 Solution

ID: past-exam-of-the-analytic-number-theory-and-elliptic-curves-course-of-the-university-of-oxford/2007/a/1/iii/solution

μ (p^{e})^{2} = 1

for

e = 0, 1

and is zero for

e \geq 2

. Therefore

H_{1} (s) = \prod_{p} (1 + p^{- s}) = \frac{ζ ( s )}{ζ ( 2 s )} .

The inverse local factor is

(1 + p^{- s})^{- 1} = \sum_{e \geq 0} (- 1)^{e} p^{- es}

. Consequently the inverse under Dirichlet convolution is the Liouville function

h_{2} (n) = (- 1)^{Ω (n)},

where

Ω (n)

p^{e}

with

e \geq 1

, their convolution is

(- 1)^{e} + (- 1)^{e - 1} = 0

, while its value at one is one. Multiplicativity then proves

h_{1} * h_{2} = I

on every positive integer. Its Dirichlet series is

H_{2} (s) = ζ (2 s) / ζ (s)

for

Re s > 1

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