# 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?

$$2c-7\equiv2c-7+c^2+8\pmod p\equiv(c+1)^2$$

• @AJ. Have you noticed my other answer. That is much more natural derivation Mar 22 '19 at 9:10

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.

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 .

• Not sure about any mistake which has caused the down-vote ! This has been put in a separate post as the approach in the other markedly different. Mar 22 '19 at 9:21
• Me neither... $(+1)$ Mar 24 '19 at 6:12