![](https://cdn.mathpix.com/cropped/2024_06_05_3ed553d927cb47e8859bg-01.jpg?height=331&width=374&top_left_y=286&top_left_x=247) $6^{\mathrm{TH}}$ European Mathematical Cup $9^{\text {th }}$ December 2017-23rd December 2017 Junior Category # Problems and Solutions Problem 1. Find all pairs $(x, y)$ of integers that satisfy the equation $$ x^{2} y+y^{2}=x^{3} $$ (Daniel Paleka) First Solution. Firstly, let us consider the case $x=0$. In that case, it is trivial to see that $y=0$, and thus we get the solution $(0,0)$. From now on, we can assume $x \neq 0$. From the trivial $x^{2} \mid x^{3}$, the equation gives $x^{2}\left|x^{2} y+y^{2} \Rightarrow x^{2}\right| y^{2}$, which means $x \mid y$. 1 point. We use the substitution $y=k x$, where $k \in \mathbb{Z}$. 1 point. The substitution gives us $$ \begin{gathered} k x^{3}+k^{2} x^{2}=x^{3} \\ k x+k^{2}=x \\ k^{2}=x(1-k) \end{gathered} $$ 2 points. Considering the greatest common divisor of $k^{2}$ and $1-k$, we get $$ G C D\left(k^{2}, 1-k\right)=G C D\left(k^{2}+k(1-k), 1-k\right)=G C D(k, 1-k)=G C D(k, 1-k+k)=G C D(k, 1)=1 $$ 3 points. That leaves us with two possibilities. a) $1-k=1 \Rightarrow k=0 \Rightarrow x=0$ which is not possible since $x \neq 0$. 1 point. b) $1-k=-1 \Rightarrow k=2 \Rightarrow x=-4, y=-8$, which gives a solution to the original equation. 1 point. Second Solution. We rearrange the equation into: $$ y^{2}=x^{2}(x-y) $$ It can easily be shown that if $y \neq 0, x-y$ must be square. 1 point. If $y=0$, from the starting equation we infer $x=0$, and we have a solution $(x, y)=(0,0)$. In the other case, we set $x=y+a^{2}$, where $a$ is a positive integer. Taking the square root of the equation gives: $$ |y|=|x| a $$ 1 point. Because $x=y+a^{2}>y$, it is impossible for $y$ to be a positive integer, because then the equation would say $y=x a>x$, which is false. That means $y<0$, and also: $$ -y=|x| a $$ 2 points. If $x$ is positive, we can write: $$ -y=x a=\left(y+a^{2}\right) a=a y+a^{3} $$ which rearranges into $$ -y(a+1)=a^{3} $$ so $a^{3}$ is divisible by $a+1$, which is not possible for positive $a$ due to $a^{3}=(a+1)\left(a^{2}-a+1\right)-1$. 2 points. We see that $x$ cannot be zero due to $y$ being negative, so the only remaining option is that $x<0$ also. We write: $$ -y=x a=-\left(y+a^{2}\right) a=-a y+a^{3} $$ which can similarly be rearranged into $$ -y(a-1)=a^{3} $$ and this time $a^{3}$ is divisible by $a-1$ 1 point. Analogously, we decompose $a^{3}=(a-1)\left(a^{2}+a+1\right)+1$, so $a-1$ divides 1 and the unique possibility is $a=2$. 2 points. The choice $a=2$ gives $y=-8$ and $x=-4$, which is a solution to the original equation. 1 point. ## Notes on marking: - Points awarded for different solutions are not additive, a student should be awarded the maximal number of points he is awarded following only one of the schemes. - Saying that $(0,0)$ is a solution is worth $\mathbf{0}$ points. The point is awarded only if the student argues that, disregarding the solution $(0,0)$, we must only consider $x \neq 0$, or a similar proposition. - Failing to check that $(0,0)$ is a solution shall not be punished. Failing to check that $(-4,-8)$ is a solution shall result in the deduction of $\mathbf{1}$ point only if a student did not use a chain of equivalences to discover the solution. Problem 2. A regular hexagon in the plane is called sweet if its area is equal to 1. Is it possible to place 2000000 sweet hexagons in the plane such that the union of their interiors is a convex polygon of area at least 1900000 ? Remark: A subset $S$ of the plane is called convex if for every pair of points in $S$, every point on the straight line segment that joins the pair of points also belongs to $S$. The hexagons may overlap. (Josip Pupic, Borna Vukorepa) Solution. It is possible to make such arrangement. 