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? 

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.
