<s>[INST] 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?
 [/INST] Hint:
As $c^2\equiv-8\pmod p,$
$$c(c+1)=c^2+c\equiv c-8\pmod p$$
$c(c+1)(c-8)\equiv?$
I believe this is how the problem naturally came into being . 
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