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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 $p \ne 3 r$?

Community
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tmaric
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    $p \in \mathbb N$.. how could it be that $p^2 = 3$? – George Aug 25 '15 at 11:08
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    anyway the fact follows because there's a theorem: if a prime number divides the square of a number it must also divide the number – George Aug 25 '15 at 11:09
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    This sort of exercise is usually given after the student is introduced to the idea that natural numbers can be uniquely factored into primes. If $p,q$ have no common factors, this amounts to saying they have no common prime factors (and 3 is a prime, so this is a particular case of unique factorization being used). – hardmath Aug 25 '15 at 11:17
  • @hardmath I edited the order of symbols, thanks. – tmaric Aug 25 '15 at 11:26

3 Answers3

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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 $3$.

chhro
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The usual articulation of the result you ask about is Euclid's lemma. Changing your notation slightly, Euclid's lemma says that if $p$ is prime, $a$ and $b$ are integers, and $p$ divides $ab$, then $p$ divides $a$ or $p$ divides $b$. Particularly, if $p$ divides $a^{2}$, then $p$ divides $a$.

Of course, if $p = 3$ it's also fine to do case by case analysis based on the division algorithm for integers.

Andrew D. Hwang
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Alright, let me explain troublesome part.

Notice, we have $$p^2=3q^2\tag 1$$ shows that the $p^2$ is divisible by $3$ which shows that $p$ is also divisible by $3$ (by fundamental theorem)

Hence, we have $$p=3m\tag 2$$ where, $m$ is some integer

Now, substituting $p=3m$ in (1), we get $$(3m)^2=3q^2$$$$\iff 3m^2=q^2$$

now, from above equation, we conclude that $q^2$ is divisible by $3$ which shows that $q$ is also divisible by $3$ (by fundamental theorem)

then we have $$q=3n\tag 3$$

Now, from (2) & (3), we find that $p$ & $q$ have a common factor $3$ which is a contradiction.

Harish Chandra Rajpoot
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