How do I prove this theorem about coprime numbers?

Question

Find all $x,y$ co-prime so that there exists a $z$ such that $x+y^2=x^2z^2+x^2+y$.

I don't see how this equation could be simplified.

Answer 1

Rewrite your equation as $$ (x-y)(x+y-1) = x^2 z^2$$

Now $\gcd(x,x-y) = \gcd(x,y) = 1$ so you must have $x^2 \mid x+y-1$ for some integer $k$, i.e. $y = 1-x+k x^2$. Conversely, if $y = 1-x+kx^2$ we do have $\gcd(x,y)=1$, and the equation becomes

$$ k^2 x^2 - 2 k x + k = z^2 $$ i.e. $$(kx+z-1)(kx-z-1) = 1-k $$

Special cases:

If $k = 0$, we have $z=0$, corresponding to $y=1-x$.

If $k=1$, we have $z = \pm (x-1)$, corresponding to $y= 1-x+x^2$.

If $k = -1$, we have $(x+1)^2 - z^2 = 2$, which has no solutions.

If $x=0$, $k=z^2$, corresponding to $y = 1$.

If $x = 1$, $k(k-1) = z^2$, which has solutions $k=0,z=0$ and $k=1,z=1$ (see the cases $k=0$ and $k=1$ above).

If $x=-1$, $k(k+3) = z^2$, which has solutions $k=-4, z=\pm 2$, $k=-3, z=0$, $k=0,z=0$ and $k=1, z = \pm 2$, corresponding to $y = -2,-1,2,3$.

If $k = x = -2$, the equation becomes $6 = z^2$ which has no solution.

Otherwise at least one of $kx+z-1$ and $kx-z-1$ has greater absolute value than $ 1-k$, and the only way the equation can be true is if both sides are $0$, but $k=1$ was already covered.