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Let $p$ be an odd prime $p > 3$. What is $\dfrac{-2}{p}$?

I need some help with this problem

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If this is from an assignment, probably $(2/p)$ was already done in class, and you know about $(-1/p)$. You can combine these two pieces of knowledge to get to $(-2/p)$. –  André Nicolas Nov 28 '12 at 20:52

1 Answer 1

This is a special case that is proved separately.

In the two cases $$p \equiv 1,3 \pmod 8, \; \; \; (-2 | p) = 1.$$

In the two cases $$p \equiv 5,7 \pmod 8, \; \; \; (-2 | p) = -1.$$

This has a concrete shape: whenever $$p \equiv 1,3 \pmod 8, \; \; \exists \; u,v \in \mathbb Z : u^2 + 2 v^2 = p.$$ These give the basic examples for alternate Pythagorean triples, $$ (u^2 - 2 v^2)^2 + 2 (2 uv)^2 = p^2. $$

Then again, whenever $p \equiv 5,7 \pmod 8,$ if $$ x^2 + 2 y^2 \equiv 0 \pmod p, \; \mbox{then} \; x,y \equiv 0 \pmod p, \; \mbox{and} \; x^2 + 2 y^2 \equiv 0 \pmod {p^2}. $$

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