# Does Fermat's Last Theorem imply $\sqrt{2} \not \in \mathbb{Q}$?

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}

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

• 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. Apr 28 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.) Apr 29 at 7:03

$$\left(18+17\sqrt{2}\right)^3 + (18-17\sqrt{2})^3 = 42^3,$$ 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.

• 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. Apr 28 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 at 13:13
• The example in this answer is also mentioned in this 2019 answer. Apr 28 at 14:05

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.

• 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? Jun 3 at 16:15
• @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. Jun 4 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! Jun 5 at 10:30
• 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. Jun 6 at 12:32
• 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) Jun 7 at 15:04