I am working on a problem, but I am not sure why I get the wrong answer. The question asks, what is, $$\sum_{d\mid n} \sigma_2(d)\cdot \mu \left(\frac{n}{d}\right) \tag{ for n=100}$$
In the previous problem, I showed that $\sigma_k(n)$, for $k\geq 1$ is a multiplicative, where, $$\sigma_{k}(n) = \sum_{d\mid n}d^k \tag{$\forall n\geq 1$}$$
So from the previous problem, it follows that for $n=100$
$$\sum_{d\mid 100} \sigma_2{(100)} \cdot \mu \left(\frac{100}{d}\right) = \left(\sum_{d\mid 100} \sigma_2(100) \right) \left(\sum_{d\mid 100}\mu\left(\frac{100}{d} \right)\right)$$
We know that $\sum_{d\mid 100} \sigma_2(100) = 100^2$
However, to solve $\require{cancel}\sum_{d\mid 100}\mu\left(\frac{100}{d}\right)$, we use $d=1,2,4,5,10,20,25,50,100$, and therefore, we have, $$\sum_{d\mid 100}\mu\left(\frac{100}{d}\right)= \cancelto{0}{\mu(100)}+\cancelto{0}{\mu(50)}+\cancelto{0}{\mu(25)}+\cancelto{0}{\mu(20)}+\cancelto{1}{\mu(10)}+\cancelto{-1}{\mu(5)}+\cancelto{0}{\mu(4)}+\cancelto{-1}{\mu(2)}+\cancelto{1}{\mu(1)}\\ =0+0+0+0+1-1-1+1 = 0$$
Finally, $$\left(\sum_{d\mid 100} \sigma_2(100) \right) \left(\sum_{d\mid 100}\mu\left(\frac{100}{d} \right) \right) = (100^2)(0)= 0$$
What did I do wrong?