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

I need some help with this problem

  • 6
    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 $(-2/p)$.2012-11-28

1 Answers 1

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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}. $