Prove that there are no integers $x$ and $y$ such that $3x^2=13+4y^2$.
From the equation, I know that $3x^2$ must be odd and therefore equal $2k + 1$ for some integer $k$.
But I am unsure what to do after that.
I have also worked out that $k = 2(y^2 + 3)$, but I don't know if that helps at all.
My instructor noted that I should look at whether $x^2$ is even or odd, but I am at a loss.
Best Answer
$3x^2=\underbrace{13+4y^2}_\text{odd}$
We know that x is odd lets' pose $x=2k+1$
\begin{align}3(2k+1)^2=13+4y^2\\ 3(4k^2+4k+1)=13+4y^2\\ 3(4k^2+4k)=10+4y^2\\ 12(k^2+k)=10+4y^2\\ \underbrace{6(k^2+k)}_\text{even}=\underbrace{5+2y^2}_\text{odd} \end{align}
but $5+2y^2$ is odd not even which contradict the factor 6 on the left side of the equation