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Find the number of integers $n$ such that the equation $xy^2 + y^2 - x - y= n$ has an infinite number of integer solutions $(x,y)$.

Firstly the equation can be rearranged into $(y-1)(y+x(y+1)) = n$. If $n$ is $0$, there's an infinite number of $(x,y)$ possible. However, I am struggling with the case when $n$ is a nonzero integer. I had a peek at the solution, and it says

There exists a divisor $k$ of $n$ such that $y-1 = k$ and $x(y+1) + y = n/k$ for infinitely many $x$. It forces $y+1 = 0$.

Yet I don't quite see how $y+1=0$... If anyone can help me understand that'd be great!

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  • $\begingroup$ $n/k-y=x(y+1)$ is an integer divisible by $x$. Since there are infinitely many such $x$ necessarily $n/k-y$ is zero. $\endgroup$ – Crostul May 26 '18 at 14:41
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Notice that for a specific $k$ (and thus $n/k$) $y$ is also fixed. Now the equation $x(y+1)+y=n/k$ implies that x is also fixed with respect to $k$ or equivalently it is a function of $k$ which for a specific $n$ can only assume a finite number of values. So in order to have infinite solution $x$ must attain infinite values and this can only happen if it doesn’t affect the equation in the first place. Thus $y+1=0$ and $x$ can assume every integer value.

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If $x(y+1)+y=\frac{n}{k}$ for infinitely many values of $x$, then for any two such values $x_0$ and $x_1$ we have $$(x_0-x_1)(y+1)=0.$$ where $x_0-x_1\neq0$, and so $y+1=0$.


The argument can be put more simply; given any integer $n$ and a solution $(x,y)$ to $$n=xy^2+y^2-x-y=(y-1)(y+x(y+1)),$$ it follows immediately that $y-1$ divides $n$ and that $$x=\frac{n+y-y^2}{y^2-1}.$$ So for there to be infinitely many solutions $n$ must have infinitely many divisors.

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