Show that $x^2+2y^2=p$ has a solution in $\mathbb{Z}\;$ if and only if $\;p \equiv 1 \; \text{or} \; 3 \mod 8$.
Can someone help on this. Thnx.
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1Relevant: The sum of a square and twice a square – MJD Jun 08 '13 at 14:17
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Have you tried showing each direction $\Rightarrow$ and $\Leftarrow$ in turn? Which direction are you stuck on? It would also help to know what you've tried, and what previous knowledge you have. – Caleb Stanford Jun 08 '13 at 14:22
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I proved that if $p=x^2+y^2$ then $(p)=(x+iy)(x-iy)$ in $\mathbb{Z}[i]$ so $p \equiv 1 \mod 4$ – Vali Jun 08 '13 at 14:37
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2@Vali You can do this question similarly but you'll need to work in $\mathbb{Z}[\sqrt{-2}]$ – Cocopuffs Jun 08 '13 at 14:41
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One direction: if $x^2+2y^2=p$ then $x^2\equiv -2y^2\pmod{p}$. But $\gcd(y,p)=1$ so $(x/y)^2\equiv -2\pmod{p}$. Hence $1=\left(\frac{-2}{p}\right)=\left(\frac{2}{p}\right)\left(\frac{-1}{p}\right)=(-1)^{\frac{(p+1)(p-1)}{8}}(-1)^{\frac{p-1}{2}}$ which implies $p\equiv 1,3\pmod{8}$.
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