Solve $3x^2-y^2=2$ for Integers 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!


For the equation: $$3y^2-x^2=2$$
Knowing the first solution: $(y,x) - (3;5)$
The equation to find the rest:
$$y_2=x+2y$$
$$x_2=2x+3y$$
