Prove $\sum_{d | n} \mu(d) (\log(d))^2=0$ If $n$ is a positive integer with more than 2 distinct prime factors, how to prove that $\sum_{d | n} \mu(d) (\log(d))^2=0$?
I struggle on how to continue from this.
Suppose $n=p_1 p_2 ... p_r$, because $\mu(d) \neq 0$ for square free $n$.
$\sum_{d | n} \mu(d) \log^2(d)=(-1)\sum_{p_i|n}\log^2(p_i)+(-1)^2\sum_{p_i|n,p_j|n,p_i\neq p_j}\log^2(p_i p_j)+...+(-1)^r\log^2(p_1 p_2 ... p_r)$
I thought of changing $\log^2(p_i p_j)$ to $(\log p_i + \log p_j)^2$, but still stuck.
How to prove this statement? If the way that I've shown won't work, feel free to suggest another way to prove this.

As you noted, it's obvious that $\prod_{i=1}^k p_i^{a_i}$ (where $p_i$ is prime and $a_i\geq 1$) and $\prod_{i=1}^k p_i$ will have the same result when the sum is evaluated, so let's just assume that $n=\prod_{i=1}^k p_i$
Denote $S_i$ as the set of all divisors of $n$ that contain exactly $i$ prime factors. For example, $S_0=\{\}$, $S_1=\{p_1,p_2,\ldots p_k\}$ and $S_n=\{p_1p_2\ldots p_k\}$.
Our sum is equivalent to
$$\sum_{i=0}^k\sum_{s\in S_i} \mu(s)\log^2(s)$$
$$=\sum_{i=0}^k (-1)^i\sum_{s\in S_i} \log^2(s)$$
While we can directly compute this inner sum, it will be better to exploit symmetry. For a fixed $i$, we will have $\binom{k}{i}$ different possibilities for $s$. We will then expand $(\log s)^2$, where $s$ is a product of $i$ primes. For any $s$, this expansion will have $i$ terms of the form $\log^2 p$ (where $p$ is a prime) and  $i(i-1)$ terms of the form $(\log p)(\log q)$ (where $p$ and $q$ are distinct primes). Using the fact that this sum is symmetric, we get that for a fixed $i$,
$$\sum_{s\in S_i} \log^2 (s)=\frac{i}{k}\binom{k}{i}\sum_{j=1}^k \log^2 (p_j)+\frac{i(i-1)}{k(k-1)}\binom{k}{i}\sum_{\substack{l,m\in [1,k]\\l\neq m}}(\log p_l)(\log p_m)$$
We can define the sums $\sum_{j=1}^k \log^2 (p_j)$ and $\sum_{\substack{l,m\in [1,k]\\l\neq m}}(\log p_l)(\log p_m)$ as constant $P_1$ and $P_2$. Hence, our original expression is equivalent to
$$=\sum_{i=0}^k (-1)^i\left(\frac{i}{k}\binom{k}{i}P_1+\frac{i(i-1)}{k(k-1)}\binom{k}{i}P_2\right)$$
$$=\sum_{i=0}^k (-1)^i\left(\binom{k-1}{i-1}P_1+\binom{k-2}{i-2}P_2\right)$$
$$=P_1\sum_{i=0}^k \binom{k-1}{i-1}(-1)^i+P_2\sum_{i=0}^k\binom{k-2}{i-2}(-1)^i$$
When $k-2>0$, using binomial theorem, both of these sums evaluate to $0$, hence our expression is equivalent to
$$=P_1(0)+P_2(0)$$
$$=\boxed{0}$$
