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 $3$.
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