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Suppose odd prime $p=a^2+b^2$, and $a$ is odd and $b$ is even. Prove that if $b\equiv2\pmod4$, then $\left(\dfrac bp\right)=-1$ and if $b\equiv0\pmod4$, then $\left(\dfrac bp\right)=1$.

What I have got is that $p\equiv 1 \pmod4$. I have tried factoring $b$ and splitting up the Legendre symbols. However, I am not sure this gets anywhere because you cannot say much about those. I have also observed that $b \equiv 0 \pmod 4$ when $p\equiv 1 \pmod 8$ and $b \equiv 2 \pmod 4$ when $p\equiv 5 \pmod 8$ if that helps.

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  • $\begingroup$ @twnly so "for each $q | b, p $ is a square $\bmod q$, then use quadratic reciprocity on $\prod_j (\frac{q_j}{p})^{e_j}$ $\endgroup$ – reuns Dec 1 '18 at 6:27
  • $\begingroup$ @reuns is right, my proposed solution has some errors. I deleted them. $\endgroup$ – twnly Dec 1 '18 at 6:27
  • $\begingroup$ is there some reason to expect there is no much simpler solution ? $\endgroup$ – reuns Dec 1 '18 at 6:31
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Write $b=2^rc$ with $c$ odd. Then $p=a^2+b^2\equiv a^2\pmod c$ and so $\left(\frac pc\right)=1$ (this is a Jacobi symbol). As $p\equiv1 \pmod 4$ then $\left(\frac cp\right)=1$ (quadratic reciprocity for Jacobi symbols).

When $p\equiv1\pmod8$ then $\left(\frac 2p\right)=1$ and so $\left(\frac bp\right)=\left(\frac 2p\right)^r\left(\frac cp\right)=1$. When $p\equiv5\pmod8$ then $\left(\frac 2p\right)=-1$ and also $r=1$. Therefore $\left(\frac bp\right)=\left(\frac 2p\right)\left(\frac cp\right)=-1$.

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