Competition problem — Number theory

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Number theory Difficulty 5.8 AIME, harder Prove it

Lemma 9 Let $p$ be a prime and $l$ a positive integer, then we have
$\sigma\left(p^{l}\right)=\frac{p^{l+1}-1}{p-1}$

Solution

Proof: Since the divisors of $p^{l}$ are $1, p, \cdots, p^{l-1}, p^{l}$ , we have
$\sigma\left(p^{2}\right)=1+p+\cdots+p^{2}$

And by Lemma 8, we know that this lemma holds.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.