17th Junior Turkish Mathematical Olympiad

Since $x^2+p^2{y}^2\equiv 0(\mathrm{mod}3)$, if $p\ne 3$ then $x^2+y^2\equiv 0(\mathrm{mod}3)$, and consequently $x\equiv y\equiv 0(\mathrm{mod}3)$. Then $9\mid x^2-3xy+p{y}^2$ but $9\nmid 12$, a contradiction. Thus, $p=3$ and $x^2-3xy+9y^2=36$. Therefore, $3\mid x$. $x=3k$, and we get a second order equation $k^2-ky+y^2-4=0$. The discriminant $\Delta =16-3y^2=m^2$. Therefore, $y^2=0$ or $y^2=4$. Solutions are: $(-6,0,3)$, $(0,-2,3)$, $(0,2,3)$, $(6,-2,3)$, $(6,0,3)$, $(6,2,3)$.