MHB Perfect Numbers: Proving Even Exponents

AI Thread Summary
A perfect number is defined as one where the sum of its divisors equals twice the number itself. The discussion centers on proving that for perfect numbers greater than six, the exponent of the smallest prime factor in its prime factorization is even. A participant shares a partial solution but expresses uncertainty about extending it to all cases. Another participant acknowledges the initial idea and offers a hint for further progress. The conversation highlights the complexity of proving properties related to perfect numbers.
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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
 
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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.​

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.​
 
Hi Bacterius, good work!

Your idea is good, if yo want a hint, open the next 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 $s(r)=\displaystyle\prod_{i=2}^{k}(1+p_{i}+p_{i}^{2}+\ldots +p_{i}^{e_{i}})$
 
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