From the text of the question posed here: "How many whole numbers less than 2010 have exactly three factors?" this statement is made:
If there is no fourth factor, then that third factor must be the square root of the number. Furthermore, that third factor must be a prime, or there would be more factors.
I don't understand this. Can you explain please.
Proof. Let $n\in \mathbb{Z}$ be a number with exactly 3 factors. Two of these factors must be 1 and $n$ since $1|n$ and $n|n$. Let $k\in \mathbb{Z}$ such that $k|n$ and $k\neq 1$, $ k\neq n$. Since $k$ is the only remaining factor of $n$ then some power of $k$ must divide $n$. However, if $k^x=n$ for $x>2$ then $k^{(x-1)}$ is also a factor of $n$. Therefore $k*k=n$. Furthermore, if $k$ is not prime then it can be represented as it's factors, $k=l*m$. We can substitute this as $l^2*m^2=n$. But this would imply $l|n$ and $m|n$. So, if a number has exactly three factors, then one of them is a prime and the square root of an integer.
Going with Henry's interpretation here. Remember that factors generally come in pairs. For example, $12=1\times 12$ or $2\times6$ or $3\times 4$. Any positive integer greater than $1$ has at least $2$ factors: $1$ and itself. If an integer $n$, greater than 1, has a factor $k$, then $\frac nk$ must also be a factor. So if $n$ has only $3$ factors, we must have $k=\frac nk$ and $n=k^2$.
It is easier than the other answers suggest.
If $k|n$ then also $\frac nk|n$, so the divisors of $n$ come in pairs unless $k=\cfrac nk$ or equivalently $n=k^2$
So the only numbers which have an odd number of divisors are squares.
Every positive integer greater than $1$ has the two divisors $1$ and $n$.
A positive integer with precisely three divisors must be a square $n=k^2$, which has divisors $1, k, n$. If $p|k$ then $p|n$ and $p$ is a further divisor unless $p=k$. So $n$ must be the square of a prime.
More generally, you may want to show that the number of divisors of $$n=p_1^{n_1}\cdot p_2^{n_2}\cdot p_3^{n_3} \dots p_r^{n_r}$$ (where this is the prime factorisation of $n$ into powers of distinct primes) $$(n_1+1)(n_2+1)(n_3+1)\dots (n_r+1)$$
The number must be the square of a prime to have exactly three factors.
The number itself, $n$, and $1$ are two of them. This leaves one more factor, which itself must not have any other factors besides $1$ and itself, which are counted already. Hence this last factor must be a prime, $p$.
The fact that there are no other factors besides $1, p, n$ means that the $n = p^2$. The value of $n$ cannot be some higher power of $p$ because then $p^2$ would be a separate (fourth) factor.