Let $\mu$ be the Möbius function, and let $\nu(n)$ be the number of distinct prime factors of $n$. Then we can define $p = \mu * \nu$, i.e. $$ p(n) = \sum_{d \mid m} \mu(d) \nu(n/d). $$ An exercise in an introductory textbook I already put back on the library shelf and can't find anymore asks:
Prove that $p$ only takes the values 0 and 1.
The obvious solution seems to be through Möbius inversion: if we define $q(n)$ to be 1 when $n$ is prime and $0$ otherwise, then $\nu = 1 * q$ so that $q = \mu * \nu = p$. However, the formulation of the question suggests that there is an easy way to prove that $p$ only takes values $0, 1$ without identifying it entirely.
How might one do that?
There is the generalization of your first solution, showing $\mu \ast f$ for additive functions is always easy to find :
Let $f(n)$ be additive that is $f(nm) =f(n)+f(m)$ for $gcd(n,m)=1$.
Let $g(p^k) = f(p^k)-f(p^{k-1})$ and $g(n) = 0$ if $n$ isn't a prime power
Then $f(n) = \sum_{p^k \| n} f(p^k) = \sum_{p^k | n} g(p^k) = \sum_{d | n} g(d)$
Whence $\sum_{d | n} \mu(d) f(n/d)= g(n)$