the gradient of $\ell$ ) such that $f(a)=k a$ for all $a$. The number $k$ is, of course, a positive integer.  Comment. Without the $a+b>C$ condition, this problem is approachable by much more naive methods. For instance, using the given divisibility for $a, b \in\{1,2,3\}$ one can prove by a somewhat tedious case-check that $f(2)=2 f(1)$ and $f(3)=3 f(1)$; this then forms the basis of an induction establishing that $f(n)=n f(1)$ for all $n$.


Let $H=\left\{\lfloor i \sqrt{2}\rfloor: i \in \mathbb{Z}_{>0}\right\}=\{1,2,4,5,7, \ldots\}$, and let $n$ be a positive integer. Prove that there exists a constant $C$ such that, if $A \subset\{1,2, \ldots, n\}$ satisfies $|A| \geqslant C \sqrt{n}$, then there exist $a, b \in A$ such that $a-b \in H$.  (Brazil) Common remarks. In all solutions, we will assume that $A$ is a set such that $\{a-b: a, b \in A\}$ is disjoint from $H$, and prove that $|A|<C \sqrt{n}$.

Solution: First, observe that if $n$ is a positive integer, then $n \in H$ exactly when  $$ \left\{\frac{n}{\sqrt{2}}\right\}>1-\frac{1}{\sqrt{2}} . $$  To see why, observe that $n \in H$ if and only if $0<i \sqrt{2}-n<1$ for some $i \in \mathbb{Z}_{>0}$. In other words, $0<i-n / \sqrt{2}<1 / \sqrt{2}$, which is equivalent to (1).  Now, write $A=\left\{a_{1}<a_{2}<\cdots<a_{k}\right\}$, where $k=|A|$. Observe that the set of differences is not altered by shifting $A$, so we may assume that $A \subseteq\{0,1, \ldots, n-1\}$ with $a_{1}=0$.  From (1), we learn that $\left\{a_{i} / \sqrt{2}\right\}<1-1 / \sqrt{2}$ for each $i>1$ since $a_{i}-a_{1} \notin H$. Furthermore, we must have $\left\{a_{i} / \sqrt{2}\right\}<\left\{a_{j} / \sqrt{2}\right\}$ whenever $i<j$; otherwise, we would have  $$ -\left(1-\frac{1}{\sqrt{2}}\right)<\left\{\frac{a_{j}}{\sqrt{2}}\right\}-\left\{\frac{a_{i}}{\sqrt{2}}\right\}<0 . $$  Since $\left\{\left(a_{j}-a_{i}\right) / \sqrt{2}\right\}=\left\{a_{j} / \sqrt{2}\right\}-\left\{a_{i} / \sqrt{2}\right\}+1$, this implies that $\left\{\left(a_{j}-a_{i}\right) / \sqrt{2}\right\}>1 / \sqrt{2}>$ $1-1 / \sqrt{2}$, contradicting (1).  Now, we have a sequence $0=a_{1}<a_{2}<\cdots<a_{k}<n$, with  $$ 0=\left\{\frac{a_{1}}{\sqrt{2}}\right\}<\left\{\frac{a_{2}}{\sqrt{2}}\right\}<\cdots<\left\{\frac{a_{k}}{\sqrt{2}}\right\}<1-\frac{1}{\sqrt{2}} . $$  We use the following fact: for any $d \in \mathbb{Z}$, we have  $$ \left\{\frac{d}{\sqrt{2}}\right\}>\frac{1}{2 d \sqrt{2}} $$  To see why this is the case, let $h=\lfloor d / \sqrt{2}\rfloor$, so $\{d / \sqrt{2}\}=d / \sqrt{2}-h$. Then  $$ \left\{\frac{d}{\sqrt{2}}\right\}\left(\frac{d}{\sqrt{2}}+h\right)=\frac{d^{2}-2 h^{2}}{2} \geqslant \frac{1}{2} $$  since the numerator is a positive integer. Because $d / \sqrt{2}+h<2 d / \sqrt{2}$, inequality (2) follows. Let $d_{i}=a_{i+1}-a_{i}$, for $1 \leqslant i<k$. Then $\left\{a_{i+1} / \sqrt{2}\right\}-\left\{a_{i} / \sqrt{2}\right\}=\left\{d_{i} / \sqrt{2}\right\}$, and we have  $$ 1-\frac{1}{\sqrt{2}}>\sum_{i}\left\{\frac{d_{i}}{\sqrt{2}}\right\}>\frac{1}{2 \sqrt{2}} \sum_{i} \frac{1}{d_{i}} \geqslant \frac{(k-1)^{2}}{2 \sqrt{2}} \frac{1}{\sum_{i} d_{i}}>\frac{(k-1)^{2}}{2 \sqrt{2}} \cdot \frac{1}{n} . $$  Here, the first inequality holds because $\left\{a_{k} / \sqrt{2}\right\}<1-1 / \sqrt{2}$, the second follows from (2), the third follows from an easy application of the AM-HM inequality (or Cauchy-Schwarz), and the fourth follows from the fact that $\sum_{i} d_{i}=a_{k}<n$.  Rearranging this, we obtain  $$ \sqrt{2 \sqrt{2}-2} \cdot \sqrt{n}>k-1 $$  which provides the required bound on $k$.


Solution: Let $\alpha=2+\sqrt{2}$, so $(1 / \alpha)+(1 / \sqrt{2})=1$. Thus, $J=\left\{\lfloor i \alpha\rfloor: i \in \mathbb{Z}_{>0}\right\}$ is the complementary Beatty sequence to $H$ (in other words, $H$ and $J$ are disjoint with $H \cup J=\mathbb{Z}_{>0}$ ). Write $A=\left\{a_{1}<a_{2}<\cdots<a_{k}\right\}$. Suppose that $A$ has no differences in $H$, so all its differences are in $J$ and we can set $a_{i}-a_{1}=\left\lfloor\alpha b_{i}\right\rfloor$ for $b_{i} \in \mathbb{Z}_{>0}$.  For any $j>i$, we have $a_{j}-a_{i}=\left\lfloor\alpha b_{j}\right\rfloor-\left\lfloor\alpha b_{i}\right\rfloor$. Because $a_{j}-a_{i} \in J$, we also have $a_{j}-a_{i}=\lfloor\alpha t\rfloor$ for some positive integer $t$. Thus, $\lfloor\alpha t\rfloor=\left\lfloor\alpha b_{j}\right\rfloor-\left\lfloor\alpha b_{i}\right\rfloor$. The right hand side must equal either $\left\lfloor\alpha\left(b_{j}-b_{i}\right)\right\rfloor$ or $\left\lfloor\alpha\left(b_{j}-b_{i}\right)\right\rfloor-1$, the latter of which is not a member of $J$ as $\alpha>2$. Therefore, $t=b_{j}-b_{i}$ and so we have $\left\lfloor\alpha b_{j}\right\rfloor-\left\lfloor\alpha b_{i}\right\rfloor=\left\lfloor\alpha\left(b_{j}-b_{i}\right)\right\rfloor$.  For $1 \leqslant i<k$ we now put $d_{i}=b_{i+1}-b_{i}$, and we have  $$ \left\lfloor\alpha \sum_{i} d_{i}\right\rfloor=\left\lfloor\alpha b_{k}\right\rfloor=\sum_{i}\left\lfloor\alpha d_{i}\right\rfloor ; $$  that is, $\sum_{i}\left\{\alpha d_{i}\right\}<1$. We also have  $$ 1+\left\lfloor\alpha \sum_{i} d_{i}\right\rfloor=1+a_{k}-a_{1} \leqslant a_{k} \leqslant n $$  so $\sum_{i} d_{i} \leqslant n / \alpha$. With the above inequalities, an argument similar to (3) (which uses the fact that $\{\alpha d\}=$ $\{d \sqrt{2}\}>1 /(2 d \sqrt{2})$ for positive integers $d)$ proves that $1>\left((k-1)^{2} /(2 \sqrt{2})\right)(\alpha / n)$, which again rearranges to give  $$ \sqrt{2 \sqrt{2}-2} \cdot \sqrt{n}>k-1 $$  Comment. The use of Beatty sequences in Solution 2 is essentially a way to bypass (1). Both Solutions 1 and 2 use the fact that $\sqrt{2}<2$; the statement in the question would still be true if $\sqrt{2}$ did not have this property (for instance, if it were replaced with $\alpha$ ), but any argument along the lines of Solutions 1 or 2 would be more complicated.


