Originally Posted by
jimmyjimmyjimmy Hi,
Let n be a positive integer.
Suppose that the sum of all positive divisors of n (besides n itself) is a multiple of n. (e.g. 1+2+3 = 6k where n=6 and k=1 in this case)
Suppose too that n is a product of two distinct primes.
Show that n must equal 6.
I started by letting n = pq for distinct primes p and q.
Then we know 1+p+q = kn = kpq for some k.
This problem just seems so simple, but I can't help but think the proof is really complicated. I would appreciate any hints. Thanks.