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.

$$4n^2+1\equiv0\pmod{13}\iff4n^2\equiv-1+65$$
As $(4,13)=1,4n^2\equiv64\iff n^2\equiv16\iff n\equiv\pm4\pmod{13} $
Similarly, $n\equiv\pm4\pmod5$
Now use Chinese Remainder Theorem
