An integer n > 0 is called perfect if the sum of all its positive divisors (including 1 and n) is 2n.
Prove that the product of two odd prime numbers is not a perfect number.

So I only know how to do this under the assumption that perfect numbers are always even. Is there another way to do it?

My solution if anyone is interested:

Let p and q be odd primes. Assume pq is a perfect number.

Sum of p's divisors = 1 + p
Sum of q's divisors = 1 + q

(p+1)(q+1) = 2pq ?
pq + p + q + 1 = 2pq
p + q + 1 = pq

If pq is a perfect number it must be even. This is a contradiction because p + q + 1 is odd (sum of three odds is odd).

Unfortunately, it has not be proven (yet ?) that a perfect number must be an even number.

Research still goes on...

If, actually, a perfect number must be even, then you can just say : pq is odd, hence, it's not a perfect number.

Let's get back to your last statement :

p + q + 1 = pq

You can assume that q > p.

p + q + 1 < q + q + 1 < 3q

and p >= 3, hence :

pq >= 3q

You have your contradiction.

Can you explain this, I don't quite follow

Suppose that q is greater than p (one of the two must be greater than or equal to the other, let it be q).

Suppose that p + q + 1 = pq

Then :

p <= q => p + q + 1 <= q + q + 1 < q + q + q = 3q

and since p is an odd prime, then p >= 3 :

pq >= 3q

Finally, p + q + 1 cannot equal pq, since the first is less than 3q, and the second is greater than or equal to 3q.

That makes sense now thank you !!