Prove: if $c^2+8 \equiv 0$ mod $p$ then $c^3-7c^2-8c$ is a quadratic residue mod $p$. I want to show:

If $c^2+8 \equiv 0$ mod $p$ for prime $p>3$, then $c^3-7c^2-8c$ is a quadratic residue mod $p$.

I have calculated that $c^3-7c^2-8c \equiv -7c^2-16c \equiv 56- 16c \equiv 8(7-2c) \equiv c^2 (2c -7)$, so it should be enough to see that $2c -7$ is a quadratic residue. What now?

We have that 
$$(c +1)^2 = c^2 + 2c + 1 = c^2 + 2c + 8 -7 = (c^2 + 8) + (2c - 7) \equiv 2c - 7 \mod p.$$
This proves the claim.
