What numbers have exactly 3 divisors?

Answer 1

An integer (call it 'x') has exactly 3 divisors if and only if it is the square of a Prime number.

In other words, to generate a list of integers with exactly 3 divisors, just keep squaring prime numbers.

A number with 3 divisors cannot be prime (a prime number has only 2 divisors, 1 and itself). So it must be a composite number, which is a number that can be factored as a product of prime numbers (Fundamental Theorem of Arithmetic) -- i.e. a composite number must have at least one prime divisor.

In the case where the number has only 3 divisors, two of them are 1 and the number itself (neither of which are prime). Therefore the third divisor must be a prime number.

So the three divisors of 'x' are: 1, p, x where p is prime.

Now since p is a divisor (or factor) of x, and the only other divisor besides 1 and x itself, x must equal p*p -- or x=p^2 . Obvious x can't equal p*x and if x = p*1, x=p so x is prime, or has only 2 divisors...

If x = p^(3) , then x = p*p* p , or p*(p^2) ... this means that p^2 would also have to be a divisor of x, and this would contradict with x having only 3 divisors. For the same reason, x = p^(greater than 3) is also not possible.

So the only possibility is that an integer with exactly 3 divisors is the square of a prime number "p". The divisors are 1, p, and p^2.

I'm sure there's a simpler, more elegant way of explaining this, but it should be clear enough.