Can this Diophantine equation have infinitely many solutions? I came across this as a intermediate step while solving another problem in analysis.
Suppose $m$, and $n$ are integers satisying,
$$\frac{1}{n} - \frac{1}{m} = \alpha$$ where $\alpha$ can be any non zero real.
I essentially need to show that only finitely many $(n,m)$ pair solutions are possible to complete my proof. How can I proceed? Clearly, we only need to check when $\alpha$ is rational.
 A: Write the equation as $\frac{1}{n} - \frac{1}{m} = \frac{r}{s}$ with $r,s\in\mathbb Z$. Clear denominators to get $rmn-sm+sn=0$. The standard procedure to solve this diophantine equations is to multiply by $r$ and factor it as $(rm+s)(rn-s)=-s^2$. This gives you only a finite possible values for $rm+s$ and $rn-s$ since they are divisors of $-s^2$.
A: This is obviously not right if $\alpha=0$, but let's say we are interested in $\alpha\ne 0$. Then, WLOG we can assume $\alpha>0$. Now, even though one of $m,n$ can be negative, by a suitable change in signs we can reduce the equation to $\frac{1}{n}\pm\frac{1}{m}=\alpha$ with $m,n>0$. Thus, we have two cases:

*

*"minus" case: $\frac{1}{n}-\frac{1}{m}=\alpha$ means that $\frac{1}{n}>\alpha$, i.e. $n<\frac{1}{\alpha}$. There are only finitely many such $n$. For each of them, the (integral) solution for $m$ is unique, if it exists (as it is given by $m=\frac{1}{\frac{1}{n}-\alpha}$).


*"plus" case: $\frac{1}{n}+\frac{1}{m}=\alpha$. In that case, WLOG let $n\le m$, so $\alpha\le\frac{2}{n}$ i.e. $n\le\frac{2}{\alpha}$. This again has finitely many solutions in $n$, each of which again having at most one integral solution in $m$, as before.
