Question 62
Let $(a_n)_{n=1}^\infty$ be a sequence of integers such that $a_1 = -5$, $a_2 = -6$ and
$$
\begin{aligned}
a_{n+1} = a_n &+ (a_1+1)(2a_2+1)(3a_3+1)\cdots\\
&\cdot((n-1)a_{n-1}+1)((n^2+n)a_n+2n+1)
\end{aligned}
$$
for all integers $n \ge 2$. Prove that if prime number $p$ divides $n a_n + 1$ for some positive integer $n$, then there exists an integer $m$ such that $m^2 \equiv 5 \pmod{p}$.
$$
\begin{aligned}
a_{n+1} = a_n &+ (a_1+1)(2a_2+1)(3a_3+1)\cdots\\
&\cdot((n-1)a_{n-1}+1)((n^2+n)a_n+2n+1)
\end{aligned}
$$
for all integers $n \ge 2$. Prove that if prime number $p$ divides $n a_n + 1$ for some positive integer $n$, then there exists an integer $m$ such that $m^2 \equiv 5 \pmod{p}$.