Given an acute triangle $A B C$ with angles $\alpha, \beta$ and $\gamma$ at vertices $A, B$ and $C$, respectively, such that $\beta>\gamma$. Point $I$ is the incenter, and $R$ is the circumradius. Point $D$ is the foot of the altitude from vertex $A$. Point $K$ lies on line $A D$ such that $A K=2 R$, and $D$ separates $A$ and $K$. Finally, lines $D I$ and $K I$ meet sides $A C$ and $B C$ at $E$ and $F$, respectively.  Prove that if $I E=I F$ then $\beta \leq 3 \gamma$.  (Iran)

Solution: We first prove that  $$ \angle K I D=\frac{\beta-\gamma}{2} $$  even without the assumption that $I E=I F$. Then we will show that the statement of the problem is a consequence of this fact.  Denote the circumcenter by $O$. On the circumcircle, let $P$ be the point opposite to $A$, and let the angle bisector $A I$ intersect the circle again at $M$. Since $A K=A P=2 R$, triangle $A K P$ is isosceles. It is known that $\angle B A D=\angle C A O$, hence $\angle D A I=\angle B A I-\angle B A D=\angle C A I-$ $\angle C A O=\angle O A I$, and $A M$ is the bisector line in triangle $A K P$. Therefore, points $K$ and $P$ are symmetrical about $A M$, and $\angle A M K=\angle A M P=90^{\circ}$. Thus, $M$ is the midpoint of $K P$, and $A M$ is the perpendicular bisector of $K P$.  ![](https://cdn.mathpix.com/cropped/2024_04_17_b8b307b9422db9068910g-50.jpg?height=1082&width=962&top_left_y=1252&top_left_x=524)  Denote the perpendicular feet of incenter $I$ on lines $B C, A C$, and $A D$ by $A_{1}, B_{1}$, and $T$, respectively. Quadrilateral $D A_{1} I T$ is a rectangle, hence $T D=I A_{1}=I B_{1}$.  Due to the right angles at $T$ and $B_{1}$, quadrilateral $A B_{1} I T$ is cyclic. Hence $\angle B_{1} T I=$ $\angle B_{1} A I=\angle C A M=\angle B A M=\angle B P M$ and $\angle I B_{1} T=\angle I A T=\angle M A K=\angle M A P=$ $\angle M B P$. Therefore, triangles $B_{1} T I$ and $B P M$ are similar and $\frac{I T}{I B_{1}}=\frac{M P}{M B}$.  It is well-known that $M B=M C=M I$. Then right triangles $I T D$ and $K M I$ are also similar, because $\frac{I T}{T D}=\frac{I T}{I B_{1}}=\frac{M P}{M B}=\frac{K M}{M I}$. Hence, $\angle K I M=\angle I D T=\angle I D A$, and  $$ \angle K I D=\angle M I D-\angle K I M=(\angle I A D+\angle I D A)-\angle I D A=\angle I A D . $$  Finally, from the right triangle $A D B$ we can compute  $$ \angle K I D=\angle I A D=\angle I A B-\angle D A B=\frac{\alpha}{2}-\left(90^{\circ}-\beta\right)=\frac{\alpha}{2}-\frac{\alpha+\beta+\gamma}{2}+\beta=\frac{\beta-\gamma}{2} . $$  Now let us turn to the statement and suppose that $I E=I F$. Since $I A_{1}=I B_{1}$, the right triangles $I E B_{1}$ and $I F A_{1}$ are congruent and $\angle I E B_{1}=\angle I F A_{1}$. Since $\beta>\gamma, A_{1}$ lies in the interior of segment $C D$ and $F$ lies in the interior of $A_{1} D$. Hence, $\angle I F C$ is acute. Then two cases are possible depending on the order of points $A, C, B_{1}$ and $E$. ![](https://cdn.mathpix.com/cropped/2024_04_17_b8b307b9422db9068910g-51.jpg?height=818&width=1486&top_left_y=907&top_left_x=317)  If point $E$ lies between $C$ and $B_{1}$ then $\angle I F C=\angle I E A$, hence quadrilateral $C E I F$ is cyclic and $\angle F C E=180^{\circ}-\angle E I F=\angle K I D$. By (1), in this case we obtain $\angle F C E=\gamma=\angle K I D=$ $\frac{\beta-\gamma}{2}$ and $\beta=3 \gamma$.  Otherwise, if point $E$ lies between $A$ and $B_{1}$, quadrilateral $C E I F$ is a deltoid such that $\angle I E C=\angle I F C<90^{\circ}$. Then we have $\angle F C E>180^{\circ}-\angle E I F=\angle K I D$. Therefore, $\angle F C E=\gamma>\angle K I D=\frac{\beta-\gamma}{2}$ and $\beta<3 \gamma$.  Comment 1. In the case when quadrilateral $C E I F$ is a deltoid, one can prove the desired inequality without using (1). Actually, from $\angle I E C=\angle I F C<90^{\circ}$ it follows that $\angle A D I=90^{\circ}-\angle E D C<$ $\angle A E D-\angle E D C=\gamma$. Since the incircle lies inside triangle $A B C$, we have $A D>2 r$ (here $r$ is the inradius), which implies $D T<T A$ and $D I<A I$; hence $\frac{\beta-\gamma}{2}=\angle I A D<\angle A D I<\gamma$.


