Find all integer solutions to the equation $$x^5 - 2x^3y^2 + xy^4 = y^2.$$
I wasn't quite sure how to start on this problem, as factoring out $y^2$ would likely give me a headache and mean that I would have to deal with nasty fractions. I also tried subtracting, but that didn't get me anywhere either.
$$x^5 - 2x^3y^2 + xy^4 = x(x^2-y^2)^2 = y^2$$
As pointed out in the comments, $x^2 - y^2$ must be a factor of $y$. Hence both $x \pm y$ are factors of $y$. Write
$$y = m(x-y), y = n(x-y)$$
where $m,n$ are integers. Then if $y \ne 0$,
$$ny = mn(x-y), my = mn(x+y)$$
$$(m-n)y = 2mny \leadsto -1 = 4mn - 2m + 2n- 1 = (2m + 1)(2n-1)$$
Thus $\{2m+1, 2n-1\} = \{\pm1\}$, i.e. $m=n=0$ or $m=-1, n=1$.
The first possibility gives $y=0$, which leads to $x=0$.
The second possibility gives $x=0$, which leads to $y=0$.
Thus $(0,0)$ is the only solution.
I use Ben's hint. Clearly, $x=0$ iff $y=0.$ Henceforth, x and y are nonzero. Since $x$ is a square, we write $x=t^2$ and get $$t{(t^4-y^2)}=y.$$ Thus, $t| y$ and, therefore, $t^2| y.$ Writing $y=t^2u$ and cancelling $t^2,$ we get $$u=t^3(1-u^2),$$ which is impossible because of at least two contradictions (using that $u$ is nonzero):
(1) $|u| +1$ divides $|u|$;
(2) $t^3\,|\, u \,|\, t^3$ because $u$ and $u^2-1$ are coprime. ($1-u^2$ would have to be $\pm 1.$)