0 points. We will stack hexagons in a triangular pattern shown below, where the first row has one hexagon, second row has two and so on. The pattern on the picture is a triangle with four rows. ![](https://cdn.mathpix.com/cropped/2024_06_05_3ed553d927cb47e8859bg-03.jpg?height=483&width=596&top_left_y=795&top_left_x=730) 3 points. Such triangle with $n$ rows has an area of $\frac{n(n+1)}{2}$ since that is the total number of hexagons used for that pattern. 1 point. Setting $n=1950$ gives us a triangle with 1950 rows. That figure has an area of 1902225 and the same number of hexagons is used. The only problem is that it is not convex. 1 point. We can use the remaining hexagons to fix the non-convex parts of the figure, as shown below. ![](https://cdn.mathpix.com/cropped/2024_06_05_3ed553d927cb47e8859bg-03.jpg?height=594&width=719&top_left_y=1739&top_left_x=674) 3 points. Every non-convex part can be fixed with two hexagons, so in total we will need $1949 \cdot 3 \cdot 2=11694$ hexagons to make the figure convex. This is because there are 1949 non-convex parts on every side of our triangular pattern. Obviously, this is much less hexagons than we have remaining. The resulting figure is now convex, so this completes the proof. 2 points. Notes on marking: - Sketches of stacking the hexagons in any pattern will not be worth any points if there is no work done on them. - No points are awarded for the claim that the construction is possible. - There are many patterns for stacking the hexagons which can give the correct solution. Each of them should be marked the same way as this one. - Work on patterns which can't produce the desired area will not be worth any points. Problem 3. Let $A B C$ be an acute triangle. Denote by $H$ and $M$ the orthocenter of $A B C$ and the midpoint of side $B C$, respectively. Let $Y$ be a point on $A C$ such that $Y H$ is perpendicular to $M H$ and let $Q$ be a point on $B H$ such that $Q A$ is perpendicular to $A M$. Let $J$ be the second point of intersection of $M Q$ and the circle with diameter $M Y$. Prove that $H J$ is perpendicular to $A M$. (Steve Dinh) Solution. We present the following diagram: ![](https://cdn.mathpix.com/cropped/2024_06_05_3ed553d927cb47e8859bg-05.jpg?height=1542&width=1019&top_left_y=565&top_left_x=333) 0 points. Since $\angle M H Y=90^{\circ}, Y$ lies on the circle with diameter $M Y$, so the quadrilateral $H M J Y$ is cyclic. 1 point. It follows that $\angle H J M=\angle H Y M$. Since $Q A \perp A M$, $$ H J \perp A M \Longleftrightarrow H J \| Q A \Longleftrightarrow \angle H J M=\angle A Q M \Longleftrightarrow \angle H Y M=\angle A Q M $$ Since $\angle Y H M=\angle Q A M=90^{\circ}$, the latter is to equivalent to $\triangle A Q M \sim \triangle H Y M$. 1 point. Now we have two different approaches to finish the solution: First Approach (Synthetic). Let $P, R$ be the reflections of $A, H$ in $M$, respectively. Then since $\angle Y H M=\angle Q A M=90^{\circ}$, i.e. $\angle Y H R=\angle Q A P=90^{\circ}$, $\triangle A Q M \sim \triangle H Y M \Longleftrightarrow \frac{A Q}{H Y}=\frac{A M}{H M} \Longleftrightarrow \frac{A Q}{H Y}=\frac{\frac{1}{2} A P}{\frac{1}{2} H R} \Longleftrightarrow \frac{A Q}{H Y}=\frac{A P}{H R} \Longleftrightarrow \triangle A Q P \sim \triangle H Y R \Longleftrightarrow \angle Q P A=\angle Y R H$. 3 points. Since $M$ is the midpoint of $B C$, the quadrilaterals $A B P C$ and $H B R C$ are parallelograms. 1 point. Since $C R \| H B$ and $H B \perp A C$, it follows that $\angle A C R=90^{\circ}$. Hence $\angle Y C R=\angle R H Y=90^{\circ}$, so the quadrilateral $Y H R C$ is cyclic. 1 point. It follows that $\angle Y R H=\angle Y C H=\angle A C F=90^{\circ}-\angle B A C$. 1 point. Since $B P \| A C$ and $A C \perp B Q$, we have $P B Q=90^{\circ}$. Hence $\angle P B Q=\angle P A Q=90^{\circ}$, so the quadrilateral $A B P Q$ is cyclic. 1 point. It follows that $\angle Q P A=\angle Q B A=\angle E B A=90^{\circ}-\angle B A C$. 