Solution: Again, define $J=\mathbb{Z}_{>0} \backslash H$, so all differences between elements of $A$ are in $J$. We start by making the following observation. Suppose we have a set $B \subseteq\{1,2, \ldots, n\}$ such that all of the differences between elements of $B$ are in $H$. Then $|A| \cdot|B| \leqslant 2 n$.  To see why, observe that any two sums of the form $a+b$ with $a \in A, b \in B$ are different; otherwise, we would have $a_{1}+b_{1}=a_{2}+b_{2}$, and so $\left|a_{1}-a_{2}\right|=\left|b_{2}-b_{1}\right|$. However, then the left hand side is in $J$ whereas the right hand side is in $H$. Thus, $\{a+b: a \in A, b \in B\}$ is a set of size $|A| \cdot|B|$ all of whose elements are no greater than $2 n$, yielding the claimed inequality.  With this in mind, it suffices to construct a set $B$, all of whose differences are in $H$ and whose size is at least $C^{\prime} \sqrt{n}$ for some constant $C^{\prime}>0$.  To do so, we will use well-known facts about the negative Pell equation $X^{2}-2 Y^{2}=-1$; in particular, that there are infinitely many solutions and the values of $X$ are given by the recurrence $X_{1}=1, X_{2}=7$ and $X_{m}=6 X_{m-1}-X_{m-2}$. Therefore, we may choose $X$ to be a solution with $\sqrt{n} / 6<X \leqslant \sqrt{n}$.  Now, we claim that we may choose $B=\{X, 2 X, \ldots,\lfloor(1 / 3) \sqrt{n}\rfloor X\}$. Indeed, we have  $$ \left(\frac{X}{\sqrt{2}}-Y\right)\left(\frac{X}{\sqrt{2}}+Y\right)=\frac{-1}{2} $$  and so  $$ 0>\left(\frac{X}{\sqrt{2}}-Y\right) \geqslant \frac{-3}{\sqrt{2 n}}, $$  from which it follows that $\{X / \sqrt{2}\}>1-(3 / \sqrt{2 n})$. Combined with (1), this shows that all differences between elements of $B$ are in $H$.  Comment. Some of the ideas behind Solution 3 may be used to prove that the constant $C=\sqrt{2 \sqrt{2}-2}$ (from Solutions 1 and 2) is optimal, in the sense that there are arbitrarily large values of $n$ and sets $A_{n} \subseteq\{1,2, \ldots, n\}$ of size roughly $C \sqrt{n}$, all of whose differences are contained in $J$.  To see why, choose $X$ to come from a sufficiently large solution to the Pell equation $X^{2}-2 Y^{2}=1$, so $\{X / \sqrt{2}\} \approx 1 /(2 X \sqrt{2})$. In particular, $\{X\},\{2 X\}, \ldots,\{[2 X \sqrt{2}(1-1 / \sqrt{2})\rfloor X\}$ are all less than $1-1 / \sqrt{2}$. Thus, by (1) any positive integer of the form $i X$ for $1 \leqslant i \leqslant\lfloor 2 X \sqrt{2}(1-1 / \sqrt{2})\rfloor$ lies in $J$.  Set $n \approx 2 X^{2} \sqrt{2}(1-1 / \sqrt{2})$. We now have a set $A=\{i X: i \leqslant\lfloor 2 X \sqrt{2}(1-1 / \sqrt{2})\rfloor\}$ containing roughly $2 X \sqrt{2}(1-1 / \sqrt{2})$ elements less than or equal to $n$ such that all of the differences lie in $J$, and we can see that $|A| \approx C \sqrt{n}$ with $C=\sqrt{2 \sqrt{2}-2}$.


