Csdrhrt

I am having some trouble trying to solve this question.

I have been learning some number theory and I have come across this question. To be honest I am kinda lost.

I have seen this being solved

Let $n in mathbb{Z}$ with $n > 0$. Let $F(n) = sum_{d mid n} lambda(d)$. Prove that $$F(n) = begin{cases}1, quad text{if }n text{ is a perfect square}\ 0, quad text{otherwise} end{cases} $$

but I have no idea how to solve the first one.

Any help would be appreciated.

By the definition that you posted, $lambda(n)=(-1)^{omega(n)}$, where $omega(n)$ is the number of prime factors of $n$, counting multiplicity. Since $omega(mn)=omega(m)+omega(n)$ for all $m,ninmathbb N$, we have that
$$lambda(mn)=(-1)^{omega(mn)}=(-1)^{omega(m)+omega(n)}=(-1)^{omega(m)}(-1)^{omega(n)}=lambda(m)lambda(n)$$
which proves that $lambda$ is completely multiplicative. As for your summation problem, I will offer you the following hint (a list of suggested steps for solving it) and let you fill in the rest:

• First prove that if $f(n)$ is a multiplicative function (not necessarily completely multiplicative), then $F(n)=sum_{d|n}f(d)$ is also multiplicative




• Then notice that if $n=prod p_i^{m_i}$ is the prime factorization of $n$ and $f$ is a multiplicative function, then $f(n)=prod f(p_i^{m_i})$




• Finally, evaluate $F(n)$ for $n=p^k$ where $p$ is some prime, and then express a general formula for $F(n)$ for any $n$ by decomposing $n$ into its prime factorization