Question
Let $p$ be a prime number. If $p$ divides $a^2$, then $p$ divides $a$, where $a$ is a positive integer.

Answer

Let the prime factorisation of $a$ be as follows :
$a=p_1 p_2 \ldots p_n$, where $p_1, p_2, \ldots, p_n$ are primes, not necessarily distinct.
Therefore, $a^2=\left(p_1 p_2 \ldots p_n\right)\left(p_1 p_2 \ldots p_n\right)=p_1^2 p_2^2 \ldots p_n^2$.
Now, we are given that $p$ divides $a^2$. Therefore, from the Fundamental Theorem of Arithmetic, it follows that $p$ is one of the prime factors of $a^2$. However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of $a^2$ are $p_1, p_2, \ldots, p_n$. So $p$ is one of $p_1, p_2, \ldots, p_n$.
Now, since $a=p_1 p_2 \ldots p_n, p$ divides $a$.
We are now ready to give a proof that $\sqrt{2}$ is irrational.
The proof is based on a technique called 'proof by contradiction'. (This technique is discussed in some detail in Appendix 1).

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