If $x$ and $y$ are integers, then solve (using elementary methods)
$$3x^2-y^2=2$$
I tried the following
If $y$ is even, then $4|y^2$ and hence $2|y^2+2$ (and $4$ doesn't divide it), but $3x^2=y^2+2$ and since for R.H.S. to be even, $2|x^2 \implies 4|x^2$, and we get a contradiction. So $x$ and $y$ both are odd. If I substitute $x=2a+1$ and $y=2b+1$ for some integers $a$ and $b$, I get a more tedious equation and can't solve further. Please Help!
Thanks!
$$y = ±1/2 ((2-\sqrt3)^n+\sqrt3 (2-\sqrt3)^n+(2+\sqrt3)^n-\sqrt3 (2+\sqrt3)^n)$$ $$x = ±1/6 (3 (2-\sqrt3)^n+\sqrt3 (2-\sqrt3)^n+3 (2+\sqrt3)^n-\sqrt3 (2+\sqrt3)^n)$$
– usethedeathstar Oct 28 '14 at 12:26