Diophantine equation $(x^2-1)^2-4y^2=0$

Question

We have the diophantine equation $(x^2-1)^2-4y^2=0$, where $x$ and $y$ are positive integers. Is the only solution $x=1$, $y=0$ or can there be infinitely many solutions?

Answer 1

Note that $$\begin{align}(x^2-1)^2-4y^2=0&\iff (x^2-1-2y)(x^2-1+2y)=0\\&\iff x^2-1-2y=0\ \ \text{or}\ \ x^2-1+2y=0.\end{align}$$

So, considering $x^2-1-2y=0\iff x^2-1=2y$ gives us, for example, $$(x,y)=(2a+1,2a^2+2a)$$ for positive integers $a$.

Hence, we know that there are infinitely many solutions.