10th Eötvös 1903

Prove that 2^p-1(2^p - 1) is perfect when 2^p - 1 is prime. [A perfect number equals the twice sum of its (positive) divisors (excluding the number itself.]

Solution

In this case sum of the divisors is 1+2+2²+ ... +2^p-1 + (2^p-1)(1+2+2²+ ... +2^p-2) = (2^p-1) + (2^p-1)(2^p-1-1) = 2^p-1(2^p-1).

10th Eötvös 1903

© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03