
Problem 2. Let $\left\{a_{n}\right\}_{n=1}^{\infty}$ be a sequence of integer numbers such that

$$
(n-1) a_{n+1}=(n+1) a_{n}-2(n-1)
$$

for any $n \geq 1$. If 2000 divides $a_{1999}$, find the smallest $n \geq 2$ such that 2000 divides $a_{n}$. Oleg Mushkarov, Nikolai Nikolov


Solution: It is obvious that $a_{1}=0$ and $a_{n+1}=\frac{n+1}{n-1} a_{n}-2$ when $n \geq 2$. Therefore the sequence is uniquely determined by its second term. Furthermore the sequence $a_{n}=(n-1)(c n+2)$ (where $c=$ $\frac{a_{2}}{2}-1$ is an arbitrary real number) satisfies the equality from the conditions of the problem. We conclude now that all sequences which satisfy this equality are of the kind given above. Since all terms are integer numbers and 2000 divides $a_{1999}$, it is easy to see that $c$ is integer and $c=1000 m+2$. Therefore 2000 divides $a_{n}$ if and only if

1000 divides $(n-1)(n+1)$. Thus $n=2 k+1$ and $k(k+1)$ is divisible by $250=5^{3} \cdot 2$. Since $k$ and $k+1$ are relatively prime, we get that the smallest $n \geq 2$ equals $2 \cdot 124+1=249$.
