How can I prove that a function $p(n)$ is multiplicative but not completely multiplicative?

A function $f\colon\mathbb N\to\mathbb C$ is called multiplicative if $f(1)=1$ and $$\gcd(a,b)=1 \implies f(ab)=f(a)f(b).$$

we have this condition only for $a$, $b$ coprime.

Completely multiplicative:

if the equality $f(ab)=f(a)f(b)$ holds for any pair of positive integers $a$, $b$.

Let $ρ(n) = (μ(n))^{2}φ(n)$.

I know that

$μ(n)$ is multiplicative so $μ(nm)=μ(n)μ(m)$ for all $(n,m)=1$

$φ(n)$ is multiplicative

I have solved that $ρ(n) = (μ(n))^{2}φ(n)$ multiplicative but I am stuck on showing that they are not completely

Any help would be appreciated it.

  • $\begingroup$ Also do you know the Euler products of $\varphi(n), \mu(n)^2, \mu(n)^2 \varphi(n)$ ? Completely multiplicative means $\sum_{n=1}^\infty f(n) n^{-s}=\prod_p \frac{1}{1-f(p)p^{-s}}$ $\endgroup$ – reuns Dec 6 '18 at 15:07
  • $\begingroup$ I know that μ(n) is multiplicative $\endgroup$ – Hidaw Dec 6 '18 at 15:32
  • $\begingroup$ So $\sum_{n=1}^\infty \mu(n) n^{-s}=\prod_p (1+\sum_{k=1}^\infty \mu(p^k) p^{-sk}) = \ldots$ $\endgroup$ – reuns Dec 6 '18 at 15:36

You would assume two cases, generally, at least if you want to go directly from the definition.

Let $a,b$ be coprime. Then show $p(a)p(b) = p(ab)$. Proving this would make $p$ multiplicative (not necessarily completely).

Then let $a,b$ be not coprime, i.e. $gcd(a,b) \neq 1$. Then, if you want to show $p$ is multiplicative but not completely so, you would show $p(a)p(b) \neq p(ab)$ in this case. How you would show this might depend on the circumstances; personally, I would do so by counterexample. For example, choose a specific $a,b$ with $gcd(a,b) \neq 1$ and then show for this given pair that $p(a)p(b) \neq p(ab)$.


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