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?
 A: 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 $y$.
A: We can prove the non-existence of solutions other than those listed in the comments by squeezing $x^2$ between two consecutive squares of integers in the remaining cases. In a sense the key is that unless $y$ is relatively small that $1+16y$ is very small in comparison to the left hand side. It turns out that more often than not it is too small to be the square of the difference between two consecutive squares, and that leads to a solution. Behold.

Clearly $y\ge0$ and if $(x,y)$ is a solution so is $(-x,y)$. So it suffices to find the solutions with $x\ge0, y\ge0$. Also, $x\neq y$.
If $x>y$ then $x^2-y^2>0$. Therefore
$$
x^2=y^2+\sqrt{1+16y}.
$$
For non-negative integers $y$ we have the upper bound $\sqrt{16y}=4\sqrt y\le 4y$ and consequently the inequality $\sqrt{1+16y}\le1+4y$. Therefore $$y^2+\sqrt{1+16y}\le y^2+4y+1<y^2+4y+4=(y+2)^2.$$ 
This estimate allows us to squeeze:
$$
y^2<x^2=y^2+\sqrt{1+16y}<(y+2)^2
$$
keading to the conclusion $x=y+1$. The equation
$$
(y+1)^2=y^2+\sqrt{1+16y}
$$
has two solutions, namely $y=0$ and $y=3$.
If $0\le x<y$, then we must have
$$
x^2=y^2-\sqrt{1+16y}.
$$
For the right hand side to be non-negative we need $y\ge3$. Assuming that we can estimate
$$
(4y-4)^2=16y^2-32y+16=16y(y-2)+16>16y+1,
$$
so $\sqrt{1+16y}<4y-4$ in the interesting range of values of $y$. Therefore
$$
y^2>x^2=y^2-\sqrt{1+16y}>y^2-(4y-4)=(y-2)^2.
$$
As above, this leaves $x=y-1$ as the only alternative. But the only solution of the equation
$$
(y-1)^2=y^2-\sqrt{1+16y}\Longleftrightarrow 2y-1=\sqrt{1+16y}
$$
is $y=5$.
A: 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 $x = \pm 4, y = 3$.