1 point. Finally, we conclude that $\angle Y R H=\angle Q P A$, as desired. Second Approach (Trigonometric). We will show that $\triangle A Q M \sim \triangle H Y M$ by proving that $$ \frac{A Q}{A M}=\frac{H Y}{H M} $$ To begin with, let $a=B C, b=C A, c=A B$ and $\alpha=\angle B A C, \beta=\angle C B A, \gamma=\angle A C B$. From right-angled $\triangle A E Q$ we get $A Q=\frac{A E}{\cos \angle E A Q}=\frac{A E}{\sin \angle M A C}$. Then from right-angled $\triangle A B E$ we obtain $A E=A B \cos \angle B A E=c \cos \alpha$, so $A Q=\frac{c \cos \alpha}{\sin \angle M A C}$, i.e. $\frac{A Q}{A M}=\frac{c \cos \alpha}{A M \sin \angle M A C}$. By the sine law applied to $\triangle A M C$, we obtain $\frac{A M}{\sin \angle A C M}=\frac{M C}{\sin \angle M A C}$, i.e. $A M \sin \angle M A C=\frac{a}{2} \sin \gamma$. It follows that $\frac{A Q}{A M}=\frac{c \cos \alpha}{\frac{a}{2} \sin \gamma}=\frac{c}{\sin \gamma} \cdot \frac{2 \cos \alpha}{a}=\frac{a}{\sin \alpha} \cdot \frac{2 \cos \alpha}{a}=2 \cot \alpha$, where we used the sine law applied to $\triangle A B C$. 4 points. To conclude, note that $\triangle A H Y \sim \triangle B M H$ since $\angle H A Y=\angle M B H=90^{\circ}-\gamma$ and $\angle Y H A=\angle H M B$ (angles with perpendicular rays). Then $\frac{H Y}{H M}=\frac{A H}{B M}=2 \cot \alpha$, so we are done. 4 points. ## Notes on marking: - The points from different approaches are not additive, a student should be awarded the maximum of points obtained from one of them. Problem 4. The real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=3$. Prove the inequality $$ x^{3}-\left(y^{2}+y z+z^{2}\right) x+y^{2} z+y z^{2} \leq 3 \sqrt{3} $$ and find all triples $(x, y, z)$ for which equality holds. (Miroslav Marinov) Solution. First let us notice the factorization of the left-hand side $$ x^{3}-\left(y^{2}+y z+z^{2}\right) x+y^{2} z+y z^{2}=(x-y)(x-z)(x+y+z) $$ 2 points. Now we get the following inequalities $$ \begin{gathered} \left(x^{3}-\left(y^{2}+y z+z^{2}\right) x+y^{2} z+y z^{2}\right)^{\frac{2}{3}} \\ =\sqrt[3]{(x-y)^{2}(x-z)^{2}(x+y+z)^{2}} \\ \stackrel{G-A}{\leqslant} \frac{1}{3}\left((x-y)^{2}+(x-z)^{2}+(x+y+z)^{2}\right) \\ =\frac{1}{3}\left(3 x^{2}+2 y^{2}+2 z^{2}+2 y z\right) \\ =\frac{1}{3}\left(6+x^{2}+2 y z\right) \\ \stackrel{G-A}{\leqslant} \frac{1}{3}\left(6+x^{2}+y^{2}+z^{2}\right) \end{gathered} $$ 1 point. 3 points. 1 point. $$ =\frac{9}{3}=3 $$ from where we get the required inequality by raising to the power of $\frac{3}{2}$. In the case of equality, expressions $|x-y|,|x-z|$ and $|x+y+z|$ are all equal to $\sqrt{3}$ which we conclude from the first G-A inequality. From the case of equality in the second G-A inequality we conclude $y=z$. Now from $|x-y|=\sqrt{3}$ we get 2 cases: a) $x-y=\sqrt{3} \Rightarrow|3 y+\sqrt{3}|=\sqrt{3}$ from where we get $y=0$ or $y=-\frac{2 \sqrt{3}}{3}$ which gives us potential solutions $(\sqrt{3}, 0,0)$ and $\left(\frac{\sqrt{3}}{3},-\frac{2 \sqrt{3}}{3},-\frac{2 \sqrt{3}}{3}\right)$. By checking only $(\sqrt{3}, 0,0)$ remains. b) $x-y=-\sqrt{3} \Rightarrow|3 y-\sqrt{3}|=\sqrt{3}$ from where we get $y=0$ or $y=\frac{2 \sqrt{3}}{3}$ which gives us potential solutions $(-\sqrt{3}, 0,0)$ and $\left(-\frac{\sqrt{3}}{3}, \frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{3}}{3}\right)$. By checking only $\left(-\frac{\sqrt{3}}{3}, \frac{2 \sqrt{3}}{3}, \frac{2 \sqrt{3}}{3}\right)$ remains. 3 points. ## Notes on marking: - Factorization from the beginning can be spotted because $y$ and $z$ are obviously roots of the polynomial equation $x^{3}-\left(y^{2}+y z+z^{2}\right) x+y^{2} z+y z^{2}=0$ in the variable $x$. - 1 point is to be deducted if potential solutions aren't checked out i.e. either $\left(\frac{\sqrt{3}}{3},-\frac{2 \sqrt{3}}{3},-\frac{2 \sqrt{3}}{3}\right)$ or $(-\sqrt{3}, 0,0)$ is stated as solutions for the case of equality and. - Proving the inequality is worth $\mathbf{7}$ points while the rest is worth $\mathbf{3}$ points non-depending on the way in which it was proved. ![](https://cdn.mathpix.com/cropped/2024_06_05_3ed553d927cb47e8859bg-08.jpg?height=334&width=380&top_left_y=284&top_left_x=244) $6^{\text {TH }}$ EUROPEAN MATHEMATICAL CUP $9^{\text {th }}$ December 2017-233rd December 2017 Senior Category ## Problems and Solutions Problem 1. