Prove $\lambda(n)=\sum_{d^2|n}\mu(n/d)^2$ and $\mu^2(n)=\sum_{d^2|n}\mu(d)$ $\lambda(n)$= $\sum_{d^2|n}$ $\mu(n/d)^2$
and $\mu^2(n)$= $\sum_{d^2|n}$ $\mu(d)$
Having a little bit of trouble here.Can I use the fact that $\sum_{d|n}\lambda(n)$ is a characteristic function for squares by $\lambda*1$.?
Edit: For the second part can I say that $\lambda^{-1}$= $\mu{^2}$ because $\mu$ is multiplicative.. and just bring in the first part then and invert the sum- if that's allowed?
 A: I assume the first one should be $\lambda(n)=\sum_{d^2\mid n}\mu(n/d)$, since otherwise it is not true.
You can't directly apply convolution identities here because the sums $\sum_{d^2\mid n}\mu(n/d)$ and $\sum_{d^2\mid n}\mu(d)$ do not run over all divisors of $n$.
Nonetheless we can prove the following: if $f$ and $g$ are multiplicative, then so is
$$n\mapsto\sum_{d^2\mid n}f(n/d)g(d).$$
The proof is almost the same as how one would do it for Dirichlet convolution. The key argument is that if $\gcd(m,n)=1$, then
$$d^2\mid mn \iff d=ef \text{ with }\gcd(e,f)=1\text{ and }e^2\mid m, f^2\mid n.$$
(Try to convince yourself that the multiplicativity of $\sum_{d^2\mid n}f(n/d)g(d)$ follows from this observation.)
Using the result above it suffices to show that
$$\lambda(n)=\sum_{d^2\mid n}\mu(n/d)\quad\text{and}\quad\mu^2(n)=\sum_{d^2\mid n}\mu(d)$$
when $n$ is a prime power.
$\lambda(n)=\sum_{d^2\mid n}\mu(n/d)$: Clearly, since when $n=p^k$ the only non-vanishing term in the RHS is $\mu(n/p^{\lfloor k/2\rfloor})=(-1)^k=\lambda(p^k)$.
$\mu^2(n)=\sum_{d^2\mid n}\mu(d)$: This is obvious for $n=1$ and for $n$ prime. If $n=p^k$, $k\geqslant2$ we have $\sum_{d^2\mid n}\mu(d)=\mu(1)+\mu(p)+0+0+\cdots=0$, as desired.
