Square factors of $a^2+1$

Question

I am trying to find $a$ such that $a^2+1$ is not squarefree, meaning that none of its factors are square numbers. I found that this is the case for $a=7$ but can't think of a way to approach this problem more generally.

Answer 1

For any prime $p\equiv 1\pmod{4}$ you have that $a^2+1\equiv 0\pmod{p}$ has two opposite solutions $\pmod{p}$, and they can be lifted to solutions $\!\!\pmod{p^2}$. For instance, by picking $p=13$ we have that the square roots of $-1$ in $\mathbb{F}_p$ are $5$ and $8$. Let us see if for some $k\in\mathbb{Z}$ $$ (13k+5)^2\equiv -1\pmod{13^2} $$ holds. Of course, since the previous equation is equivalent to $$ 10k+2 \equiv 0\pmod{13} $$ which is solved by $k\equiv 5\pmod{13}$.
It follows that $(5\cdot 13+5)^2+1=70^2+1$ is a multiple of $13^2$.