Solution: We give a different proof for (1). Then the solution can be finished in the same way as above.  Define points $M$ and $P$ again; it can be proved in the same way that $A M$ is the perpendicular bisector of $K P$. Let $J$ be the center of the excircle touching side $B C$. It is well-known that points $B, C, I, J$ lie on a circle with center $M$; denote this circle by $\omega_{1}$.  Let $B^{\prime}$ be the reflection of point $B$ about the angle bisector $A M$. By the symmetry, $B^{\prime}$ is the second intersection point of circle $\omega_{1}$ and line $A C$. Triangles $P B A$ and $K B^{\prime} A$ are symmetrical with respect to line $A M$, therefore $\angle K B^{\prime} A=\angle P B A=90^{\circ}$. By the right angles at $D$ and $B^{\prime}$, points $K, D, B^{\prime}, C$ are concyclic and  $$ A D \cdot A K=A B^{\prime} \cdot A C . $$  From the cyclic quadrilateral $I J C B^{\prime}$ we obtain $A B^{\prime} \cdot A C=A I \cdot A J$ as well, therefore  $$ A D \cdot A K=A B^{\prime} \cdot A C=A I \cdot A J $$  and points $I, J, K, D$ are also concyclic. Denote circle $I D K J$ by $\omega_{2}$.  ![](https://cdn.mathpix.com/cropped/2024_04_17_b8b307b9422db9068910g-52.jpg?height=997&width=1145&top_left_y=798&top_left_x=433)  Let $N$ be the point on circle $\omega_{2}$ which is opposite to $K$. Since $\angle N D K=90^{\circ}=\angle C D K$, point $N$ lies on line $B C$. Point $M$, being the center of circle $\omega_{1}$, is the midpoint of segment $I J$, and $K M$ is perpendicular to $I J$. Therefore, line $K M$ is the perpendicular bisector of $I J$ and hence it passes through $N$.  From the cyclic quadrilateral $I D K N$ we obtain  $$ \angle K I D=\angle K N D=90^{\circ}-\angle D K N=90^{\circ}-\angle A K M=\angle M A K=\frac{\beta-\gamma}{2} $$  Comment 2. The main difficulty in the solution is finding (1). If someone can guess this fact, he or she can compute it in a relatively short way.  One possible way is finding and applying the relation $A I^{2}=2 R\left(h_{a}-2 r\right)$, where $h_{a}=A D$ is the length of the altitude. Using this fact, one can see that triangles $A K I$ and $A I D^{\prime}$ are similar (here $D^{\prime}$ is the point symmetrical to $D$ about $T$ ). Hence, $\angle M I K=\angle D D^{\prime} I=\angle I D D^{\prime}$. The proof can be finished as in Solution 1.


Point $P$ lies on side $A B$ of a convex quadrilateral $A B C D$. Let $\omega$ be the incircle of triangle $C P D$, and let $I$ be its incenter. Suppose that $\omega$ is tangent to the incircles of triangles $A P D$ and $B P C$ at points $K$ and $L$, respectively. Let lines $A C$ and $B D$ meet at $E$, and let lines $A K$ and $B L$ meet at $F$. Prove that points $E, I$, and $F$ are collinear.  (Poland)

Solution: Let $\Omega$ be the circle tangent to segment $A B$ and to rays $A D$ and $B C$; let $J$ be its center. We prove that points $E$ and $F$ lie on line $I J$.  ![](https://cdn.mathpix.com/cropped/2024_04_17_b8b307b9422db9068910g-53.jpg?height=682&width=920&top_left_y=647&top_left_x=611)  Denote the incircles of triangles $A D P$ and $B C P$ by $\omega_{A}$ and $\omega_{B}$. Let $h_{1}$ be the homothety with a negative scale taking $\omega$ to $\Omega$. Consider this homothety as the composition of two homotheties: one taking $\omega$ to $\omega_{A}$ (with a negative scale and center $K$ ), and another one taking $\omega_{A}$ to $\Omega$ (with a positive scale and center $A$ ). It is known that in such a case the three centers of homothety are collinear (this theorem is also referred to as the theorem on the three similitude centers). Hence, the center of $h_{1}$ lies on line $A K$. Analogously, it also lies on $B L$, so this center is $F$. Hence, $F$ lies on the line of centers of $\omega$ and $\Omega$, i. e. on $I J$ (if $I=J$, then $F=I$ as well, and the claim is obvious).  Consider quadrilateral $A P C D$ and mark the equal segments of tangents to $\omega$ and $\omega_{A}$ (see the figure below to the left). Since circles $\omega$ and $\omega_{A}$ have a common point of tangency with $P D$, one can easily see that $A D+P C=A P+C D$. So, quadrilateral $A P C D$ is circumscribed; analogously, circumscribed is also quadrilateral $B C D P$. Let $\Omega_{A}$ and $\Omega_{B}$ respectively be their incircles. ![](https://cdn.mathpix.com/cropped/2024_04_17_b8b307b9422db9068910g-53.jpg?height=754&width=1714&top_left_y=1962&top_left_x=223)  Consider the homothety $h_{2}$ with a positive scale taking $\omega$ to $\Omega$. Consider $h_{2}$ as the composition of two homotheties: taking $\omega$ to $\Omega_{A}$ (with a positive scale and center $C$ ), and taking $\Omega_{A}$ to $\Omega$ (with a positive scale and center $A$ ), respectively. So the center of $h_{2}$ lies on line $A C$. By analogous reasons, it lies also on $B D$, hence this center is $E$. Thus, $E$ also lies on the line of centers $I J$, and the claim is proved.  Comment. In both main steps of the solution, there can be several different reasonings for the same claims. For instance, one can mostly use Desargues' theorem instead of the three homotheties theorem. Namely, if $I_{A}$ and $I_{B}$ are the centers of $\omega_{A}$ and $\omega_{B}$, then lines $I_{A} I_{B}, K L$ and $A B$ are concurrent (by the theorem on three similitude centers applied to $\omega, \omega_{A}$ and $\omega_{B}$ ). Then Desargues' theorem, applied to triangles $A I_{A} K$ and $B I_{B} L$, yields that the points $J=A I_{A} \cap B I_{B}, I=I_{A} K \cap I_{B} L$ and $F=A K \cap B L$ are collinear.  For the second step, let $J_{A}$ and $J_{B}$ be the centers of $\Omega_{A}$ and $\Omega_{B}$. Then lines $J_{A} J_{B}, A B$ and $C D$ are concurrent, since they appear to be the two common tangents and the line of centers of $\Omega_{A}$ and $\Omega_{B}$. Applying Desargues' theorem to triangles $A J_{A} C$ and $B J_{B} D$, we obtain that the points $J=A J_{A} \cap B J_{B}$, $I=C J_{A} \cap D J_{B}$ and $E=A C \cap B D$ are collinear.


