what r the integer solutions( if any) to this equation
$\displaystyle x^2-18y^2=12$
Show that $\displaystyle x^2$, and hence also $\displaystyle x$, is a multiple of 3, say $\displaystyle x=3z$. Then $\displaystyle 3z^2-6y^2 = 4$. But the left side is a multiple of 3 and the right side is not ... .