Is this solution correct for equation $3x^2-4y^2=13$? Prove equation $3x^2-4y^2=13$ has no integer solution.
Solution: Suppose (y, 3)=1, We have:
$y≡( 1, 2) \mod (3)$
⇒ $4y^2≡( 1, 2) \ mod(3)$
$3x^2≡0 \mod (3)$
The common remainder between $3x^2$ and $4y^2$ is 0 , so we may write:
⇒ $3x^2-4y^2≡ 0 \ mod(3)$
But, $13≡1 \mod (3)$
Hence this equation has no integer solution. Similar result comes out when we consider remainder of nomials on 4.
Is this solution correct? 
 A: If $3 \mid y$, then $3 \mid -4y^2$ and since $3 \mid 3x^2$, you would have $3 \mid 13$, which is not true. Thus, as you stated, $3 \not\mid y$, which means $y^2 \equiv 1 \pmod 3$.
As such, $3x^2 - 4y^2 \equiv -y^2 \equiv 2 \pmod 3$. Note, you made a mistake when you stated "common remainder between $3x^2$ and $4y^2$ is $0$" (I'm not quite sure how you determined this) giving $3x^2 - 4y^2 \equiv 0 \pmod 3$. Also, $4y^2 \equiv 1 \pmod 3$ only, i.e., $4y^2 \not\equiv 2 \pmod 3$, which you indicated is a possible solution. Regardless, you have $13 \equiv 1 \pmod 3$. Since this doesn't match the left hand side, it means there are no integer solutions to $3x^2 - 4y^2 = 13$. This is basically what J. W. Tanner's question comment states.
As you state, you can also prove there are no solutions by using modulo $4$ instead. In this case, $4 \not\mid x$ (since, otherwise, you would need $4 \mid 13$ which is not true), so $x^2 \equiv 1 \pmod 4 \implies 3x^2 \equiv 3 \pmod 4$, so $3x^2 - 4y^2 \equiv 3 \pmod 4$, but $13 \equiv 1 \pmod 4$. As these congruence values don't match, it means there's no integer solution.
A: I think you are trying to say:
If you assume $\gcd(y,3) =1$ (Why are you assuming this?  What if $\gcd(y,3) = 3$?) or in other words if $3\not\mid y$ then $y\equiv 1 \pmod 3$ or $y\equiv 2 \pmod 3$.
ANd therefore $y^2\equiv 1 \pmod 3$ or $y^2\equiv 4\equiv 1\pmod 3$. and so $4y^2 \equiv 1\pmod 3$.  I'm not sure why you didn't eliminate $4y^2 \equiv 2\pmod 3$.
And $3x^2 \equiv 0\pmod 3$.
Then you said "The common remainder between $3x^2$ and $4y^2$ is $0$" and I can not figure out what you mean and I can't think of any interpretation that would be compatible with you discovery that $3x^2 \not \equiv 4y^2$.
You claim $3x^2 - 4y^2 \equiv 0\pmod 3$ which is completely wrong as $4y^2 \equiv 1\pmod 3$ we have $3x^2 - 4y^2 \equiv -1 \equiv 2 \pmod 3$.
Which is enough to show $13\equiv 1\not \equiv 2\pmod 3$.  So there is not integer solution.
IF we assume $\gcd(3,y)=1$.  But you didn't explain why we are assuming $\gcd(3,y) = 1$.
Which you never considered.
Better to simply note:
$3x^2 - 4y^2 \equiv \begin{cases}3*0^2\equiv 0& x\equiv 0 \\3*1^2\equiv3\equiv 0 &x\equiv 1\pmod 3\\3*2^2\equiv 3*4\equiv 0&x\equiv 2\pmod 3\end{cases}-\begin{cases}4*0^2\equiv 0& y\equiv 0\pmod 3\\4*1^2\equiv 1&y\equiv 1\pmod 3\\4*2^2\equiv 16\equiv 1&y\equiv 2\pmod 3\end{cases}\pmod 3\equiv$
$0 - \begin{cases}0& \text{if }y\equiv 0\pmod 3\\1&\text{otherwise}\end{cases}\equiv$
$\begin{cases}0\\-1\end{cases}\equiv \begin{cases}0\\2\end{cases}\pmod 3$
$\not \equiv 1 \equiv 13 \pmod 3$.
And it's enough to prove it for just modulo $3$.
FWIW if you tried modulo $4$ you'd get.
$3x^2 - 4y^2 \equiv 3x^2\equiv 3*\begin{cases}0^2\\1^2\\2^2\\3^2\end{cases}\equiv$
$\begin{cases}0\\3\\0\\3\end{cases}\not \equiv 1\equiv 13\pmod 4$.
But there was utterly no reason to do those as we were done.
A: Here is another method:
$3x^2-4y^2=13$
Clearly x must be odd; let $x=2k+1$, then:
$3x^2-4y^2=12k^2+12k+3-4y^2=4(3k^2+3k-y^2)=10$
⇒$2(3k^2+3k-y^2)=5$
LHS is even but RHS is odd, so the equation can not have integer solution.
