Prove that there are infinitely many integers n such that $4n^2+1$ is divisible by both $13$ and $5$ 
Prove that there are infinitely many integers n such that $4n^2+1$ is divisible by both $13$ and $5$.

This can be written as:
$$65k = (2n)^2 + 1$$
It's clear that $k$ will always be odd.
Now I am stuck. I wrote a program to find the first few $n$'s. They are $4, 9$.
$n$ can only end with $4, 6, 8, 9$ if I'm correct in my deductions.
I have made no further progress. Please help me find the solution. Thanks.

As soon as you find a single $n$ such that $4n^2+1\equiv0$ mod  $65$, then $4m^2+1\equiv0$ mod $65$ for all $m\equiv n$ mod $65$. And it's easy to see that $n=4$ is such a value.
Remark: This answer differs from robjohn's answer mainly in presentation. The key idea, from modular arithmetic, is that a residue class is an infinite set of integers. Note, we have not found all numbers $n$ such that $65$ divides $4n^2+1$, just one infinite family. But that's all the problem asks for.
