### 33rd Putnam 1972

**Problem A5**

Show that n does not divide 2^{n} - 1 for n > 1.

**Solution**

Suppose that n does divide 2^{n} - 1. Then n must be odd. Let p be the smallest prime dividing n. Then 2^{p-1} = 1 (mod p). Let m be the smallest divisor of p - 1 such that 2^{m} = 1 (mod p). Since m is smaller than p it must be coprime to n, so n = qm + r with 0 < r < m. Hence 2^{r} = 1 (mod p). Contradiction.

33rd Putnam 1972

© John Scholes

jscholes@kalva.demon.co.uk

27 Jan 2001