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that the inequality $$ f(x)+y f(f(x)) \leq x(1+f(y)) $$ holds for all positive integers $x, y$. (Adrian Beker) Solution. We claim that $f(x)=x$ is the only function that satisfies the inequality for all positive integers $x, y$. We can see that it is indeed the solution because $x+y x=x(1+y)$. Setting $x=1$ and $y=1$, we obtain: $$ f(1)+f(f(1)) \leq 1+f(1) $$ which implies $f(f(1)) \leq 1$, so $f(f(1))=1$ because it must be a positive integer. Setting $x=1$ and $y=f(1)$, we obtain: $$ f(1)+f(1) \leq 1+f(1) $$ which similarly implies $f(1)=1$. 3 points. Now, setting $x=1$ gives: $$ 1+y \leq 1+f(y) $$ so $f(y) \geq y$ for all positive integers $y$. Setting $y=1$ and using the previous fact, we write: $$ f(x)+f(f(x)) \leq 2 x \leq 2 f(x)=f(x)+f(x) \leq f(x)+f(f(x)) $$ so equality holds on each step. In particular, $f(x)=x$ for every positive integer $x$ 6 points. Notes on marking: - Checking that $f(x)=x$ satisfies the inequality is worth 0 points. If a student shows that a solution, if it exists, must be the identity function ("solves the problem"), but fails to show that the identity function is indeed the solution, he or she shall be deducted 1 point. A sentence saying something along the lines of "it is trivial to show that the identity function satisfies the inequality", due to the sentence being true, shall not be deducted the point. Problem 2. A friendly football match lasts 90 minutes. In this problem, we consider one of the teams, coached by Sir Alex, which plays with 11 players at all times. a) Sir Alex wants for each of his players to play the same integer number of minutes, but each player has to play less than 60 minutes in total. What is the minimum number of players required? b) For the number of players found in a), what is the minimum number of substitutions required, so that each player plays the same number of minutes? Remark: Substitutions can only take place after a positive integer number of minutes, and players who have come off earlier can return to the game as many times as needed. There is no limit to the number of substitutions allowed. (Athanasios Kontogeorgis, Demetres Christofides) Solution. a) Since exactly 11 players play at all times, the total number of minutes played by all of the players combined is $11.90=990$. Let $n$ be the number of Sir Alex's players that have participated in the match and let $k$ be the number of minutes which each of them spent playing, with $k<60$ and $k \in \mathbb{Z}$. Now the equality $n k=990$ holds. 1 point. From that fact combined with $k<60$ we get $n \geqslant 17$ and $n \mid 990$ as well. Finally, it is easy to conclude that the minimal such $n$ is 18 . 1 point. Construction. 1 point. b) We can formulate the problem by using graphs. Let us construct a graph with 18 vertices that represent the players. Two vertices are connected by an edge if one of the corresponding players substituted the other. 1 point. Suppose that less than 17 substitutions were made. Then the graph isn't connected and the smallest connected component consists of $k \leqslant 9$ players among which all of their substitutions were made. 1 point. Let us suppose that exactly $r$ of them are on the pitch at all times. It is easy to determine that each of the 18 players will play exactly 55 minutes. So the players from the smallest connected component will spend the combined total of $55 k$ minutes playing. But, from the same conclusion as earlier, we get the equality $55 k=90 r$. It follows that $11 \mid r$ which implies $r>9$ and so we reach a contradiction. $\Rightarrow$ The graph is connected and at least 17 substitutions are required. 