Solution: As in   Choose $Y$ to be a solution to the Pell-like equation $X^{2}-2 Y^{2}= \pm 1$; such solutions are given by the recurrence $Y_{1}=1, Y_{2}=2$ and $Y_{m}=2 Y_{m-1}+Y_{m-2}$, and so we can choose $Y$ such that $n /(3 \sqrt{2})<Y \leqslant n / \sqrt{2}$. Furthermore, it is known that for such a $Y$ and for $1 \leqslant x<Y$,  $$ \{x \sqrt{2}\}+\{(Y-x) \sqrt{2}\}=\{Y / \sqrt{2}\} $$  if $X^{2}-2 Y^{2}=1$, and  $$ \{x \sqrt{2}\}+\{(Y-x) \sqrt{2}\}=1+\{Y / \sqrt{2}\} $$  if $X^{2}-2 Y^{2}=-1$. Indeed, this is a statement of the fact that $X / Y$ is a best rational approximation to $\sqrt{2}$, from below in the first case and from above in the second.  Now, consider the sequence $\{\sqrt{2}\},\{2 \sqrt{2}\}, \ldots,\{(Y-1) \sqrt{2}\}$. The Erdős-Szekeres theorem tells us that this sequence has a monotone subsequence with at least $\sqrt{Y-2}+1>\sqrt{Y}$ elements; if that subsequence is decreasing, we may reflect (using (4) or (5)) to ensure that it is increasing. Call the subsequence $\left\{y_{1} \sqrt{2}\right\},\left\{y_{2} \sqrt{2}\right\}, \ldots,\left\{y_{t} \sqrt{2}\right\}$ for $t>\sqrt{Y}$.  Now, set $B=\left\{\left\lfloor y_{i} \sqrt{2}\right\rfloor: 1 \leqslant i \leqslant t\right\}$. We have $\left\lfloor y_{j} \sqrt{2}\right\rfloor-\left\lfloor y_{i} \sqrt{2}\right\rfloor=\left\lfloor\left(y_{j}-y_{i}\right) \sqrt{2}\right\rfloor$ for $i<j$ (because the corresponding inequality for the fractional parts holds by the ordering assumption on the $\left\{y_{i} \sqrt{2}\right\}$ ), which means that all differences between elements of $B$ are indeed in $H$. Since $|B|>\sqrt{Y}>\sqrt{n} / \sqrt{3 \sqrt{2}}$, this is the required set.  Comment. Any solution to this problem will need to use the fact that $\sqrt{2}$ cannot be approximated well by rationals, either directly or implicitly (for example, by using facts about solutions to Pelllike equations). If $\sqrt{2}$ were replaced by a value of $\theta$ with very good rational approximations (from below), then an argument along the lines of Solution 3 would give long arithmetic progressions in $\{\lfloor i \theta\rfloor: 0 \leqslant i<n\}$ (with initial term 0 ) for certain values of $n$.


