@TonyK does that not fail in the last step though, where we check for all integers between $1$ and the number in question ($4$ in that case)?

"it can be trivially proven that if $a^2/b^2$ is an integer then $a/b$ is an integer" : can you explain why ? (the flaw is very often in the "trivial" part)

The proof appears multiple times even on this site alone, see e.g. here

@Trebor I'm sure it's my understanding here that's at fault, but that proof seems a bit more complex to me and includes details like the unique factorization theorem and other suppositions that this proof doesn't rely on. If it's not entirely obvious, could you please create an answer to clarify how these two proofs are the same?

"if $a^2/b^2$ is an integer then $a/b$ is an integer" (where $a$, $b$ are integers) is exactly the desired claim that $\sqrt n$ is rational only when $n$ is a perfect square,

Why does $\operatorname{gcd}(a,b)=1$ imply $\operatorname{gcd}(a^2,b^2)=1$, and why does $\operatorname{gcd}(x,y)=1\land x/y\in\Bbb Z$ force $y=1\lor y=-1$ ?

@zkutch we're still working with $a$ and $b$ in the proof, which are coprime integers, so the counterexample does not seem to apply.

@SassatelliGiulio for the latter, we're working with $x$ and $y$ coprime, otherwise $y=\pm1$ would indeed not be a requirement. As for the former, this was something I thought was fair to be assumed true; a proof can be found here.

@HagenvonEitzen So, just to clarify, this is easily generalised to a valid proof of $\sqrt{\mathbb{N}}$ is either an integer or irrational then?

@TonyK sorry, the only edits/additions I have made are after the bar. But thank you for raising the idea, I was sure that'd prove it wrong too.

Sorry, I missed that. I have deleted my comments.

This common proof, as well as many closely related proofs - are already presented in many prior answers here, e.g. see the inked dupes and the posts linked there. Please search for answers before asking questions. $\ \ $

There is already much prior discussion of the common gap / circularity mentioned by @Hagen above, including even further discussion on meta.

"If $b^2/a^2$ is an integer, then $b/a$ is an integer." Really?? Let $b = \sqrt{3}$ and $a=1$....

@BillDubuque I wish the link you added was one of the dupe targets. I did find both the current dupe targets before writing this question; it still seems to me that the second doesn't contain the proof in my question, and though I do see that the first is correct now, it contained some extra details that confused me. Only now after your link specifically do I see the circular nature of the argument. I don't know if you have the valid permissions (via gold badge or whatever) but if you do, please add your link to the list of duplicates.

@DavidG.Stork We're working with coprime integers here, so your counterexample doesn't apply.