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A well-known overkill proof of the irrationality of $2^{1/n}$ ($n \geqslant 3$ an integer) using Fermat's Last Theorem goes as follows: If $2^{1/n} = a/b$, then $2b^n = b^n + b^n = a^n$, which contradicts FLT. (See this, and see this comment for the reason this is a circular argument when using Wiles' FLT proof)

The same method of course can't be applied to prove the irrationality of $\sqrt{2}$, since FLT doesn't say anything about the solutions of $x^2 + y^2 = z^2$. Often this fact is stated humorously as, "FLT is not strong enough to prove that $\sqrt{2} \not \in \mathbb{Q}$." But clearly, the failure of one specific method that works for $n \geqslant 3$ does not rule out that some other argument could work in the case $n = 2$ in which the irrationality of $\sqrt{2}$ is related to a Fermat-type equation.

(For example, if we knew that there are integers $x,y,z$ such that $4x^4 + 4y^4 = z^4$, then with $\sqrt{2} = a/b$, we would have $a^4 x^4 / b^4 + a^4 y^4 / b^4 = z^4$ and hence

\begin{align} X^4 + Y^4 = Z^4, \quad \quad (X, Y, Z) = (ax, ay, bz) \in \mathbb{Z}^3, \end{align}

a contradiction to FLT.)

Is there a proof along these lines that $\sqrt{2} \not \in \mathbb{Q}$ using Fermat's Last Theorem?

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  • $\begingroup$ If $x,y,z\in \Bbb Z$ and $4x^4+4y^4=z^4$ then $z$ is even so let $z=2w.$ And let $v=2w^2.$ Then $x^4+y^4=v^2$... This has no solution except $x=y=v=0$. I gave an elementary proof on this site once. It can also be found in textbooks. $\endgroup$ Apr 28 at 20:58
  • $\begingroup$ Alright! I have read about the solutions to $x^4 + y^4 = z^2$, but I didn't see the connection to $4x^4 + 4y^4 = z^4$. (Of course, $4x^4 + 4y^4 = z^4$ was just an example of an equation which FLT doesn't say anything about, but which can be reduced to $X^4 + Y^4 = Z^4$ provided that $\sqrt{2}$ is rational.) $\endgroup$ Apr 29 at 7:03

2 Answers 2

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$$ \left(18+17\sqrt{2}\right)^3 + (18-17\sqrt{2})^3 = 42^3, $$ so $\sqrt{2}\in \mathbb{Q}$ would contradict FLT (once you know that $\sqrt{2}\not\in\{\pm 18/17\}$ of course).

Source: this article, which also show that this is 'the only way' to show $\sqrt{2}$ is irrational using FLT, because FLT is almost true in $\mathbb{Q}(\sqrt{2})$ -- only in exponent $3$ do we get counterexamples and all of them are 'generated' (see Lemma $2.1$ and the discussion immediately following its proof at the bottom half of page $4$) by the counterexample given above.

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  • $\begingroup$ Thanks for the answer and the article! I'll definitely check it out. I'm a bit surprised that that this is essentially the only way to prove it using FLT. $\endgroup$ Apr 28 at 12:15
  • $\begingroup$ As for a relatively low-calculation proof $z_\pm:=18\pm17\sqrt{2}\implies z_+^3+z_-^3=42^3$, since $z_++z_-=6^2$ and $z_+z_-=-254$,$$z_+^2-z_+z_-+z_-^2=(z_++z_-)^2-3z_+z_-=1296+762=2058=6\times7^3,$$so $z_+^3+z_-^3=(6\times7)^3$. $\endgroup$
    – J.G.
    Apr 28 at 13:13
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    $\begingroup$ The example in this answer is also mentioned in this 2019 answer. $\endgroup$ Apr 28 at 14:05
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One can generalize this beyond $\sqrt{2}$, showing that $2$ is not special at all. For example, for rational $k$ other than $0$ and $-1$, we have the identity $$\left(3+\sqrt{-3(1+4k^3)}\right)^3+\left(3-\sqrt{-3(1+4k^3)}\right)^3+(6k)^3=0$$ Obviously it is not trivial to know that FLT is not valid in quadratic fields as it is in the reals (since for all $a,b∈ℝ$ and $n∈ℕ^+$ we have $a^n+b^n=c^n$ for $c=\sqrt[n]{a^n+b^n}$), but as the above identity shows, it is not hard either and essentially the same for all quadratic fields.

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  • $\begingroup$ I do not understand why you disagree. If $\sqrt{2}∈ℚ$ then the stated identity in Mastrem's answer is of the form "$(a/d)^3+(b/d)^3 = c^3$" with $a,b,c,d ≠ 0$, contradicting FLT on multiplying by $d^3$. So what's your point? $\endgroup$
    – user21820
    Jun 3 at 16:15
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    $\begingroup$ @user21820: The same can be "a proof" that if $-3(1+4k^3)=t^2r$ (see above) where $r$ is square-free then $\sqrt r$ is irrational. And this proof would be using FLT. No, the formule used above is exactly a proof that FLT is not valid in quadratic fields. $\endgroup$
    – Piquito
    Jun 4 at 20:12
  • $\begingroup$ Maybe you are not a native English speaker, but my point is simply that you are wrong to say "I do not agree with [Mastrem's] proof", since there is nothing wrong with it. You may want to say additional things, but saying you disagree with a proof when it is correct is just wrong! $\endgroup$
    – user21820
    Jun 5 at 10:30
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    $\begingroup$ Maybe you are right: it is true that I am not a native English speaker and maybe what I have did is to show that the asked proof is trivial to do keeping in mind the irrational involved can be parameterized by the identity above. Regards. $\endgroup$
    – Piquito
    Jun 6 at 12:32
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    $\begingroup$ I don't remember what i have said but i assume you are not malicious with me. So i am satisfied with you have written whitout knowing what it is. Regards. (My English is deficient) $\endgroup$
    – Piquito
    Jun 7 at 15:04

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