Question

Find all numbers p such that p and p^2 are both prime. prove that you found them all

Answer 1

There are no such numbers. No prime number is a square of any integer.

If $p$ is a prime number, then it's divisible only by 1 and $p$ (itself).

If $p=1\cdot p$

then

$p^2=(1\cdot p)^2 =1\cdot p^2=1\cdot p \cdot p$

As we can see $p^2$ is divisible not only by 1 and $p^2$(itself), but also by $p$, which means it can't be a prime number.

Answer 2

0 and 1 are neither prime nor composite. A prime is any number greater than 1 that has just 1 and itself as factors. Primes can only start at x > 1

When that happens (when you start with numbers greater than one) p^2 is a composite consisting of 2 primes, so any composite will obey the law that he number will have at least 3 factors making it up -- in this case p p^2 and 1.

So the answer to the question by definition is that 0 numbers can have the property of both p and p^2 to be prime.