# Question 3.11(b) of Apostol's Introduction to analytic number theory

The following question is from Apostol's Introduction to analytic number theory on page 71.

It's image:

How can I prove (b) part of 11 . I have proved it to be multiplicative but I don't know how how to prove the identity for $$p^{k}$$.

Kindly give hints.

Thank you!!

• Please don't use pictures. Oct 19, 2020 at 7:59

It is not that difficult for $$p^k \geq p$$. $$\varphi_1(p^k) = \sum_{d^2| p^k } \mu(d) \sigma(p^k/d^2) = \mu(1)\sigma(p^k) + \mu(p) \sigma(p^k/p^2)$$ $$\sigma(p^{k}) - \sigma(p^{k-2}) = (1 + p + \dots + p^{k-2} + p^{k - 1} + p^k) - (1 + p + \dots + p^{k-2}) = p^{k - 1} + p^k = p^k\prod_{p|p^k}(1 + 1/p) = \varphi_1(p^k)$$. Therefore it suffices to prove that the right hand side is multiplicative. This can be done as follows. Let $$n = ab$$ where $$(a,b) = 1$$ $$\sum_{d^2|ab} \mu(ab) \sigma(ab/(d^2))$$ One can also split the divisor $$d$$ in to two relatively prime divisors $$d_a$$ and $$d_b$$ such that $$d_a|a$$ and $$d_b|b$$. Then the sum becomes $$\sum_{d_b^2 d_a^2|ab} \mu(ab) \sigma(ab/(d_a^2 d_b^2))$$ Fixing $$d_b$$ and summing over $$d_a^2$$ one can get the following result:
$$\sum_{d_b^2|b} \sum_{d_a^2|a}\mu(a) \mu(b) \sigma\bigg(\dfrac{a}{d_a^2}\bigg) \sigma\bigg(\dfrac{b}{d_b^2}\bigg)$$. since $$\mu$$ and $$\sigma$$ are multiplicative. Finally you will get
$$\varphi_1(ab) = \sum_{d_b^2|b}\mu(b) \sigma\bigg(\dfrac{b}{d_b^2}\bigg) \sum_{d_a^2|a}\mu(a) \sigma\bigg(\dfrac{a}{d_a^2}\bigg)$$
$$\varphi_1(ab) =\varphi_1(a) \sum_{d_b^2|b}\mu(b)\sigma\bigg(\dfrac{b}{d_b^2}\bigg) = \varphi_1(a)\varphi_1(b)$$. Hence the given equation is true.