Find all pairs $(k, n)$ of positive integers for which $7^{k}-3^{n}$ divides $k^{4}+n^{2}$.  (Austria)  Answer. $(2,4)$.

Solution: Suppose that a pair $(k, n)$ satisfies the condition of the problem. Since $7^{k}-3^{n}$ is even, $k^{4}+n^{2}$ is also even, hence $k$ and $n$ have the same parity. If $k$ and $n$ are odd, then $k^{4}+n^{2} \equiv 1+1=2(\bmod 4)$, while $7^{k}-3^{n} \equiv 7-3 \equiv 0(\bmod 4)$, so $k^{4}+n^{2}$ cannot be divisible by $7^{k}-3^{n}$. Hence, both $k$ and $n$ must be even.  Write $k=2 a, n=2 b$. Then $7^{k}-3^{n}=7^{2 a}-3^{2 b}=\frac{7^{a}-3^{b}}{2} \cdot 2\left(7^{a}+3^{b}\right)$, and both factors are integers. So $2\left(7^{a}+3^{b}\right) \mid 7^{k}-3^{n}$ and $7^{k}-3^{n} \mid k^{4}+n^{2}=2\left(8 a^{4}+2 b^{2}\right)$, hence  $$ 7^{a}+3^{b} \leq 8 a^{4}+2 b^{2} . $$  We prove by induction that $8 a^{4}<7^{a}$ for $a \geq 4,2 b^{2}<3^{b}$ for $b \geq 1$ and $2 b^{2}+9 \leq 3^{b}$ for $b \geq 3$. In the initial cases $a=4, b=1, b=2$ and $b=3$ we have $8 \cdot 4^{4}=2048<7^{4}=2401,2<3$, $2 \cdot 2^{2}=8<3^{2}=9$ and $2 \cdot 3^{2}+9=3^{3}=27$, respectively.  If $8 a^{4}<7^{a}(a \geq 4)$ and $2 b^{2}+9 \leq 3^{b}(b \geq 3)$, then  $$ \begin{aligned} 8(a+1)^{4} & =8 a^{4}\left(\frac{a+1}{a}\right)^{4}<7^{a}\left(\frac{5}{4}\right)^{4}=7^{a} \frac{625}{256}<7^{a+1} \quad \text { and } \\ 2(b+1)^{2}+9 & <\left(2 b^{2}+9\right)\left(\frac{b+1}{b}\right)^{2} \leq 3^{b}\left(\frac{4}{3}\right)^{2}=3^{b} \frac{16}{9}<3^{b+1}, \end{aligned} $$  as desired.  For $a \geq 4$ we obtain $7^{a}+3^{b}>8 a^{4}+2 b^{2}$ and inequality (1) cannot hold. Hence $a \leq 3$, and three cases are possible.  Case 1: $a=1$. Then $k=2$ and $8+2 b^{2} \geq 7+3^{b}$, thus $2 b^{2}+1 \geq 3^{b}$. This is possible only if $b \leq 2$. If $b=1$ then $n=2$ and $\frac{k^{4}+n^{2}}{7^{k}-3^{n}}=\frac{2^{4}+2^{2}}{7^{2}-3^{2}}=\frac{1}{2}$, which is not an integer. If $b=2$ then $n=4$ and $\frac{k^{4}+n^{2}}{7^{k}-3^{n}}=\frac{2^{4}+4^{2}}{7^{2}-3^{4}}=-1$, so $(k, n)=(2,4)$ is a solution.  Case 2: $a=2$. Then $k=4$ and $k^{4}+n^{2}=256+4 b^{2} \geq\left|7^{4}-3^{n}\right|=\left|49-3^{b}\right| \cdot\left(49+3^{b}\right)$. The smallest value of the first factor is 22 , attained at $b=3$, so $128+2 b^{2} \geq 11\left(49+3^{b}\right)$, which is impossible since $3^{b}>2 b^{2}$.  Case 3: $a=3$. Then $k=6$ and $k^{4}+n^{2}=1296+4 b^{2} \geq\left|7^{6}-3^{n}\right|=\left|343-3^{b}\right| \cdot\left(343+3^{b}\right)$. Analogously, $\left|343-3^{b}\right| \geq 100$ and we have $324+b^{2} \geq 25\left(343+3^{b}\right)$, which is impossible again.  We find that there exists a unique solution $(k, n)=(2,4)$.


