Team Selection Test for IMO 2019

a. By putting $b_1=1$ and $b_{n+1}=a_n^2+n{a}_n$ for all $n\ge 1$ we get $$a_{n+2}-a_{n+1}=a_{n+1}^2+(n+1)a_{n+1}-a_n^2-n{a}_n=b_{n+2}-b_{n+1}.$$ We also have $a_1=b_1=1$, $a_2=b_2=2$. Hence, we conclude that $a_n=b_n$ for all $n\ge 1$. Then we get $a_{n+1}=b_{n+1}=a_n(a_n+n)$ for all $n\ge 1$.

Assume that the set of all prime numbers dividing at least one element of the sequence are finite. Denote these primes by $p_1,p_2,\dots ,p_k$. Since $a_n\mid a_{n+1}$, we have $a_n\mid a_m$ for all $m\ge n$. This means that if $p_i\mid a_n$ for some $i,n$, then $p_i\mid a_m$ for all $m\ge n$. Therefore, there exists an index $N$ for which $p_1\cdot p_2\cdots p_k\mid a_n$ for all $n>N$.

Take an index $\ell$ satisfying $\ell >N+1$ and $\ell \equiv 2(\mathrm{mod}{p}_1\cdot p_2\cdots p_k)$. In this case, we get $a_{\ell }=a_{\ell -1}(a_{\ell -1}+\ell -1)$ and the expression $a_{\ell -1}+\ell -1$ is not divisible by $p_i$ for all $i=1,2,\dots ,k$. It is clear that this expression is larger than 1 and hence $a_{\ell }$ should have a prime divisor different from $p_i$'s, which is a contradiction.

b. We will show that the primes 3, 5, 19 do not divide any term of $(a_n)$. We use the fact that for a prime number $p$ if $a_m\equiv a_{m+p}(\mathrm{mod}p)$ for some index $m$, then $a_{\ell }\equiv a_{\ell +p}(\mathrm{mod}p)$ for all $\ell \ge m$. Hence, after the index $m$, the sequence becomes periodic modulo $p$. Hence, if for some index $m$, $p\nmid a_n$ for all $n<m+p$, then we conclude that $p\nmid a_n$ for all $n$.

Let us consider the sequence $(a_n)$ modulo 3, 5 and 19. We get $a_1,a_2,a_3,a_4\equiv 1,2,2,1(\mathrm{mod}3)$ so $m=1,p=3$ holds. $a_1,a_2,a_3,a_4,a_5,a_6\equiv 1,2,3,3,1,1(\mathrm{mod}5)$ so $m=1,p=5$ holds. $a_1,a_2,\dots ,a_{25}\equiv 1,2,8,12,2,14,14,9,1,10,10,1,13,15,17,12,13,10,14,6,4,5,2,12,14(\mathrm{mod}19)$ so $m=6,p=19$ holds. We are done.