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In Brent and Cohen's paper about odd perfect numbers, they show this inequality.

$N \ge p^a\sigma(p^a) \gt p ^ {2a}$ where a is even.

I understand the next second half of this: $p^a\sigma(p^a) \gt p ^ {2a}$. The component * sum of its divisors is always larger than the square of the component, , since the sum of a component's divisors is always larger than the component itself.

However, I do not understand the first half of this: $N \ge p^a\sigma(p^a)$. I believe this is saying that for our hypothetical odd perfect number, N, it must be larger than one of it's even-exponent components times the sum of divisors of that component.

I do not understand this. Say, for example, N was $4.96*10^{13} = 5^9 * 71^4$.

$71^8 = 6.45 * 10^{14}$, so in this case, $N \lt p^{2a}$. Obviously this N is not an odd perfect number, so what extra criteria do odd perfect numbers need to satisfy to make the equation shown in the paper true?

Thank you!

louis
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1 Answers1

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The key observation is that $\sigma(p^a)=p^a+p^{a-1} \cdots+p+1$ is relatively prime to $p^a$. Note that $\sigma(p^a)|N$ since $\sigma(p^a)|\sigma(N)$ and $\sigma(N)=2N$. Since $p$ is a prime which does not divide $\sigma(p^a)$, one must have $\sigma(p^a)p^a|N$, which implies the desired result.

JoshuaZ
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  • Ah, thank you Joshua! I understand now. What I failed to put together initially was that $sigma(p^a) \perp p^a$, and that $sigma(p^a)|N$. Also, for anyone else reading, one other thing I noticed that I think is important is that the paper specifies that $a$ is even, AKA $p^a$ is odd. Because if it wasn't odd, then it could divide $sigma(N)$ but not $N$. – louis Mar 23 '24 at 08:51
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    I think in your last bit you mean that $\sigma(p^a)$ is odd. – JoshuaZ Mar 23 '24 at 12:56
  • Oops, yes. Thanks again. – louis Mar 23 '24 at 21:19