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This problem also had me prove that $(\frac pq) = (\frac aq)$, but I've already managed to do that. I've tried messing around with the Law of Quadratic Reciprocity but can't get anything. I've also tried using the first identity by proving that $(\frac pq) = (\frac ap)$, but again I've got nothing.

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marked as duplicate by no identity, Old John, Thomas Andrews, Daniel Rust, EuYu Dec 16 '13 at 22:44

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@DietrichBurde, this version seems to have an easier proof. –  Barry Cipra Dec 16 '13 at 21:21
    
I disagree with the votes to close. Yes, this question is a corollary of the other question, but it's a special case, not a duplicate -- and in this case it's a special case with a simpler proof. –  Barry Cipra Dec 16 '13 at 21:35

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If $p=q+4a$, then $p\equiv q$ mod $4$, in which case the Law of Quadratic Reciprocity says

$$\left({p\over q}\right)=\left({-q\over p}\right)$$ But this gives

$$\left({a\over q}\right)=\left({q+4a\over q}\right)=\left({p\over q}\right)=\left({-q\over p}\right)=\left({4a-p\over p}\right)=\left({a\over p}\right)$$

(using the fact that $4$ is a square, hence a quadratic residue for both $p$ and $q$).

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But $(\dfrac {p}{q}) = (\dfrac {-q}{p})$ only if $p \equiv q \equiv 3 \space (mod \space 4).$ If $p \equiv q \equiv 1 \space (mod \space 4)$ then $(\dfrac {p}{q}) = (\dfrac {q}{p})$. –  clocks Dec 16 '13 at 21:26
    
@clocks, if $p\equiv q\equiv1$ mod $4$, then $({-1\over p})=({-1\over q})=1$, so the equality I asserted holds in both cases. –  Barry Cipra Dec 16 '13 at 21:29

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