<s>[INST] How to solve in integers the equation, $(x^2-y^2)^2=1+16y$? How to solve in integers the equation,
$$(x^2-y^2)^2=1+16y$$
By observation we can take $y=3,5$
Is there another method to solve the equation?
 [/INST] Notice that $y\ge 0$, $y\neq 1$ and $x \neq 0$. If $(x, y)$ is a solution then $(-x, y)$ is a solution too. Hence W.L.O.G $x>0$.
It seems that $x$ and $y$ cannot be too far from each other, otherwise LHS would become larger than RHS. We'll find out this. Suppose that $z=|x-y|$. Notice that $z\neq 0$. Now, we have  
$$z^2(x+y)^2=|x-y|^2 (x+y)^2=(x^2-y^2)^2=16y+1$$ 
And since $y\ge 0$, $y\neq 1$ and $x > 0$, then
$$z^2=\frac{16y+1}{(x+y)^2}<\frac{4(1+y)^2}{(1+y)^2}=4  \ \ \Longrightarrow \ \ z<2$$Therefore $z=1$. It follows that $x=y \pm 1$ and equation turns into a simple quadratic for $y$.
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