divisibility - Prove that $sqrt{3}$ is not a rational number

There is a similar question however that question asks why $3 |p^2$. Here the question is about $3 | p^2 \rightarrow 3 | p$.

It is a simple exercise (1.2.1) from Abbot's "Understanding Analysis".

$\nexists p,q \in \mathbb{N} : \left(\dfrac{p}{q}\right)^2 = 3$

$p$ and $q$ have no common factors, otherwise they would cancel each other out.

Contradiction:

$p^2 = 3q^2$

The troublesome part for me is

From this, we can see that $p^2$ is a multiple of 3 and hence $p$ must
also be a multiple of 3.

Why is that?

The proof goes on with

$p = 3 r$

$q^2 = 3r^2 \rightarrow q = 3 \lambda$.

It concluses now that $q$ and $p$ have common factors, which invalidates the original statement.

How can I prove $p^2 = 3q^2 \rightarrow p = 3 r$? If this is not true, then I don't see how the contradiction proves the statement, since it only invalidates it for such $p,q$ that have common factors.

What if $p^2 = 3q^2$ and $p \ne 3 r$?

Answer

If $p^2=3m$ for some $m\in \mathbb Z$, then $p=3k$ for some $k\in \mathbb Z$. Suppose otherwise. By Division Algorithm, $p=3k+1$ or $p=3k+2$. Squaring $p$, we get either $9k^2+6k+1$ or $9k^2+12k+4$ which are both not multiples of $3$, contradicting the assumption. Thus, $p$ has to be a multiple of $3$.