In another thread I have ended up with this equation

$\displaystyle y=\dfrac{(3v_1+u_1)\pm\sqrt{u_1^2+6u_1v_1+v_1^2}}{ 2}$

I need y to be an integer <> 0, so the first requirement is that

$\displaystyle u_1^2+6u_1v_1+v_1^2$ be a perfect square

If I go at it like this

$\displaystyle x^2=u^2+6uv+v^2$

$\displaystyle x^2=(u+v)^2+4uv$

$\displaystyle x^2-(u+v)^2=4uv$

$\displaystyle (x+(u+v) )(x-(u+v) )=4uv$

$\displaystyle let\ \ \ \ \ x+(u+v)=r$

$\displaystyle and\ \ \ \ x-(u+v)=s$

$\displaystyle x=(r+s)/2$

$\displaystyle u+v=(r-s)/2$

$\displaystyle uv=rs/4$

But I have no individual values of $\displaystyle u$ and $\displaystyle v$ now so cannot calculate

$\displaystyle 3v_1+u_1$

Anyone have any other ideas how I can ensure $\displaystyle u_1^2+6u_1v_1+v_1^2$ is a perfect square?