<s>[INST] Diophantine equation $(x^2-1)^2-4y^2=0$ 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?
 [/INST] 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.
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