Prove that the Liouvilles function is given by the formula $\lambda(n)= \sum\limits_{d^{2} \mid n} \mu \Bigl( \frac{n}{d^2} \Bigr)$
2. First prove that $F(n) = \sum_{d|n}\lambda(d)$ is 1 if n is a square and 0 otherwise. (you may prove that factoring, notice that Liouville's function is multiplicative)
Then, by Möbius' Inversion Formula: $\lambda(n)=\sum_{d|n}F(d)\cdot \mu\left(\frac{n}{d}\right)$ but remember what I said above, then $\lambda(n)=\sum_{d^2|n}F(d^2)\cdot \mu\left(\frac{n}{d^2}\right)=\sum_{d^2|n}\mu\left (\frac{n}{d^2}\right)$ since all the other terms are 0.