real analysis - What is the most rigorous proof of the irrationality of the square root of 3?

I am currently trying to self-study Stephen Abbott's Understanding Analysis. The first exercise asks to prove the irrationality of √3, and I understand the general idea of the contradiction by finding that the relatively prime integers p and q have a common factor. However, I am stuck on the idea that if p^2 is divisible by 3, then p is divisible by 3. Abbott's solution assumes this, but I have also seen proofs that analyze the situations where a and b are even or odd (such as NASA's). Even or odd really is just saying multiple of 2, which confuses me as to why the even/odd method (which is much less concise) would be used.

Sorry for the block of rambling text, I just want to start writing proofs the right way. I guess my real questions are:

If p^2 is divisible by a prime number, is p also divisible by that prime number? Can this just be assumed, or is there a theorem I have to mention in the proof?
Why do some proofs analyze the even/odd situations of a and b? Are they more rigorous, and if they are not, why are they used, considering their added length and complexity? Finally, am I simply over thinking the idea of being rigorous and missing the big picture?

Answer

personally I prefer to prove these results by contrapositive. If $p,q$ are coprime positive integers with $q>1$ then
$$p^{k}/q^{k}$$
is not an integer for any $k>0.$ This immediately implies the irrationality of all roots of $3.$ In fact, it proves that any root of an integer that is not an integer is irrational.

The key is, as everyone has already said, is that $p,q$ coprime implies that $p^{k}$ and $q^k$ are also coprime, and so $p^k/q^k$ is not an integer.

This is implied by that if a prime $m$ divides $ab$ then it divides $a$ or $b.$ So if it divides $p^k$ it divides $p$ and if it divides $q^k$ it divides $q.$ So no prime will divide both of them unless it divides $p$ and $q$ which we ruled out.

How do we prove the result that $m$ must divide $a$ or $b$? Either it divides $a$ and we are done or $m$ and $a$ are co-prime since $m$ is prime. We then have that there exist $\alpha$ and $\beta$ such that
$$\alpha m + \beta a =1.$$
(this follows from the Euclidean algorithm.)
So
$$b\alpha m + \beta ab =b$$

Since $m$ divides $ab$ it divides the LHS, so it divides the RHS too.

(See my book "proof patterns" for more discussion.)