Let $b, n>1$ be integers. Suppose that for each $k>1$ there exists an integer $a_{k}$ such that $b-a_{k}^{n}$ is divisible by $k$. Prove that $b=A^{n}$ for some integer $A$.  (Canada)

Solution: Let the prime factorization of $b$ be $b=p_{1}^{\alpha_{1}} \ldots p_{s}^{\alpha_{s}}$, where $p_{1}, \ldots, p_{s}$ are distinct primes. Our goal is to show that all exponents $\alpha_{i}$ are divisible by $n$, then we can set $A=p_{1}^{\alpha_{1} / n} \ldots p_{s}^{\alpha_{s} / n}$.  Apply the condition for $k=b^{2}$. The number $b-a_{k}^{n}$ is divisible by $b^{2}$ and hence, for each $1 \leq i \leq s$, it is divisible by $p_{i}^{2 \alpha_{i}}>p_{i}^{\alpha_{i}}$ as well. Therefore  $$ a_{k}^{n} \equiv b \equiv 0 \quad\left(\bmod p_{i}^{\alpha_{i}}\right) $$  and  $$ a_{k}^{n} \equiv b \not \equiv 0 \quad\left(\bmod p_{i}^{\alpha_{i}+1}\right) $$  which implies that the largest power of $p_{i}$ dividing $a_{k}^{n}$ is $p_{i}^{\alpha_{i}}$. Since $a_{k}^{n}$ is a complete $n$th power, this implies that $\alpha_{i}$ is divisible by $n$.  Comment. If $n=8$ and $b=16$, then for each prime $p$ there exists an integer $a_{p}$ such that $b-a_{p}^{n}$ is divisible by $p$. Actually, the congruency $x^{8}-16 \equiv 0(\bmod p)$ expands as  $$ \left(x^{2}-2\right)\left(x^{2}+2\right)\left(x^{2}-2 x+2\right)\left(x^{2}+2 x+2\right) \equiv 0 \quad(\bmod p) $$  Hence, if -1 is a quadratic residue modulo $p$, then congruency $x^{2}+2 x+2=(x+1)^{2}+1 \equiv 0$ has a solution. Otherwise, one of congruencies $x^{2} \equiv 2$ and $x^{2} \equiv-2$ has a solution.  Thus, the solution cannot work using only prime values of $k$.


Let $X$ be a set of 10000 integers, none of them is divisible by 47 . Prove that there exists a 2007-element subset $Y$ of $X$ such that $a-b+c-d+e$ is not divisible by 47 for any $a, b, c, d, e \in Y$.  (Netherlands)

Solution: Call a set $M$ of integers good if $47 \nmid a-b+c-d+e$ for any $a, b, c, d, e \in M$.  Consider the set $J=\{-9,-7,-5,-3,-1,1,3,5,7,9\}$. We claim that $J$ is good. Actually, for any $a, b, c, d, e \in J$ the number $a-b+c-d+e$ is odd and  $$ -45=(-9)-9+(-9)-9+(-9) \leq a-b+c-d+e \leq 9-(-9)+9-(-9)+9=45 $$  But there is no odd number divisible by 47 between -45 and 45 .  For any $k=1, \ldots, 46$ consider the set  $$ A_{k}=\{x \in X \mid \exists j \in J: \quad k x \equiv j(\bmod 47)\} . $$  If $A_{k}$ is not good, then $47 \mid a-b+c-d+e$ for some $a, b, c, d, e \in A_{k}$, hence $47 \mid k a-k b+$ $k c-k d+k e$. But set $J$ contains numbers with the same residues modulo 47 , so $J$ also is not good. This is a contradiction; therefore each $A_{k}$ is a good subset of $X$.  Then it suffices to prove that there exists a number $k$ such that $\left|A_{k}\right| \geq 2007$. Note that each $x \in X$ is contained in exactly 10 sets $A_{k}$. Then  $$ \sum_{k=1}^{46}\left|A_{k}\right|=10|X|=100000 $$  hence for some value of $k$ we have  $$ \left|A_{k}\right| \geq \frac{100000}{46}>2173>2007 . $$  This completes the proof.  Comment. For the solution, it is essential to find a good set consisting of 10 different residues. Actually, consider a set $X$ containing almost uniform distribution of the nonzero residues (i. e. each residue occurs 217 or 218 times). Let $Y \subset X$ be a good subset containing 2007 elements. Then the set $K$ of all residues appearing in $Y$ contains not less than 10 residues, and obviously this set is good.  On the other hand, there is no good set $K$ consisting of 11 different residues. The CauchyDavenport theorem claims that for any sets $A, B$ of residues modulo a prime $p$,  $$ |A+B| \geq \min \{p,|A|+|B|-1\} . $$  Hence, if $|K| \geq 11$, then $|K+K| \geq 21,|K+K+K| \geq 31>47-|K+K|$, hence $\mid K+K+K+$ $(-K)+(-K) \mid=47$, and $0 \equiv a+c+e-b-d(\bmod 47)$ for some $a, b, c, d, e \in K$.  From the same reasoning, one can see that a good set $K$ containing 10 residues should satisfy equalities $|K+K|=19=2|K|-1$ and $|K+K+K|=28=|K+K|+|K|-1$. It can be proved that in this case set $K$ consists of 10 residues forming an arithmetic progression. As an easy consequence, one obtains that set $K$ has the form $a J$ for some nonzero residue $a$.


