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?
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 $x&gt;0$.

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 $x &gt; 0$, then

$$z^2=\frac{16y+1}{(x+y)^2}&lt;\frac{4(1+y)^2}{(1+y)^2}=4  \ \ \Longrightarrow \ \ z&lt;2$$Therefore $z=1$. It follows that $x=y \pm 1$ and equation turns into a simple quadratic for $y$.
The integers solution I found are: $(\pm1,0),(\pm4,3),(\pm4,5)$.
Here's how I did it:
\begin{eqnarray}
(x^2-y^2)^2&amp;=&amp;1+16y\\ &amp;=&amp; 1\textrm{ mod }16 \qquad \textrm{(means when you divide a number by 16, remainder is 1)}\\x^2-y^2 &amp;=&amp; \pm\sqrt{1\textrm{ mod }16},
\end{eqnarray}
The only values of $y$ that give an integer square root for the above are 0,3 and 5.
For example, $y=3$ works because, $16\times3 + 1 = 49, \pm\sqrt{49} = \pm7$.
Now, $x^2-3^2 = \pm 7 \implies x^2 = \pm7+9 \implies x = \pm4$ or $x = \pm \sqrt{2}$. But since we are ol,y interested in integer solutions we only take the integer value of $x$. Hence, one possible integer solution is $x = \pm 4, y = 3$.