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}
a contradiction to FLT.)
Is there a proof along these lines that $\sqrt{2} \not \in \mathbb{Q}$ using Fermat's Last Theorem?
 A: $$
\left(18+17\sqrt{2}\right)^3 + (18-17\sqrt{2})^3 = 42^3,
$$
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.
A: 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.