For every integer $k \geq 2$, prove that $2^{3 k}$ divides the number  $$ \left(\begin{array}{c} 2^{k+1} \\ 2^{k} \end{array}\right)-\left(\begin{array}{c} 2^{k} \\ 2^{k-1} \end{array}\right) $$  but $2^{3 k+1}$ does not.  (Poland)

Solution: We use the notation $(2 n-1) ! !=1 \cdot 3 \cdots(2 n-1)$ and $(2 n) ! !=2 \cdot 4 \cdots(2 n)=2^{n} n$ ! for any positive integer $n$. Observe that $(2 n) !=(2 n) ! !(2 n-1) ! !=2^{n} n !(2 n-1) !$ !.  For any positive integer $n$ we have  $$ \begin{aligned} & \left(\begin{array}{l} 4 n \\ 2 n \end{array}\right)=\frac{(4 n) !}{(2 n) !^{2}}=\frac{2^{2 n}(2 n) !(4 n-1) ! !}{(2 n) !^{2}}=\frac{2^{2 n}}{(2 n) !}(4 n-1) ! ! \\ & \left(\begin{array}{c} 2 n \\ n \end{array}\right)=\frac{1}{(2 n) !}\left(\frac{(2 n) !}{n !}\right)^{2}=\frac{1}{(2 n) !}\left(2^{n}(2 n-1) ! !\right)^{2}=\frac{2^{2 n}}{(2 n) !}(2 n-1) ! !^{2} . \end{aligned} $$  Then expression (1) can be rewritten as follows:  $$ \begin{aligned} \left(\begin{array}{c} 2^{k+1} \\ 2^{k} \end{array}\right) & -\left(\begin{array}{c} 2^{k} \\ 2^{k-1} \end{array}\right)=\frac{2^{2^{k}}}{\left(2^{k}\right) !}\left(2^{k+1}-1\right) ! !-\frac{2^{2^{k}}}{\left(2^{k}\right) !}\left(2^{k}-1\right) ! !^{2} \\ & =\frac{2^{2^{k}}\left(2^{k}-1\right) ! !}{\left(2^{k}\right) !} \cdot\left(\left(2^{k}+1\right)\left(2^{k}+3\right) \ldots\left(2^{k}+2^{k}-1\right)-\left(2^{k}-1\right)\left(2^{k}-3\right) \ldots\left(2^{k}-2^{k}+1\right)\right) . \end{aligned} $$  We compute the exponent of 2 in the prime decomposition of each factor (the first one is a rational number but not necessarily an integer; it is not important).  First, we show by induction on $n$ that the exponent of 2 in $\left(2^{n}\right)$ ! is $2^{n}-1$. The base case $n=1$ is trivial. Suppose that $\left(2^{n}\right) !=2^{2^{n}-1}(2 d+1)$ for some integer $d$. Then we have  $$ \left(2^{n+1}\right) !=2^{2^{n}}\left(2^{n}\right) !\left(2^{n+1}-1\right) ! !=2^{2^{n}} 2^{2^{n}-1} \cdot(2 d+1)\left(2^{n+1}-1\right) ! !=2^{2^{n+1}-1} \cdot(2 q+1) $$  for some integer $q$. This finishes the induction step.  Hence, the exponent of 2 in the first factor in $(2)$ is $2^{k}-\left(2^{k}-1\right)=1$.  The second factor in (2) can be considered as the value of the polynomial  $$ P(x)=(x+1)(x+3) \ldots\left(x+2^{k}-1\right)-(x-1)(x-3) \ldots\left(x-2^{k}+1\right) . $$  at $x=2^{k}$. Now we collect some information about $P(x)$.  Observe that $P(-x)=-P(x)$, since $k \geq 2$. So $P(x)$ is an odd function, and it has nonzero coefficients only at odd powers of $x$. Hence $P(x)=x^{3} Q(x)+c x$, where $Q(x)$ is a polynomial with integer coefficients.  Compute the exponent of 2 in $c$. We have  $$ \begin{aligned} c & =2\left(2^{k}-1\right) ! ! \sum_{i=1}^{2^{k-1}} \frac{1}{2 i-1}=\left(2^{k}-1\right) ! ! \sum_{i=1}^{2^{k-1}}\left(\frac{1}{2 i-1}+\frac{1}{2^{k}-2 i+1}\right) \\ & =\left(2^{k}-1\right) ! ! \sum_{i=1}^{2^{k-1}} \frac{2^{k}}{(2 i-1)\left(2^{k}-2 i+1\right)}=2^{k} \sum_{i=1}^{2^{k-1}} \frac{\left(2^{k}-1\right) ! !}{(2 i-1)\left(2^{k}-2 i+1\right)}=2^{k} S \end{aligned} $$  For any integer $i=1, \ldots, 2^{k-1}$, denote by $a_{2 i-1}$ the residue inverse to $2 i-1$ modulo $2^{k}$. Clearly, when $2 i-1$ runs through all odd residues, so does $a_{2 i-1}$, hence  $$ \begin{aligned} & S=\sum_{i=1}^{2^{k-1}} \frac{\left(2^{k}-1\right) ! !}{(2 i-1)\left(2^{k}-2 i+1\right)} \equiv-\sum_{i=1}^{2^{k-1}} \frac{\left(2^{k}-1\right) ! !}{(2 i-1)^{2}} \equiv-\sum_{i=1}^{2^{k-1}}\left(2^{k}-1\right) ! ! a_{2 i-1}^{2} \\ &=-\left(2^{k}-1\right) ! ! \sum_{i=1}^{2^{k-1}}(2 i-1)^{2}=-\left(2^{k}-1\right) ! ! \frac{2^{k-1}\left(2^{2 k}-1\right)}{3} \quad\left(\bmod 2^{k}\right) . \end{aligned} $$  Therefore, the exponent of 2 in $S$ is $k-1$, so $c=2^{k} S=2^{2 k-1}(2 t+1)$ for some integer $t$.  Finally we obtain that  $$ P\left(2^{k}\right)=2^{3 k} Q\left(2^{k}\right)+2^{k} c=2^{3 k} Q\left(2^{k}\right)+2^{3 k-1}(2 t+1), $$  which is divisible exactly by $2^{3 k-1}$. Thus, the exponent of 2 in $(2)$ is $1+(3 k-1)=3 k$.  Comment. The fact that (1) is divisible by $2^{2 k}$ is known; but it does not help in solving this problem.


