For the number of divisors function, $d(n)$, deduce all positive integers when $d(n)$ is odd

Question

From the definition I have deduced $d(n)$ is multiplicative by letting $n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$, where

$$d(p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k})= (a_1+1)\cdots(a_k+1)=d(p_1^{a_1})\cdot d(p_2^{a_2})\cdots d(p_k^{a_k})$$

as we know the divisors of $n$ are $d=p_1^{b_1}p_2^{b_2}\cdots p_k^{b_k}$ where each $b_k$ gives a distinct divisor in the form of $(a_k+1)$. However, I am unsure how to find the given integers, $n$, for the two points below;

• $d(n)$ is odd
• $d(n)= p_0,$ where $p_0$ is a fixed prime.

Answer 1

For $d(n)$ to be odd all the expressions $(a_j+1)$ must be odd, thereby implying each exponent $a_j$ must be even. So the number $n$ must be a perfect square.

Likewise for $d(n)$ to be prime, note that we can only have at most one term in the product representation of $d(n)$.

Answer 2

Hint 1: In order for $d(n)$ to be odd, each of $a_1+1, a_2+1, \ldots, a_k+1$ must be odd.

Hint 2: In order for $d(n)$ to be odd, prove that $k=1$ and that $a_1+1$ must be prime.