Perfect numbers, please help

You can say :

• even perfect numbers are of the form $n=2^{p−1}(2^p − 1)$ where $p$ and $2^p-1$ are prime

• $n$ then has $2p-1$ divisors other than $n$, and the product of those divisors is $n^{p-1}$

The basis for the first of these is the Euler-Euclid theorem and the second is a simple derivation from the first

That's not a coincidence. The divisors of any (non-perfect-square) number $n$ are paired together: any time you have a divisor $d$, you have $n/d$ which is the other divisor, in some sense, his brother.

For example, for $n=496$ $$1 \cdot 2 \cdot 4 \cdot 8 \cdot 16 \cdot 31 \cdot 62 \cdot 124 \cdot 248 =$$ $$=(2 \cdot 248) \cdot (4 \cdot 124) \cdot (8 \cdot 62) \cdot (16 \cdot 31)= 496^4$$

How is that the power is $4$? That's because there are $9$ divisors, and you throw away the trivial divisor $1$ (whose brother is $496$). The remaining $8$ divisors are paired, and you get $4$ pairs of numbers whose product is $496$.