3 points. The following table shows a match in which each of the 18 players played 55 minutes and exactly 17 substitutions were made (the shaded regions correspond to the time intervals played by each player). Time ![](https://cdn.mathpix.com/cropped/2024_06_05_3ed553d927cb47e8859bg-09.jpg?height=756&width=1157&top_left_y=1912&top_left_x=538) ## Notes on marking: - An example for construction in a) is the construction from b) but there are far easier examples than that and any correct one can bring that 1 point. If a student didn't make the construction in a) but finds one in b), he or she shall be awarded both the 1 point from $\mathbf{a}$ ) and 2 points from $\mathbf{b}$ ) for it but if the student doesn't write a construction neither in a) nor $\mathbf{b}$ ), all 3 mentioned points are to be deducted. - There can be an argument made for $\mathbf{b}$ ) without observing graphs and has to be evaluated accordingly. If a student reaches the conclusion equivalent to the smallest connected component, $\mathbf{2}$ points have to be given, one for that conclusion and one that is intended for observing the graph in the official solution. Problem 3. Let $A B C$ be a scalene triangle and let its incircle touch sides $B C, C A$ and $A B$ at points $D, E$ and $F$ respectively. Let line $A D$ intersect this incircle at point $X$. Point $M$ is chosen on the line $F X$ so that the quadrilateral $A F E M$ is cyclic. Let lines $A M$ and $D E$ intersect at point $L$ and let $Q$ be the midpoint of segment $A E$. Point $T$ is given on the line $L Q$ such that the quadrilateral $A L D T$ is cyclic. Let $S$ be a point such that the quadrilateral TFSA is a parallelogram, and let $N$ be the second point of intersection of the circumcircle of triangle $A S X$ and the line $T S$. Prove that the circumcircles of triangles $T A N$ and $L S A$ are tangent to each other. (Andrej Ilievski) Solution. We present the following diagram: ![](https://cdn.mathpix.com/cropped/2024_06_05_3ed553d927cb47e8859bg-11.jpg?height=1136&width=1388&top_left_y=634&top_left_x=337) 0 points. Let $P$ be the midpoint of segment $A F$ and let $R$ be the second intersection of $E X$ and the circumcircle of $\triangle A F E$. Let $K$ denote the intersection of lines $A R$ and $D F$. By the tangent-chord theorem, we have $\angle E D X=\angle A E X$. Since $D E X F$ is cyclic, we have $\angle E D X=\angle E F X=\angle E F M$. Since $A F E M$ is cyclic, we have $\angle E F M=E A M=\angle E A L$. Hence, $\angle A E X=E A L$, so $A L \| E X$, and analogously $A K \| F X$. Furthermore, we have $\angle E D X=\angle E A L$, i.e. $\angle E D A=\angle E A L$, so by the converse of the tangent-chord theorem, $L A$ is tangent to the circumcircle of $\triangle A E D$. 2 points. By power of a point, we have $L A^{2}=L E \cdot L D$. If we denote the radical axis of the incircle of $\triangle A B C$ and the degenerate circle $A$ by $\ell$, this means that $L$ lies on $\ell$. Analogously, $K$ lies on $\ell$. 1 point. Now, since $Q A^{2}=Q E^{2}, P A^{2}=P F^{2}$ and $Q E, P F$ are tangents to the incircle of $\triangle A B C$, it follows that $P, Q$ both lie on $\ell$. Since $P Q$ is the midline of $\triangle A E F$, we have $\ell \| E F$. 1 point. Since $A L\|E X, A K\| F X$ and $X F D E$ is cyclic, it follows that $A K D L$ is also cyclic. Hence, $K$ is the second intersection of $L Q$ and the circumcircle of $\triangle A L D$, so we must have $K \equiv T$, i.e. $T, F, D$ are collinear. 