<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] 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&=&1+16y\\ &=& 1\textrm{ mod }16 \qquad \textrm{(means when you divide a number by 16, remainder is 1)}\\x^2-y^2 &=& \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 $x = \pm 4, y = 3$.
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