Problem
NT-B1-M02-P018 Sum of Squares and Sum
#18
★★★☆☆ Level 3 of 5
Let \(\gcd(a,b)=1\). Prove that \(\gcd(a^2+b^2,a+b)\mid2\).
Modulo \(a+b\), replace \(a\) by \(-b\).
Let \(d=\gcd(a^2+b^2,a+b)\). Then \(a\equiv -b\pmod d\), so \(a^2+b^2\equiv2b^2\pmod d\). Hence \(d\mid2b^2\). But \(\gcd(a+b,b)=\gcd(a,b)=1\), so \(\gcd(d,b)=1\). Therefore \(d\mid2\).
The problem has two ideas: reduction modulo the GCD and cancellation using coprimality.