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