But note: each solution \((x, y)\) is counted once for each factor pair. Since \( (a,b) \) and \( (b,a) \) yield potentially different \((x,y)\), but in our setup \(a = x - y\), \(b = x + y\), so order matters in assignment. However, fixing \(a, b\) as both even and same parity already covers all valid factorizations.

Tue, 05 May 2026 17:34:14 +0000

Content under review