Find all surjective functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for every $m, n \in \mathbb{N}$ and every prime $p$, the number $f(m+n)$ is divisible by $p$ if and only if $f(m)+f(n)$ is divisible by $p$.  ( $\mathbb{N}$ is the set of all positive integers.)  (Iran)  Answer. $f(n)=n$.

Solution: Suppose that function $f: \mathbb{N} \rightarrow \mathbb{N}$ satisfies the problem conditions.  Lemma. For any prime $p$ and any $x, y \in \mathbb{N}$, we have $x \equiv y(\bmod p)$ if and only if $f(x) \equiv f(y)$ $(\bmod p)$. Moreover, $p \mid f(x)$ if and only if $p \mid x$.  Proof. Consider an arbitrary prime $p$. Since $f$ is surjective, there exists some $x \in \mathbb{N}$ such that $p \mid f(x)$. Let  $$ d=\min \{x \in \mathbb{N}: p \mid f(x)\} . $$  By induction on $k$, we obtain that $p \mid f(k d)$ for all $k \in \mathbb{N}$. The base is true since $p \mid f(d)$. Moreover, if $p \mid f(k d)$ and $p \mid f(d)$ then, by the problem condition, $p \mid f(k d+d)=f((k+1) d)$ as required.  Suppose that there exists an $x \in \mathbb{N}$ such that $d \not x$ but $p \mid f(x)$. Let  $$ y=\min \{x \in \mathbb{N}: d \not\{x, p \mid f(x)\} . $$  By the choice of $d$, we have $y>d$, and $y-d$ is a positive integer not divisible by $d$. Then $p \nmid\{(y-d)$, while $p \mid f(d)$ and $p \mid f(d+(y-d))=f(y)$. This contradicts the problem condition. Hence, there is no such $x$, and  $$ p|f(x) \Longleftrightarrow d| x . $$  Take arbitrary $x, y \in \mathbb{N}$ such that $x \equiv y(\bmod d)$. We have $p \mid f(x+(2 x d-x))=f(2 x d)$; moreover, since $d \mid 2 x d+(y-x)=y+(2 x d-x)$, we get $p \mid f(y+(2 x d-x))$. Then by the problem condition $p|f(x)+f(2 x d-x), p| f(y)+f(2 x d-x)$, and hence $f(x) \equiv-f(2 x d-x) \equiv f(y)$ $(\bmod p)$.  On the other hand, assume that $f(x) \equiv f(y)(\bmod p)$. Again we have $p \mid f(x)+f(2 x d-x)$ which by our assumption implies that $p \mid f(x)+f(2 x d-x)+(f(y)-f(x))=f(y)+f(2 x d-x)$. Hence by the problem condition $p \mid f(y+(2 x d-x))$. Using (1) we get $0 \equiv y+(2 x d-x) \equiv y-x$ $(\bmod d)$.  Thus, we have proved that  $$ x \equiv y \quad(\bmod d) \Longleftrightarrow f(x) \equiv f(y) \quad(\bmod p) $$  We are left to show that $p=d$ : in this case (1) and (2) provide the desired statements.  The numbers $1,2, \ldots, d$ have distinct residues modulo $d$. By (2), numbers $f(1), f(2), \ldots$, $f(d)$ have distinct residues modulo $p$; hence there are at least $d$ distinct residues, and $p \geq d$. On the other hand, by the surjectivity of $f$, there exist $x_{1}, \ldots, x_{p} \in \mathbb{N}$ such that $f\left(x_{i}\right)=i$ for any $i=1,2, \ldots, p$. By (2), all these $x_{i}$ 's have distinct residues modulo $d$. For the same reasons, $d \geq p$. Hence, $d=p$.  Now we prove that $f(n)=n$ by induction on $n$. If $n=1$ then, by the Lemma, $p \not \nmid f(1)$ for any prime $p$, so $f(1)=1$, and the base is established. Suppose that $n>1$ and denote $k=f(n)$. Note that there exists a prime $q \mid n$, so by the Lemma $q \mid k$ and $k>1$.  If $k>n$ then $k-n+1>1$, and there exists a prime $p \mid k-n+1$; we have $k \equiv n-1$ $(\bmod p)$. By the induction hypothesis we have $f(n-1)=n-1 \equiv k=f(n)(\bmod p)$. Now, by the Lemma we obtain $n-1 \equiv n(\bmod p)$ which cannot be true.  Analogously, if $k<n$, then $f(k-1)=k-1$ by induction hypothesis. Moreover, $n-k+1>1$, so there exists a prime $p \mid n-k+1$ and $n \equiv k-1(\bmod p)$. By the Lemma again, $k=f(n) \equiv$ $f(k-1)=k-1(\bmod p)$, which is also false. The only remaining case is $k=n$, so $f(n)=n$.  Finally, the function $f(n)=n$ obviously satisfies the condition.


