Annals of Mathematics

Completing squares, the given equation can be rewritten as

$(m - \frac{3}{2})^{2} - \frac{5}{4} = (n + \frac{1}{2})^{2} - \frac{5}{4} \Longrightarrow (2m - 3)^{2} = (2n + 1)^{2} \Longrightarrow 2m - 3 = \pm (2n + 1)$ .

Now if $m = 1$ then $(2n + 1)^{2} = 1 \Longrightarrow n = 0$ , so along with the given condition we know that $m \ge 2$ , and thus

$2m - 3 = 2n + 1 \Longrightarrow 2(m - n) = 4 \Longrightarrow m - n = \boxed{2}$ .

This can be solved as a Diophantine equation treating one side of the equation like a single number:

${ m }^{ 2 }-3m+(-{ n }^{ 2 }-n+2)=0\\ \frac { 3\pm \sqrt { (-3)^{ 2 }-4(-n^{ 2 }-n+2) } }{ 2 } =\\ \frac { 3\pm \sqrt { 4n^{ 2 }+4n+1 } }{ 2 } =\\ \frac { 3\pm \sqrt { (2n+1)^{ 2 } } }{ 2 } =\frac { 3\pm (2n+1) }{ 2 } \\ \\ \\$

We take the positive solution $m=n+2$ so all the solutions can be written as $(n,m=n+2)$ so

$\left| (n+2)-n \right| =2$