As we know, every Pythagorean prime $p$ is expressible as $p=x^2+y^2$ such that $x$ is an odd integer and $y$ is an even integer. Prove or disprove: $x$ is a quadratic residue modulo $p$.
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$\begingroup$ This seems to be true up until $p=149$. I don't have a computer with me right now, though. $\endgroup$– S.C.B.Commented Feb 5, 2017 at 12:31
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$\begingroup$ And 157. hmmmm.... $\endgroup$– Oscar LanziCommented Feb 5, 2017 at 12:47
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1 Answer
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It's true.
Divide $p$ by $x$ polynomially to get a perfect square remainder, $y^2$. Then $p$ is a quadratic resudue mod $q$ where $q$ is any odd prime factor of $x$. By quadratic reciprocity with $p$ a $4k+1$ prime, $q$ is a quadratic residue mod $p$. By multiplication $x$ is a quadratic residue mod $p$.
answered Feb 5, 2017 at 13:28