Peter wrote: Find all pairs $ (m,n)$ of integers that satisfy the equation \[ (m - n)^{2} = \frac {4mn}{m + n - 1}. \] Case 1 $ m=n$ then $ m=n=0$ $ m,n$ distint. Suppose $ m>n$ $ (m-n)^2|4mn$ Let $ d=gcd(m,n)$ then $ m=da,n=db$ $ \Rightarrow (a-b)^2|4ab$ But $ \gcd(a-b,ab)=1$ so $ a-b|2$ So $ a-b=1$ or $ a-b=2$ 1. $ a-b=1$ so $ a=b+1$ $ d^2(d(a+b)-1)=4d^2ab$ $ \Rightarrow d(a+b)-1=4ab$ $ d(2b+1)-1=4b(b+1)$ $ d=2b+1$ So $ m=(2b+1)(b+1),n=(2b+1)b$ Case 2 $ a=b+2$ $ 4d^2(d(a+b)-1)=4d^2ab$ $ d(2b+2)-1=(b+2)b$ $ \Leftrightarrow d=\frac{b+1}{2}$ So $ m=\frac{(b+2)(b+1)}{2},n=\frac{b(b+2)}{2}$

$ (m - n)^{2} = \frac {4mn}{m + n - 1}$ $ (m - n)^2 + 4mn = 4mn(\frac {1}{m + n -1}) + 4mn$ $ (m + n)^2 = 4mn (\frac {1}{m + n - 1} + 1)$ $ (m + n)^2 = 4mn (\frac {m + n}{m + n -1})$ $ (m + n) = \frac {4mn}{m + n - 1}$ $ (m + n) = (m - n)^2$ check it.