Prove that there are no integers $x$ and $y$ such that $3x^2=13+4y^2$ 
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.
 A: You can look mod4.
Your equation is equivalent to
$1=3x^{2} \pmod 4$.
$3=x^{2}\pmod 4$. 
But 3 is not a square mod 4.
A: Square of an integer is always of the form $4k$ or $4k+1$.
$$\text{RHS} =4(y^2+3)+1=4n+1$$
While, on the other hand (left one! :P)
$$\text{LHS}=3(4m) ~\text{or} ~3(4m+1) \equiv 4l ~ \text{or} ~4l+3$$
LHS and RHS both leave different remainders while dividing with $4$, therefore they can never be equal.
A: 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.
A: $3x^2=\underbrace{13+4y^2}_\text{odd}$
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
A: Welp, you know that $x$ must be odd.  Let's chase our tail.
So let $x = 2k+1$ and
$12k^2 + 12k + 3 = 13 + 4y^2$
$12k^2 + 12k = 10 + 4y^2$.
$6k^2 + 6k = 5 + 2y^2$ which isn't possible.
