TITLE: Prove that there are no integers $x$ and $y$ such that $3x^2=13+4y^2$
QUESTION [7 upvotes]: 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.

REPLY [3 votes]: Ignore $4y^2$ for a minute.
The equation $3x^2 = 13$ has no solutions in integers either. If it had a solution, then $x$ would have to be odd, obviously. Then $3x^2 \equiv 3 \bmod 4$ but $13 \equiv 1 \bmod 4$.
Since $4y^2 \equiv 0 \bmod 4$, the equation $3x^2 = 4y^2 + 13$ is likewise hopelessly insoluble (is that the right word? you know what I mean) in integers.