with equality on the right exactly when A = 1. Suppose A /= 1. Then the preceding
inequalities show that σ(B) f 2B, and so there is a prime p1 dividing σ(B) to a
higher power than that to which it divides 2B; for definiteness, fix p1 as the least such prime. It now follows from (2) that p1 | A. Let e1 ≥ 1 be such that pe1 I A