Problem
NT-B1-M05-P021 An Infinite Family
#21
★★★★☆ Level 4 of 5
Find all positive integer solutions of \(x^2+y^2=2xy+x+y\).
Move \(2xy\) to the left and look at \((x-y)^2\).
The equation is equivalent to \((x-y)^2=x+y\). If \(x=y\), then \(0=2x\), impossible for positive \(x\). Suppose \(x>y\) and let \(d=x-y\). Then \(d^2=2y+d\), so \(y=\frac{d^2-d}{2}\), \(x=\frac{d^2+d}{2}\). We need \(d\ge2\). By symmetry, all solutions are \(\left(\frac{d(d+1)}{2},\frac{d(d-1)}{2}\right)\) and \(\left(\frac{d(d-1)}{2},\frac{d(d+1)}{2}\right)\), where \(d\ge2\) is an integer.
This is not a finite list but a parametrised family.