Prove that $\displaystyle \sum_{d|n}\lambda(d)= \left\{\begin{array}{ll}1,&\mbox{ if } n=m^2 \mbox{ for some m}\\0,&\mbox{ otherwise } \end{array}\right.$

where $\displaystyle \lambda(n)= (-1)^{\Omega(n)}$ and $\displaystyle \Omega(n)$ is the number of prime divisors of n counting multiplicity.