, 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 $\ell \equiv 2 \pmod{p_1 \cdot p_2 \cdots p_k}$. 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 $a_m \equiv a_{m+p} \pmod{p}$ for some index $m$, then $a_\ell \equiv a_{\ell+p} \pmod{p}$ 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
$a_1, a_2, a_3, a_4 \equiv 1, 2, 2, 1 \pmod 3$ so $m=1, p=3$ holds.
$a_1, a_2, a_3, a_4, a_5, a_6 \equiv 1, 2, 3, 3, 1, 1 \pmod 5$ so $m=1, p=5$ holds.
$a_1, a_2, \dots, a_{25} \equiv 1, 2, 8, 12, 2, 14, 14, 9, 1, 10, 10, 1, 13, 15, 17, 12, 13, 10, 14, 6, 4, 5, 2, 12, 14 \pmod{19}$ so $m=6, p=19$ holds. We are done.