A perfect number is equal to the sum of all its positive divisors other than itself. Show that if a perfect number > 6 is divisible by 3, then it is divisible by 9. Show that a perfect number > 28 divisible by 7 must be divisible by 49.
Solution
Suppose n is perfect and divisible by 3 but not 9. Put n = 3m. Write the sum of the positive divisors of k as s(k). Then 6m = s(n) = s(3m) = s(3)s(m) = 4s(m), so s(m) = 3m/2. Hence, in particular m is even. So it certainly has divisors 1, m, m/2 (which are all distinct for n > 6, but not for n = 6, when m/2 = 1) with sum > 3m/2. Contradiction.
Similarly, if n is divisible by 7 but not 49, but n = 7k. Then 14k = s(n) = s(7k) = s(7)s(k) = 8s(k), so s(k) = 7/4 k. Hence k must be divisible by 4, so it certainly has factors 1, k, k/2, k/4 (which are all distinct for n > 28, but not for n = 28, when k/4 = 1) with sum 7k/4 + 1. Contradiction.
Thanks to Stefanos Aretakis
© John Scholes
jscholes@kalva.demon.co.uk
20 December 2003