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

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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!!

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  • $\begingroup$ Please don't use pictures. $\endgroup$
    – Dietrich Burde
    Oct 19, 2020 at 7:59

1 Answer 1

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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.

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