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.

Okay.  The simplest way to "see" this is to realize that $n = 65m + k$ will yield $4(65m + k)^2 + 1 = 65^2*4m^2 + 65*2*4mk + 4k^2 + 1 = 65*M + (4k^2 + 1)$ which is divisible by 65 whenever $4k^2 + 1$ is divisible by 65.
Setting $4k^2 + 1 = 65$ we get $k = 4$ so any $n=65m + 4$ will yield $4n^2 + 1$ is divisible by 65.
Thus for $m = 0,1,2...$ we have $n = 4, 69, 134....$