Let $k$ be a positive integer. Prove that the number $\left(4 k^{2}-1\right)^{2}$ has a positive divisor of the form $8 k n-1$ if and only if $k$ is even.  (United Kingdom)

Solution: The statement follows from the following fact.  Lemma. For arbitrary positive integers $x$ and $y$, the number $4 x y-1$ divides $\left(4 x^{2}-1\right)^{2}$ if and only if $x=y$.  Proof. If $x=y$ then $4 x y-1=4 x^{2}-1$ obviously divides $\left(4 x^{2}-1\right)^{2}$ so it is sufficient to consider the opposite direction.  Call a pair $(x, y)$ of positive integers bad if $4 x y-1$ divides $\left(4 x^{2}-1\right)^{2}$ but $x \neq y$. In order to prove that bad pairs do not exist, we present two properties of them which provide an infinite descent.  Property (i). If $(x, y)$ is a bad pair and $x<y$ then there exists a positive integer $z<x$ such that $(x, z)$ is also bad.  Let $r=\frac{\left(4 x^{2}-1\right)^{2}}{4 x y-1}$. Then  $$ r=-r \cdot(-1) \equiv-r(4 x y-1)=-\left(4 x^{2}-1\right)^{2} \equiv-1 \quad(\bmod 4 x) $$  and $r=4 x z-1$ with some positive integer $z$. From $x<y$ we obtain that  $$ 4 x z-1=\frac{\left(4 x^{2}-1\right)^{2}}{4 x y-1}<4 x^{2}-1 $$  and therefore $z<x$. By the construction, the number $4 x z-1$ is a divisor of $\left(4 x^{2}-1\right)^{2}$ so $(x, z)$ is a bad pair.  Property (ii). If $(x, y)$ is a bad pair then $(y, x)$ is also bad.  Since $1=1^{2} \equiv(4 x y)^{2}(\bmod 4 x y-1)$, we have  $$ \left(4 y^{2}-1\right)^{2} \equiv\left(4 y^{2}-(4 x y)^{2}\right)^{2}=16 y^{4}\left(4 x^{2}-1\right)^{2} \equiv 0 \quad(\bmod 4 x y-1) $$  Hence, the number $4 x y-1$ divides $\left(4 y^{2}-1\right)^{2}$ as well.  Now suppose that there exists at least one bad pair. Take a bad pair $(x, y)$ such that $2 x+y$ attains its smallest possible value. If $x<y$ then property (i) provides a bad pair $(x, z)$ with $z<y$ and thus $2 x+z<2 x+y$. Otherwise, if $y<x$, property (ii) yields that pair $(y, x)$ is also bad while $2 y+x<2 x+y$. Both cases contradict the assumption that $2 x+y$ is minimal; the Lemma is proved.  To prove the problem statement, apply the Lemma for $x=k$ and $y=2 n$; the number $8 k n-1$ divides $\left(4 k^{2}-1\right)^{2}$ if and only if $k=2 n$. Hence, there is no such $n$ if $k$ is odd and $n=k / 2$ is the only solution if $k$ is even.  Comment. The constant 4 in the Lemma can be replaced with an arbitrary integer greater than 1: if $a>1$ and $a x y-1$ divides $\left(a x^{2}-1\right)^{2}$ then $x=y$.


For a prime $p$ and a positive integer $n$, denote by $\nu_{p}(n)$ the exponent of $p$ in the prime factorization of $n$ !. Given a positive integer $d$ and a finite set $\left\{p_{1}, \ldots, p_{k}\right\}$ of primes. Show that there are infinitely many positive integers $n$ such that $d \mid \nu_{p_{i}}(n)$ for all $1 \leq i \leq k$.  (India)

