11-00-ResultOnQuadraticResidues.tex

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\pmtitle{result on quadratic residues}
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\begin{document}
\textbf{Theorem.}
Let $p$ be an odd prime. Then $-3$ is a quadratic residue modulo $p$ if and only if $p \equiv 1\pmod{3}$.

\emph{Proof.}
Preliminary to the proof, we remark first that $-1$ is a quadratic residue modulo $p$, where $p$ is an odd prime, if and only if $p \equiv 1\pmod{4}$.

If $p \equiv 1\pmod{4}$ then
\[\left(\frac{-3}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{3}{p}\right) = \left(\frac{p}{3}\right).\]
Now if $p \equiv 3\pmod{4}$, then 
\[ \left(\frac{-3}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{3}{p}\right) = (-1)(-1)\left(\frac{p}{3}\right) = \left(\frac{p}{3}\right).\] 
Thus, $\displaystyle \left(\frac{-3}{p}\right) = \left(\frac{p}{3}\right)$, and $\displaystyle \left(\frac{p}{3}\right)=1$ if and only if $p \equiv 1\pmod{3}$. $\Box$


 
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