Show that the equation $x^2+xy-y^2=3$ does not have integer solutions. Show that the equation $$x^2+xy-y^2=3$$ does not have integer solutions. 
I solved the equation for $x$:
$x=\displaystyle \frac{-y\pm\sqrt{y^2+4(y^2+3)}}{2}$
$\displaystyle =\frac{-y\pm\sqrt{5y^2+12}}{2}$
I was then trying to show that $\sqrt{5y^2+12}$ can not be an integer using $r^2\equiv 12 \pmod{5y^2}$.
I got stuck here.

Just note that $5y^2$ is a multiple of $5$ and hence it ends on $0$ or $5$. Thus, $5y^2+12$ ends in $2$ or $7$, but there are no perfect squares that have this endings. 