Solution: For arbitrary prime $p$ and positive integer $n$, denote by $\operatorname{ord}_{p}(n)$ the exponent of $p$ in $n$. Thus,  $$ \nu_{p}(n)=\operatorname{ord}_{p}(n !)=\sum_{i=1}^{n} \operatorname{ord}_{p}(i) $$  Lemma. Let $p$ be a prime number, $q$ be a positive integer, $k$ and $r$ be positive integers such that $p^{k}>r$. Then $\nu_{p}\left(q p^{k}+r\right)=\nu_{p}\left(q p^{k}\right)+\nu_{p}(r)$.  Proof. We claim that $\operatorname{ord}_{p}\left(q p^{k}+i\right)=\operatorname{ord}_{p}(i)$ for all $0<i<p^{k}$. Actually, if $d=\operatorname{ord}_{p}(i)$ then $d<k$, so $q p^{k}+i$ is divisible by $p^{d}$, but only the first term is divisible by $p^{d+1}$; hence the sum is not.  Using this claim, we obtain  $$ \nu_{p}\left(q p^{k}+r\right)=\sum_{i=1}^{q p^{k}} \operatorname{ord}_{p}(i)+\sum_{i=q p^{k}+1}^{q p^{k}+r} \operatorname{ord}_{p}(i)=\sum_{i=1}^{q p^{k}} \operatorname{ord}_{p}(i)+\sum_{i=1}^{r} \operatorname{ord}_{p}(i)=\nu_{p}\left(q p^{k}\right)+\nu_{p}(r) . $$  For any integer $a$, denote by $\bar{a}$ its residue modulo $d$. The addition of residues will also be performed modulo $d$, i. e. $\bar{a}+\bar{b}=\overline{a+b}$. For any positive integer $n$, let $f(n)=\left(f_{1}(n), \ldots, f_{k}(n)\right)$, where $f_{i}(n)=\overline{\nu_{p_{i}}(n)}$.  Define the sequence $n_{1}=1, n_{\ell+1}=\left(p_{1} p_{2} \ldots p_{k}\right)^{n_{\ell}}$. We claim that  $$ f\left(n_{\ell_{1}}+n_{\ell_{2}}+\ldots+n_{\ell_{m}}\right)=f\left(n_{\ell_{1}}\right)+f\left(n_{\ell_{2}}\right)+\ldots+f\left(n_{\ell_{m}}\right) $$  for any $\ell_{1}<\ell_{2}<\ldots<\ell_{m}$. (The addition of $k$-tuples is componentwise.) The base case $m=1$ is trivial.  Suppose that $m>1$. By the construction of the sequence, $p_{i}^{n_{\ell_{1}}}$ divides $n_{\ell_{2}}+\ldots+n_{\ell_{m}}$; clearly, $p_{i}^{n_{\ell_{1}}}>n_{\ell_{1}}$ for all $1 \leq i \leq k$. Therefore the Lemma can be applied for $p=p_{i}, k=r=n_{\ell_{1}}$ and $q p^{k}=n_{\ell_{2}}+\ldots+n_{\ell_{m}}$ to obtain  $$ f_{i}\left(n_{\ell_{1}}+n_{\ell_{2}}+\ldots+n_{\ell_{m}}\right)=f_{i}\left(n_{\ell_{1}}\right)+f_{i}\left(n_{\ell_{2}}+\ldots+n_{\ell_{m}}\right) \quad \text { for all } 1 \leq i \leq k $$  and hence  $$ f\left(n_{\ell_{1}}+n_{\ell_{2}}+\ldots+n_{\ell_{m}}\right)=f\left(n_{\ell_{1}}\right)+f\left(n_{\ell_{2}}+\ldots+n_{\ell_{m}}\right)=f\left(n_{\ell_{1}}\right)+f\left(n_{\ell_{2}}\right)+\ldots+f\left(n_{\ell_{m}}\right) $$  by the induction hypothesis.  Now consider the values $f\left(n_{1}\right), f\left(n_{2}\right), \ldots$ There exist finitely many possible values of $f$. Hence, there exists an infinite sequence of indices $\ell_{1}<\ell_{2}<\ldots$ such that $f\left(n_{\ell_{1}}\right)=f\left(n_{\ell_{2}}\right)=\ldots$ and thus  $$ f\left(n_{\ell_{m+1}}+n_{\ell_{m+2}}+\ldots+n_{\ell_{m+d}}\right)=f\left(n_{\ell_{m+1}}\right)+\ldots+f\left(n_{\ell_{m+d}}\right)=d \cdot f\left(n_{\ell_{1}}\right)=(\overline{0}, \ldots, \overline{0}) $$  for all $m$. We have found infinitely many suitable numbers.


Solution: We use the same Lemma and definition of the function $f$.  Let $S=\{f(n): n \in \mathbb{N}\}$. Obviously, set $S$ is finite. For every $s \in S$ choose the minimal $n_{s}$ such that $f\left(n_{s}\right)=s$. Denote $N=\max _{s \in S} n_{s}$. Moreover, let $g$ be an integer such that $p_{i}^{g}>N$ for each $i=1,2, \ldots, k$. Let $P=\left(p_{1} p_{2} \ldots p_{k}\right)^{g}$.  We claim that  $$ \{f(n) \mid n \in[m P, m P+N]\}=S $$  for every positive integer $m$. In particular, since $(\overline{0}, \ldots, \overline{0})=f(1) \in S$, it follows that for an arbitrary $m$ there exists $n \in[m P, m P+N]$ such that $f(n)=(\overline{0}, \ldots, \overline{0})$. So there are infinitely many suitable numbers.  To prove (1), let $a_{i}=f_{i}(m P)$. Consider all numbers of the form $n_{m, s}=m P+n_{s}$ with $s=\left(s_{1}, \ldots, s_{k}\right) \in S$ (clearly, all $n_{m, s}$ belong to $[m P, m P+N]$ ). Since $n_{s} \leq N<p_{i}^{g}$ and $p_{i}^{g} \mid m P$, we can apply the Lemma for the values $p=p_{i}, r=n_{s}, k=g, q p^{k}=m P$ to obtain  $$ f_{i}\left(n_{m, s}\right)=f_{i}(m P)+f_{i}\left(n_{s}\right)=a_{i}+s_{i} \text {; } $$  hence for distinct $s, t \in S$ we have $f\left(n_{m, s}\right) \neq f\left(n_{m, t}\right)$.  Thus, the function $f$ attains at least $|S|$ distinct values in $[m P, m P+N]$. Since all these values belong to $S, f$ should attain all possible values in $[m P, m P+N]$.  Comment. Both solutions can be extended to prove the following statements.  Claim 1. For any $K$ there exist infinitely many $n$ divisible by $K$, such that $d \mid \nu_{p_{i}}(n)$ for each $i$.  Claim 2. For any $s \in S$, there exist infinitely many $n \in \mathbb{N}$ such that $f(n)=s$.