± 1 Solutions

Solution. (a) Since φ(3n) = 2 · 3n−1, the problem amounts to showing that 3n 23 − 1 and 3n 22·3 − 1 (when n ≥ 2). The first claim follows from reduction mod 3, and the second claim follows from the exponent lifting trick, as 3 ‖ 22−1, so that 3n−1 ‖ 22·3−1. (b) Since φ(pn) = φ(2pn) = (p − 1)pn−1, it suffices to show pn g(p−1)p − 1 and pn gdp − 1 for any divisor d of p − 1 with d < p − 1. The first claim follows from pn−1 ‖ g(p−1)p − 1 by the exponent lifting trick as p ‖ gp−1 − 1 by assumption, and the second claim follows from the fact that p | gm − 1 if and only if (p − 1) | m as g is a primitive root mod p.