I have been given a question which begins with me deducing $d(n)$ is multiplicative, I know $d(n)= \sum_{d|n} 1,$ and obtain a formula of $d(n)$ in terms of prime decomposition (once again I know is $p_1^{a_1}p_2^{a_2}\cdot\cdot p_k^{a_k}$). I am aware that in order for a function to be multiplicative $f(nm)=f(n)f(m)$ where $(m,n)=1$ and f is non zero. The final part is deduce all positive integers $n$ for when;
$\bullet$ $d(n)$ is odd
$\bullet$ $d(n)= p_0,$ where $p_0$ is a fixed prime.
I have deduced $d(n)$ is multiplicative by letting $n=p_1^{a_1}p_2^{a_2}\cdot\cdot p_k^{a_k}$, where
$$d(p_1^{a_1}p_2^{a_2}\cdot\cdot p_k^{a_k})= (a_1+1)\cdot \cdot(a_k+1)=d(p_1^{a_1})\cdot d(p_2^{a_2})\cdot \cdot \cdot d(p_k^{a_k})$$
as we know the divisors of $n$ are $d=p_1^{b_1}p_2^{b_2}\cdot\cdot p_k^{b_k} $ where eack $b_k$ gives a distinct divisor in the form of $(a_k+1)$. It is the next bit I do not understand.