# for $n=x^2+3y^2$ , $n=\prod p^{a(p)}$ , $a(p)$ is even for all $p \equiv 2 \pmod 3$ where $p$ is prime

I'd really like your help with the following Number Theory question:

I need to show that if I can write an integer $n=x^2+3y^2$ so in the factorization of $n$ to primes, every $p \equiv 2\pmod 3$ would be with a even power, I mean if $n=\prod p^{a(p)}$ so $a(p)$ is even for all $p \equiv 2 \pmod 3$, where $p$ is prime.

I don't rally know how to start this one.

Thank you.

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Hint: If $p\neq 3$ is a prime and $p|x^2+3y^2$ and $p\not\mid x$, $p\not\mid y$, then $-3$ is a square mod $p$.
Let me try to understand:I'll assume that $n=x^2+y^2$ for an odd power for $p$ with residue 2 modulo 3,then I'm dividing $x^2+3y^2$ with $p$, then I actually divided $n$ with $p^2$ so p's power is still odd. I'll keep doing that until either $y$ or $x$ wouldn't be divided by $p$, but $n$ does, so I'll get that $y$ is inverse modulp $p$ I have $n^*y'^2=(y'y^*)^2+(x^*y')^2=(y'y^*)^2+1=0 \pmod p$, now back to your hint I get there's a solution for $z^2=-3 \pmod p$ which yields that $p \equiv 1 \pmod 3$, and here a contradiction. Is this good enough? Thanks! – Jozef Jul 5 '12 at 15:51
Where did you get $x^2+y^2$? Did you mean $x^2+3y^2$? But otherwise, yes. – Thomas Andrews Jul 5 '12 at 15:59
You can think of this as a proof by descent. Assume there is a least $n=x^2+3y^2$ which is a counter-example. Then there is some $p\equiv 2\pmod 3$ which goes into $n$ an odd number of times. But then, by the above, $p|x$ and $p|y$, so $n/p^2 = (x/p)^2+3(y/p)^2$ is a smaller counter-example. – Thomas Andrews Jul 5 '12 at 16:08