$p|(20a^2+16b^2+2559)$

Question

Show that for a prime $p \geq 7$, there always exist integers $a, b$ such that $p|(20a^2+16b^2+2559)$

I still cannot figure out how to solve this problem.

My thought :

$x^2 \equiv -1(\bmod p) \rightarrow x^4 \equiv 1(\bmod p)$ ---[1]

By Fermat's little theorem, $ x^{p-1} \equiv 1(\bmod p)$ ---[2]

From[1],[2], $4|p-1 \rightarrow p \equiv 1(\bmod 4)$

Answer 1

You want to show there are $a$, $b$ with $$20a^2\equiv-16b^2-2559\pmod p.$$ When $p\ge7$, $20a^2$ takes $\frac12(p+1)$ distinct values modulo $p$. So does $-16b^2-2550$....