## Task 4.

Find (one) integer $a$ such that for the polynomial $P(x)=x^{5}+a x$ the statement

$$
\text { " if } n \mid P(k)-P(l) \text { then } n \mid k-l \text {, for all } k, l \in \mathbb{Z} "
$$

holds only for finitely many natural numbers $n$, among which is $n=95$.
## Solution.

One such number is $a=-95^{4}$.

For this $a$, we have $P(95)=P(0)=0$, so $n$ divides $P(95)-P(0)$ for any natural number $n$, and $n$ does not divide $95-0$ if $n$ is not a divisor of 95. Therefore, the statement in the problem holds only for a finite number of natural numbers $n$.

Let's show that the statement holds for $n=95$. Let $k, l \in \mathbb{Z}$ be such that $95 \mid P(k)-P(l)$. Since $95 \mid$ and we conclude that $95 \mid k^{5}-l^{5}$. We want to show that $95 \mid k-l$, so it is enough to show that $5 \mid k-l$ and $19 \mid k-l$.

By Fermat's Little Theorem, we have $k^{5} \equiv k(\bmod 5)$ and $l^{5} \equiv l(\bmod 5)$, so we conclude that $5 \mid k-l$ because $5 \mid k^{5}-l^{5}$.

Also, by Fermat's Little Theorem, if $k$ is coprime with 19, then $k^{18} \equiv 1 (\bmod 19)$, so $k^{54} = (k^{18})^{3} \equiv 1 (\bmod 19)$. Therefore, $k^{55} \equiv k (\bmod 19)$ regardless of whether $k$ is divisible by 19 or not. Similarly, $l^{55} \equiv l (\bmod 19)$. We conclude that $k \equiv (k^{5})^{11} \equiv (l^{5})^{11} \equiv l (\bmod 19)$, i.e., $19 \mid k-l$, which is what we needed to prove.

Ministry of Science, Education and Sports of the Republic of Croatia

Agency for Education and Upbringing

Croatian Mathematical Society

## CROATIAN MATHEMATICAL OLYMPIAD  final test for selecting the MEMO team

May 14, 2011.
15. Determine all sequences $a: \mathbb{N} \rightarrow \mathbb{N}$ such that for all $n \in \mathbb{N}$, the following holds:

$$
a_{n}+a_{n+1}=a_{n+2} a_{n+3}-200
$$

2. Let $n \geqslant 3$ be a natural number. Determine the minimum number of points that need to be marked inside any convex $n$-gon so that every triangle whose vertices are also vertices of that $n$-gon contains at least one marked point in its interior.
3. Inside an acute-angled triangle $A B C$, there is a point $S$ such that $\varangle S A B=\varangle S B C=\varangle S C A$. The lines $A S, B S, C S$ intersect the circumcircles of triangles $S B C, S C A$, $S A B$ at points $A_{1}, B_{1}, C_{1}$, respectively. Prove the inequality

$$
P\left(A_{1} C B\right)+P\left(B_{1} A C\right)+P\left(C_{1} B A\right) \geqslant 3 P(A B C)
$$

4. For a natural number $n$, consider the set

$$
S=\{0,1,1+2,1+2+3, \ldots, 1+2+3+\ldots+(n-1)\}
$$

a) If $n$ is a power of 2, prove that all elements of $S$ give different remainders when divided by $n$.

b) If $n$ is not a power of 2, prove that there exist two elements of $S$ that give the same remainder when divided by $n$.

## CROATIAN MATHEMATICAL OLYMPIAD  final test for selecting the MEMO team  Solutions to the problems