Via my friend Reid Atcheson, I find the following “funny” (for some definition of “funny”) proof of the irrationality of the cube root of 2.
First, “recall” the standard proof of the irrationality of any root of 2. Take, for instance, the square root of 2. For purposes of contradiction, assume that the square root of 2 is rational. Then we can write sqrt(2) = p/q for some integers p/q. Moreover, we can define p/q as the irreducible form of the fraction, so p and q don’t share any divisors. Then we can write:
sqrt(2) = p/q
2 = p2 / q2
p2 = 2 q2
This means p2 is even, which means p is even, so we can write p = 2k for some k. Now rewrite the previous:
(2k)2 = 2 q2
4 k2 = 2 q2
2 k2 = q2
This means q2 is even, which means q is even, which contradicts that p and q share no divisors.
You can do this with any root of 2: 2 ^ (1/n) = p/q implies 2 = p^n / q^n, so p^n = 2 q^n, so p^n is even, so p is even, and so on.
Now let’s instead do the “funny” proof. Consider the cube root of 2:
2 ^ (1/3) = p/q
2 = p3 / q3
2 q3 = p3
q3 + q3 = p3
Whoops, this violates Fermat’s Last Theorem! You can do this for all roots 3 and above, but unfortunately Fermat’s Last Theorem isn’t strong enough to prove it for square roots.
Note: This proof is discussed here. Whenever you use something super-complicated to prove something super-simple, you have to check that the super-complicated thing doesn’t rely on the super-simple thing in some subtle way. One of the commenters on the thread cited above says there is circularity here:
This argument is essentially circular. Indeed, we can assume n is prime (just like for FLT) and then the proof of FLT first passes from a hypothetical nontrivial solution to a^n+b^n=c^n for prime n>2 to a suitable “Frey curve” y2=x(x−an)(x+bn) where one has to rig certain congruential and gcd conditions on (a,b,c), including that a, b, and c are pairwise coprime. Yet that step applied to (p,q,q) is exactly what would be the “Euclid-style” proof that 2 is not a rational nth power. Hmm, another disguised version of a Euclid proof....
But it seems from this discussion that what’s going is not circularity, but rather that Fermat’s Last Theorem is, down in the details, essentially doing a traditional-style proof, so using Fermat’s Last Theorem isn’t really solving the problem from a different direction.