Problem
NT-B2-M10-P012 Divisibility of Prime-Power Sums
#12
★★★★☆ Level 4 of 5
Let \(p\) be prime and \(a,b\ge 0\). Prove that \(\sigma(p^a)\mid \sigma(p^b)\) if and only if \(a+1\mid b+1\).
1001 Problems in Classical Number Theory (method inspiration) · Problem 498
Use \(\sigma(p^a)=\frac{p^{a+1}-1}{p-1}\).
We have \(\sigma(p^a)=1+p+\cdots+p^a=\frac{p^{a+1}-1}{p-1}\). By the standard fact \(p^u-1\mid p^v-1\) if and only if \(u\mid v\), we get that \(\sigma(p^a)\mid\sigma(p^b)\) if and only if \(p^{a+1}-1\mid p^{b+1}-1\), that is \(a+1\mid b+1\). Dividing by the common factor \(p-1\) is valid because both expressions are the corresponding geometric quotients.
It may help to recall the gcd of \(p^u-1\) type expressions first.