Problem
ALG-B1-M02-P007 Odd Sum of Powers
#7
★★☆☆☆ Level 2 of 5
Prove that \(a+b\) divides \(a^{2n+1}+b^{2n+1}\) for every integer \(n\ge0\).
Put \(m=2n+1\) and use the factorisation of a sum of odd powers.
For odd \(m\), \(a^m+b^m=(a+b)(a^{m-1}-a^{m-2}b+\cdots-ab^{m-2}+b^{m-1})\).
With \(m=2n+1\), the factor \(a+b\) appears, so the divisibility is proved.
The idea is inspired by classical divisibility of powers, but the statement is a training version.