Perfect Numbers: Proving Even Exponents

[Fallen Angel]

Hi,

Let $n >6$ be a perfect number (A number $n$ is called perfect if $s(n)=2n$ where $s(n)$ is the sum of the divisors of $n$) with prime factorization $n=p_{1}^{e_{1}}p_{2}^{e_{2}}\cdots p_{k}^{e_{k}}$ where $1<p_{1}<p_{2}<\ldots <p_{k}$. Prove that $e_{1}$ is even

 

[Nono713]

Great challenge! I have a partial solution below, but I am not sure how to extend it to the remaining cases. I might come back to it later.​

Spoiler

Proof for $n$ even:

All even perfect numbers are of the form $n = 2^{p - 1} (2^p - 1)$ whenever $2^p - 1$ is prime by the Euclid-Euler theorem. If $p > 2$, that is, $n > 6$, $p$ must be odd so $2^{p - 1}$ is an even power and the result immediately follows.

Proof for $n$ odd:

If $n$ is odd, then it can be written as $p_1^{e_1} r$ where $p_1$ is the smallest odd prime factor of $n$ and $r$ is an odd number not divisible by $p_1$. Assume $n$ is perfect, then it can be written as:
$$2n = s(r) + p_1 s(r) + \cdots + p_1^{e_1} s(r)$$
This is equivalent to:
$$2n = s(r) \left ( 1 + p_1 + \cdots + p_1^{e_1} \right )$$
Now suppose $r$ is not a square, so that $s(r)$ is even. It follows $n$ can be perfect only if the $1 + p_1 + \cdots + p_1^{e_1}$ term is odd, and it is odd if and only if $e_1$ is even. It remains to check the case for $r$ square, but I don't know how to do that; possibly this is not the right approach.​

 

[Fallen Angel]

Hi Bacterius, good work!

Your idea is good, if yo want a hint, open the next spoiler

Spoiler

From $2n=s(r)(1+p_{1}+p_{1}^{2}+\ldots +p_{1}^{e_{1}})$

Try to determine what happens if $(p+1)$ divides $(1+p_{1}+p_{1}^{2}+\ldots +p_{1}^{e_{1}})$ taking into account that $s(r)=\displaystyle\prod_{i=2}^{k}(1+p_{i}+p_{i}^{2}+\ldots +p_{i}^{e_{i}})$