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For which prime numbers $p$ there exists $x,y\in \Bbb{Z}$ such that $p=x^2+2y^2$? I guess I am to use continued fraction, but I am not sure how. I know how to find solutions for defined numbers but I can't find sets of $p$ that satisfy the above. It will be much appreciated if you have any lead or hint as for how to do it.

Meitar
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    Since the class number of $\mathbb{Q}(\sqrt{-2})$ is one, any prime for which $\left(\frac{2}{p}\right)=+1$, i.e. any prime $p\equiv \pm 1\pmod{8}$, can be represented in such a way. You can prove this fact also by Fermat's descent, by exploting the set of $a^2+2b^2$ being a semigroup. – Jack D'Aurizio Jun 13 '15 at 17:27
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    A good reference is "Cox, primes of the form $x^2+ny^2$." – Jack D'Aurizio Jun 13 '15 at 17:27
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    see here http://math.stackexchange.com/questions/236087/numbers-representable-as-x2-2y2 – Dr. Sonnhard Graubner Jun 13 '15 at 17:32
  • @JackD'Aurizio, it is actually $(-2|p) = 1$ and primes $p \equiv 1,3 \pmod 8.$ You did $x^2 - 2 y^2;$ there may have been an edit I cannot see, of course. – Will Jagy Jun 13 '15 at 20:24
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    @WillJagy: no, you are right, I messed up a little. It is $\left(\frac{-2}{p}\right)=+1$ and $p\equiv 1,3\pmod{8}$, true. – Jack D'Aurizio Jun 13 '15 at 20:30

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