TITLE: Does Fermat's Last Theorem imply $\sqrt{2} \not \in \mathbb{Q}$?
QUESTION [38 upvotes]: 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?

REPLY [2 votes]: COMMENT(not an answer!).-Very cordially I do not agree at all with this proof because in my opinion it does not use FLT but exposes a counterexample that contradicts FLT in quadratic fields which has been known very well for almost a century. 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 where for all $a,b$ we have for all $n$ that $a^n+b^n=c^n$ for $c=\sqrt[n]{a^n+b^n}$ but using this fact is not using FLT.