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.

Note that
$$x^2+xy-y^2=(x-2y)^2+5(xy-y^2)=(x-2y)^2\qquad({\rm mod}\>5)\ .$$
But 
$$0^2=0,\quad(\pm1)^2=1,\quad(\pm2)^2=-1\qquad({\rm mod}\>5)\ ,$$
which implies that $3$ is not a quadratic residue modulo $5$.
