1. ## the number 6

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.

2. If $\displaystyle p$ and $\displaystyle q$ are distinct primes, then the sum of the positive divisors of $\displaystyle pq$ other than $\displaystyle pq$ is clearly $\displaystyle 1+p+q.$ Hence, if $\displaystyle 1+p+q=kpq,$ we would have

$\displaystyle p\mid q+1$ and $\displaystyle q\mid p+1$

The only pair of primes satisfying the above condition are $\displaystyle 2$ and $\displaystyle 3.$

3. 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.

$\displaystyle p \mid kpq \,\,on\,\, the \,\, RH \,\, and \,\,p \mid p \,\,\ on \,\ the \,\, LH \,\, \Longrightarrow p \mid q+1$

In the same way $\displaystyle q \mid p+1$

So you've a pair of primes s.t. each of them divides the succesor of the other one....there aren't many of those, I believe, specially if we take into account that there's only one even prime.

Tonio