Exercise 745Number theory › Divisibility and gcdL1Added 2 Nov 2026

Divisibility of $a^n-b^n$

#divisibility#factorization#powers

Statement

Show that for all integers $a,b$ and all $n\ge1$ , $a-b$ divides $a^n-b^n$ .

Hint

Factor: $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+b^{n-1})$ .

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Factorization.
$a^n-b^n=(a-b)\sum_{k=0}^{n-1}a^{n-1-k}b^k.$
The sum is an integer since $a,b$ are integers. Hence $a-b\mid a^n-b^n$ .

Remark. This is the algebraic identity behind many divisibility results, for example $n^3-1=(n-1)(n^2+n+1)$ .