Q:

How can I prove this? For every integer m with 2 divides m and 4 does not divide m, there are no integers x and y that satisfy x^2+3y^2=m.

I know that m is even and m/2 is odd, but I don't know where/how I can use this. Also, 3y^2 is odd and the sum is odd when x^2 is even. I'm trying to prove that its always odd, but I'm stuck.
Can someone please help?
Thanks

A:

Suppose : $m=x^2+3y^2$ is even. 
Case $1$ : $x$ is even. Then $y$ must be even as well. So, $m$ is divisble by $4$.
Case $2$ : $x$ is odd. Then, $\ x^2\equiv 1 \mod 4\ $. Since $y$ must be odd as well, we also have $\ y^2\equiv 1\mod 4\ $. Hence, $x^2+3y^2\equiv 0\mod 4$. 
So, again, $m$ is divisble by $4$

