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!
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.