<s>[INST] 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? 
 [/INST] 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.
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