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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?

Bill Dubuque
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primes.against.humanity
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  • 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. – DanielWainfleet Apr 28 '22 at 20:58
  • 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.) – primes.against.humanity Apr 29 '22 at 07:03

2 Answers2

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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.

Mastrem
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    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. – primes.against.humanity Apr 28 '22 at 12:15
  • 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 $z_+^3+z_-^3=(6\times7)^3$. – J.G. Apr 28 '22 at 13:13
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    The example in this answer is also mentioned in this 2019 answer. – Bill Dubuque Apr 28 '22 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.

user21820
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Piquito
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  • 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? – user21820 Jun 03 '22 at 16:15
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    @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. – Piquito Jun 04 '22 at 20:12
  • 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! – user21820 Jun 05 '22 at 10:30
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    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. – Piquito Jun 06 '22 at 12:32
  • I have edited your post to attempt to capture what you wanted to say. Please check. If you're satisfied, we can delete all our comments. =) – user21820 Jun 06 '22 at 13:58
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    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) – Piquito Jun 07 '22 at 15:04
  • You can check the history (including your original version) by clicking on the "edited ..." link. So you can take a look to see if I miswrote anything. If you're satisfied, you can delete your comments and I'll delete mine. If you're not satisfied, you can either edit it further or rollback to your own version. Thanks for your reply! – user21820 Jun 07 '22 at 15:26