I'm reading Cox's "Primes of the form $x^2+ny^2$", and Lemma $2.5$ reads: "...Conversely, suppose $D=b^2$ mod $m$. Since $m$ is odd, we can assume that $D$ and $b$ have the same parity (replace $b$ by $b+m$ if necessary) and then $D=0,1$ mod $4$ implies $D=b^2$ mod $4m$." Can someone explain how he arrives at $D=b^2$ mod $4m$? Thanks in advance!
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2$D\equiv0,1\pmod 4$ implies $D\equiv b^2 \pmod 4$ since $D, b$ have the same parity. Now apply Chinese Remainder Theorem with $D\equiv b^2 \pmod m$. – player3236 Jun 24 '21 at 14:42
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@player3236 Thank you! – JBuck Jun 24 '21 at 14:54