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This question already has an answer here:

I ran in to this problem today, and I seem to have trouble.

Suppose $p$ is prime, $p = 1\mod4$, and that $a^2+b^2=p$ with $a$ odd and positive. Show that $(\frac{a}{p}) = 1$.

How can I show that this holds true?

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marked as duplicate by Dietrich Burde, Did, Namaste, Leucippus, Shailesh Nov 5 '17 at 1:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ What do you get from quadratic reciprocity applied to each $q\ |\ a$ ? $\endgroup$ – reuns Nov 4 '17 at 14:58
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We have the following chain of equalities of Jacobi symbols, where the first equality follows from quadratic reciprocity ($p\equiv 1\pmod{4}$):

$$ \left(\frac{a}{p}\right) = \left(\frac{p}{a}\right)= \left(\frac{a^2+b^2}{a}\right) = \left(\frac{b^2}{a}\right) = 1 $$ The LHS is also a Legendre symbol, hence $a$ is a quadratic residue $\!\!\pmod{p}$.

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