# Prove that f(pa)=pa=paf(p^a)=p^a=p^a for an arithmetic function

Let f(n)=∑d|n,d>0ϕ(d)f(n)=\sum_{d|n, d>0}\phi(d).
If p is prime, prove that f(pa)=paf(p^a)=p^a. Deduce that f(n)=nf(n)=n

So far I figured out that the divisors of pap^a are 1,p,…,pa1, p, …, p^a

so f(pa)f(p^a)=ϕ(1)+ϕ(p)+…+ϕ(pa)\phi(1)+\phi(p)+…+\phi(p^a)=(1)+(p−1)+…+(pa−1)=(1)+(p-1)+…+(p^{a}-1)

I got stuck here. How can I go from here to pap^a?

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That really looks like a geometric sum.
– YoTengoUnLCD
2 days ago

Actually it isn’t, it just telescopes nicely.
– xyzzyz
2 days ago

1

Possible duplicate of Prove that ∑d|nϕ(d)=n\sum_{d|n}\phi(d)=n where ϕ\phi is the Euler’s phi function, n,c∈Nn,c\in\mathbb{N}
– Dietrich Burde
2 days ago

It’s both a telescoping sum, and (aside from the initial 11) a geometric sum.
– Greg Martin
2 days ago

1

@DietrichBurde: I’d say this isn’t a duplicate of the other question, since this question asks about a very specific proof method, whereas the other one uses other proofs.
– Greg Martin
2 days ago

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