Odd number of divisors

General case:

For a number $ab$ , where $a,b,c$ are prime, its factors are $1, a, b, ab$ , and for a number $c^2$ , its factors are $1,c,c^2$ .

This is because since $ab \ | \ b$ and $ab \ | \ a$ (which is trivial), $a,b$ are factors of $ab$ , and by definition, $1$ and $ab$ are factors of $ab$ . We can confirm there are no more factors by calculating $(a+1)(b+1)$ . By definition of $a,b$ being prime, the index of $a,b$ is $1$ , so there are $(1+1)(1+1)$ factors, which equals $4$ .

We can observe the second case as a special case of the first, where $a=b$ . Therefore, $a$ and $b$ are double-counted, and there is only one factor which is not $1$ or $ab$ , which is $c$ . Again, finding the number of factors in $c^2$ requires finding the index, which is $2$ . Therefore, there are $2+1$ or $3$ factors, which makes $1,c,c^2$ the only factors.