Infinite natural number to form $4n^2+1$ can be divided by $5$ and $13$

Question

Can someone proof with contradiction that there are infinite natural numbers $n$ that $4n^2+1$ can be divided by $5$ and $13$?

Answer 1

$4n^2 \equiv -1 \mod 65$ if and only if $4n^2\equiv 64 \mod 65$ if and only if $n^2\equiv 16 \mod 65$. Then for example any $n \equiv 4 \mod 65$ will give infinitely many solutions. I realize this is not a proof by contradiction, but then you could proceed as Daniel Fischer suggested in the comments, but as he indicated it is an odd thing to do.