Let $a$ and $b$ be two positive integers. Prove that the integer  $$ a^{2}+\left\lceil\frac{4 a^{2}}{b}\right\rceil $$  is not a square. (Here $\lceil z\rceil$ denotes the least integer greater than or equal to z.) (Russia)

Solution: Arguing indirectly, assume that  $$ a^{2}+\left\lceil\frac{4 a^{2}}{b}\right\rceil=(a+k)^{2}, \quad \text { or } \quad\left\lceil\frac{(2 a)^{2}}{b}\right\rceil=(2 a+k) k $$  Clearly, $k \geqslant 1$. In other words, the equation  $$ \left\lceil\frac{c^{2}}{b}\right\rceil=(c+k) k $$  has a positive integer solution $(c, k)$, with an even value of $c$. Choose a positive integer solution of (1) with minimal possible value of $k$, without regard to the parity of $c$. From  $$ \frac{c^{2}}{b}>\left\lceil\frac{c^{2}}{b}\right\rceil-1=c k+k^{2}-1 \geqslant c k $$  and  $$ \frac{(c-k)(c+k)}{b}<\frac{c^{2}}{b} \leqslant\left\lceil\frac{c^{2}}{b}\right\rceil=(c+k) k $$  it can be seen that $c>b k>c-k$, so  $$ c=k b+r \quad \text { with some } 0<r<k $$  By substituting this in (1) we get  $$ \left\lceil\frac{c^{2}}{b}\right\rceil=\left\lceil\frac{(b k+r)^{2}}{b}\right\rceil=k^{2} b+2 k r+\left\lceil\frac{r^{2}}{b}\right\rceil $$  and  $$ (c+k) k=(k b+r+k) k=k^{2} b+2 k r+k(k-r) $$  so  $$ \left\lceil\frac{r^{2}}{b}\right\rceil=k(k-r) $$  Notice that relation (2) provides another positive integer solution of (1), namely $c^{\prime}=r$ and $k^{\prime}=k-r$, with $c^{\prime}>0$ and $0<k^{\prime}<k$. That contradicts the minimality of $k$, and hence finishes the solution.


Solution: Suppose that  $$ a^{2}+\left\lceil\frac{4 a^{2}}{b}\right\rceil=c^{2} $$  with some positive integer $c>a$, so  $$ \begin{aligned} & c^{2}-1<a^{2}+\frac{4 a^{2}}{b} \leqslant c^{2} \\ & 0 \leqslant c^{2} b-a^{2}(b+4)<b \end{aligned} $$  Let $d=c^{2} b-a^{2}(b+4), x=c+a$ and $y=c-a$; then we have $c=\frac{x+y}{2}$ and $a=\frac{x-y}{2}$, and (3) can be re-written as follows:  $$ \begin{aligned} \left(\frac{x+y}{2}\right)^{2} b-\left(\frac{x-y}{2}\right)^{2}(b+4) & =d \\ x^{2}-(b+2) x y+y^{2}+d=0 ; \quad 0 & \leqslant d<b \end{aligned} $$  So, by the indirect assumption, the equation (4) has some positive integer solution $(x, y)$. Fix $b$ and $d$, and take a pair $(x, y)$ of positive integers, satisfying (4), such that $x+y$ is minimal. By the symmetry in (4) we may assume that $x \geqslant y \geqslant 1$.  Now we perform a usual "Vieta jump". Consider (4) as a quadratic equation in variable $x$, and let $z$ be its second root. By the Vieta formulas,  $$ x+z=(b+2) y, \quad \text { and } \quad z x=y^{2}+d, $$  so  $$ z=(b+2) y-x=\frac{y^{2}+d}{x} $$  The first formula shows that $z$ is an integer, and by the second formula $z$ is positive. Hence $(z, y)$ is another positive integer solution of (4). From  $$ \begin{aligned} (x-1)(z-1) & =x z-(x+z)+1=\left(y^{2}+d\right)-(b+2) y+1 \\ & <\left(y^{2}+b\right)-(b+2) y+1=(y-1)^{2}-b(y-1) \leqslant(y-1)^{2} \leqslant(x-1)^{2} \end{aligned} $$  we can see that $z<x$ and therefore $z+y<x+y$. But this contradicts the minimality of $x+y$ among the positive integer solutions of (4).  The activities of the Problem Selection Committee were supported by Trinity College, Cambridge University of Cambridge