Could anyone tell me how to show $x^2+3y^2=n$ has an integral solution iff the prime factor of $n$ which are of the form $3k-1$ has even exponents?
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2What have you tried so far? – Adina Goldberg Jul 28 '20 at 20:25
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Well, do you know the Legendre symbol and quadratic reciprocity??? – Will Jagy Jul 28 '20 at 20:29
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https://en.wikipedia.org/wiki/Legendre_symbol – Will Jagy Jul 28 '20 at 20:32
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The key idea is that $,x^2+3y^2=(x+\sqrt{-3}y)(x-\sqrt-{3}y),$ and you factor elements in the ring of "numbers" the form $,x+\sqrt{-3}y.,$ This is similar to the Gaussian integers. – Somos Jul 28 '20 at 21:31
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1Every prime $p\equiv 1\pmod{3}$ can be written as $x^2+3y^2$ (proof: https://math.stackexchange.com/a/2300356/44121) and since $(x^2+3y^2)(X^2+3Y^2)=(xX+3yY)^2+3(Xy-Yx)^2$ every number with prime factors of the form $3k+1$ can be written as $x^2+3y^2$. If $x$ and $y$ are coprime and we assume that $p=3k-1$ is a divisor of $n=x^2+3y^2$, we get that $z^2+3\equiv 0\pmod{p}$ has a solution, which is absurd since $-3$ is a non-quadratic residue $\pmod{p}$. – Jack D'Aurizio Jul 29 '20 at 10:51