If and are distinct primes, then the sum of the positive divisors of other than is clearly Hence, if we would have
The only pair of primes satisfying the above condition are and
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.