Let $\omega(n)$ denote the number of distinct prime divisors of $n>1$, with $\omega(1)=0$.
(a) Show that $2^{\omega(n)}$ is a multiplicative function.
(b) Prove that $$\sigma(n^2)=\sum_{d\mid n} 2^{\omega(d)}.$$
I have done the part (a) and I am stuck by (b).
First, I set $d=p_1^{e_1}\cdots p_k^{e_{k}}$ be a factor of $n$. Then I don't know what's the next step.
Best Answer
Suppose we seek to show that $$\tau(n^2) = \sum_{d|n} 2^{\omega(d)}.$$
This can be done using Dirichlet series and Euler products. We have for the RHS and $$\sum_{n\ge 1} \frac{1}{n^s} 2^{\omega(n)}$$ the Euler product $$\prod_p \left(1 + \frac{2}{p^s} + \frac{2}{p^{2s}} + \frac{2}{p^{3s}} +\cdots\right).$$ which is $$\prod_p \left(-1 + 2\frac{1}{1-1/p^s}\right) = \prod_p \frac{-1+1/p^s+2}{1-1/p^s} \\ = \prod_p \frac{1+1/p^s}{1-1/p^s} = \prod_p \frac{1-1/p^{2s}}{(1-1/p^s)^2} = \frac{\zeta(s)^2}{\zeta(2s)}.$$
Therefore $$\sum_{n\ge 1} \frac{1}{n^s} \sum_{d|n} 2^{\omega(d)} = \frac{\zeta(s)^3}{\zeta(2s)}.$$
On the other hand we have $$\sum_{n\ge 1} \frac{1}{n^s} \tau(n^2) \\= \prod_p \left(1 + (2+1) \frac{1}{p^s} + (4+1) \frac{1}{p^{2s}} + (6+1) \frac{1}{p^{3s}} + (8+1) \frac{1}{p^{4s}} + \cdots\right).$$
This is $$\prod_p \left(1+\frac{1/p^s}{1-1/p^s} + \sum_{k\ge 1} \frac{2k}{p^{ks}} \right) \\ = \prod_p \left(1+\frac{1/p^s}{1-1/p^s} + 2 \frac{1/p^s}{(1-1/p^s)^2} \right).$$
To aid in simplification we put $z=1/p^s$ to get for the inner term
$$1 + \frac{z}{1-z} + \frac{2z}{(1-z)^2}$$
This simplifies to $$\frac{1+z}{(1-z)^2}.$$
On the other hand $$\frac{\zeta(s)^3}{\zeta(2s)} = \prod_p \frac{1-z^2}{(1-z)^3} = \prod_p \frac{1+z}{(1-z)^2}.$$
We have equality, QED.