1 point. Since $T F S A$ is a parallelogram, $T S$ bisects the segment $A F$, i.e. $T, P, S$ are collinear, which means that $S$ lies on $\ell$. Moreover, since $A T \| F S$ and $A T \| F X$, it follows that $F, X, S$ are collinear. Then since $A N X S$ is cyclic, $\angle N A X=\angle N S X$. Since $N S \| F E, \angle N S X=\angle E F X$. Since $A L D T$ and $X E D F$ are both cyclic, we have $\angle N T A=\angle L T A=\angle L D A=\angle E D X=\angle E F X$, so $\angle N A X=\angle N T A$. Hence, by the converse of the tangent-chord theorem, $A X$ is tangent to the circumcircle of $\triangle T A N$. 2 points. Finally, since $A S \| T F$, i.e. $A S \| F D$, we have $\angle X A S=\angle X D F$. Again, using the cyclicity of $A L D T$ and $X E D F$, we have $\angle A L S=\angle A L T=\angle A D T=\angle X D F$, so $\angle X A S=\angle A L S$. Hence, by the converse of the tangent-chord theorem, $A X$ is tangent to the circumcircle of $\triangle L S A$. 2 points. Since $A X$ is the common tangent of the circumcircles of triangles $T A N$ and $L S A$, it follows that they are tangent to each other at $A$, as desired. ## Notes on marking: - There are many different ways to finish the solution once the collinearities of $T, F, D$ and $F, X, S$ are established. One can, for example, show that $E, X, N$ are also collinear by noting that $\angle X N S=\angle X A S=\angle X D F=\angle X E F$ and using the fact that $S N \| F E$, for which a student should be awarded 2 points. Then one can establish the result by introducing the tangent to the circumcircle of $\triangle T A N$ at $A$ and using the tangent-chord theorem and its converse together with the fact that $\angle N T A+\angle A L S=\angle N A S$ holds. This part is worth 2 points. However, points from different approaches are not additive, a student should be awarded the maximum of points obtained from one of them. Problem 4. Find all polynomials $P$ with integer coefficients such that $P(0) \neq 0$ and $$ P^{n}(m) \cdot P^{m}(n) $$ is a square of an integer for all nonnegative integers $n, m$. Remark: For a nonnegative integer $k$ and an integer $n, P^{k}(n)$ is defined as follows: $P^{k}(n)=n$ if $k=0$ and $P^{k}(n)=P\left(P^{k-1}(n)\right)$ if $k>0$. (Adrian Beker) Solution. Let $Q(n, m)$ denote the assertion " $P^{n}(m) \cdot P^{m}(n)$ is a square of an integer". We claim that $P(x)=x+1$ is the unique polynomial with integer coefficients such that $P(0) \neq 0$ and $Q(n, m)$ is true for all $n, m \in \mathbb{N}_{0}$. First we check that this polynomial indeed satisfies the conditions. An easy induction on $k$ shows that $P^{k}(n)=n+k$ for all $n, k \in \mathbb{N}_{0}$. Then $P^{n}(m) \cdot P^{m}(n)=(m+n)^{2}$, which is clearly a square of an integer, hence $Q(n, m)$ is true for all $n, m \in \mathbb{N}_{0}$. 1 point. Now we show that $P(x)=x+1$ is the only polynomial satisfying all the conditions. Consider the sequence $\left(a_{n}\right)_{n \geqslant 0}$ defined by $a_{n}=P^{n}(0)$ for all $n \geqslant 0$. Then $Q(n, 0)$ implies that $n \cdot a_{n}$ is a square of an integer for all $n \in \mathbb{N}_{0}$. 1 point. Lemma 1. For all sufficiently large primes $p$, the sequence $\left(a_{n}\right)$ modulo $p$ is periodic with minimal period of length exactly $p$. In particular, for all sufficiently large primes $p, P$ is bijective when considered modulo $p$. Proof: Fix a prime $p>\max \{|P(0)|, 2\}$. Let $t$ be the smallest positive integer for which there exists a nonnegative integer $s2$. Finally, we conclude that $t-s=p$ must hold, i.e. the length of the minimal period is $p$. In particular, $P$ is surjective and hence bijective when considered modulo $p$. 2 points. Alternative proof: Again, fix a prime $p>|P(0)|$. Since $p \cdot a_{p}$ is a perfect square, $a_{p}$ must be divisible by $p$. It follows that for all $n \geqslant 0, a_{n+p}=P^{n}\left(a_{p}\right) \equiv P^{n}(0) \equiv a_{n}(\bmod p)$, hence $\left(a_{n}\right)$ modulo $p$ is periodic with period of length $p$. 1 point. Now suppose there exist $i, j \in\{0,1, \ldots, p-1\}$ with $i