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

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.

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

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 $$p(a)p(b) \neq p(ab)$$.