$\sqrt[n]{2}$ is irrational for $n\ge 3$ (proof using Fermat’s last theorem)

Theorem 1.

If $n\mathrm{\ge }\mathrm{3}$, then $\sqrt[n]{\mathrm{2}}$ is irrational.

The below proof can be seen as an example of a pathological proof.It gives no information to “why” the result holds, orhow non-trivial the result is.Yet, assuming Wiles’ proof does not use the above theorem anywhere,it proves the statement. Otherwise, the below proof would be anexample of a circular argument.

Proof.

Suppose $\sqrt[n]{2}=a/b$ for some positive integers $a,b$. It follows that$2=a^n/b^n$, or

$b^n+b^n$$=$$a^n.$(1)

We can now apply a recent result of Andrew Wiles [1],which states that there are no non-zero integers $a$, $b$ satisfying equation (1).Thus $\sqrt[n]{2}$ is irrational.∎

The above proof is given in [2], where it is attributed to W.H. Schultz.