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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?

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

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

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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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    $\begingroup$ 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. $\endgroup$ Mar 22 '19 at 9:21
  • $\begingroup$ Me neither... $(+1)$ $\endgroup$
    – Mr Pie
    Mar 24 '19 at 6:12

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