In my lecture notes, we have been given the theorem
If $N \in \mathbb{Z}_{+} \setminus \{1\}$ is odd and perfect, and written $\prod_{i = 1}^k p_i^{n_i}$ as shown, then $k \geq 3$, that is $N$ has at least $3$ distinct prime divisors.
The proof for this has been given by using the formula
$$2 \prod_{i = 1}^{k} \left(1 – \frac{1}{p_i} \right)$$
and my lecturer put in $k = 1$ and $k = 2$ and showed they were both $> 1$ and somehow this proves it.
Firstly, how does this prove that there are atleast 3 dividers? I thought the divider had to be an integer and subbing in $k = 1,2$ gives you two fractions which are then not dividers?
Also, how can you can the particular theorem exist and be proven if there is a chance that no odd perfect numbers exist? I don't really understand that
Best Answer
The theorems are of the form, "If any odd perfect numbers exist, then they must be of the form..."
If there are no odd perfect numbers, then these theorems are vacuously true (and indeed, one of the ways of eventually proving there are no odd perfect numbers would be to prove that if any existed, then some contradictory properties would have to be true about them.)
If there are any odd perfect numbers, then these theorems narrow down where we should look.
If $p_1^{a_1}, p_2^{a_2}, \ldots, p_m^{a_m}$ are the distinct prime divisors of an integer $n$, then $$2n = \prod_i (1 + p_i + \ldots + p_i^{a_i}),$$ which you can check by expanding the RHS. The terms in the product are partial geometric series, so $$2n = \prod_i \frac{p_i^{a_i+1}-1}{p_i-1},$$ and shuffling around the terms, $$2 \prod_i (1-p_i^{-1}) = \prod_i (1-p_i^{-a_i-1}).$$ Notice that the right-hand side is less than 1. Moreover, the LHS is larger the larger the $p_i$ are, so for $n$ odd and $m=2$, $$2\prod_{i=1}^2 (1-p_i^{-1}) \geq 2(1-3^{-1})(1-5^{-1})=\frac{16}{15} > 1,$$ so $m$ cannot be $2$.