Explain the proof that the root of a prime number is an irrational number

Though the proof of this is done in a previous question, I have some doubt about a certain concept. So I ask to clarify it.

In the proof we say that $\sqrt{p} = \frac{a}{b}$ (In their lowest form).
Now
$$p = a^2 / b^2\\p\cdot b^2 = a^2.$$

Hence $p$ divides $a^2$ so $p$ divides $a$. We say that the above mentioned condition ("Hence $p$ divides $a^2$ so $p$ divides $a$") is valid as $p$ is a prime number. I didn't get the fact that why this is only true for prime numbers. Could someone please me this?