(1)... I like this proof
(2)... Fermat's Last Theorem is not strong enough!
(3)... circular argument ... not actually valid.

The circularity is non-trivial.

I've always like Euclid's proof of the infinitude of primes, and Cantor's diagonal argument that there are more real numbers than there are integers.

But my favourite would have to be:

Seems so
But the circularity is not only non-trivial, but non-existent, too.

Pardon me: I don't have to go into the details of the proof
of any theorem I use for my proof. If the proposition, I am going
to prove, is already a trivial subcase of the theorem, then so
what? That is no problem and there is no circulus vitiosus.

If you want to convince me of having used a circular argument
you should point to a certain line in my proof and say:
"Ha, here you are - that is something you use as if already
proved!"

For your convenience and easier pointing-out here are the
lines of my proof, numbered:

1. Assume natural numbers p and q such that (p/q)^3 = 2.
2. Then p^3 / q^3 = 2,
3. i.e. p^3 = 2 * q^3 = q^3 + q^3.
4. Let a = q, b = q, c = p.
5. Then a^3 + b^3 = c^3 for some natural numbers a, b and c.
6. That contradicts Fermat.
7. Thus the assumption led to a contradiction.
8. q.e.d.

The first error is in line ....?

Cheers,
Rainer

I always liked the integral of a gaussian over the reals.

Integral(e^(-x^2)) from -Inf to Inf

1) Square the integral,
2) Then change from cartesian to polar,
3) Solve integral
4) Take the square root
5) Square again
6) Multiply by 2
7) Have dessert

6. (Or are you being pedantic about the fact that no step in the proof assumes Fermat, so 6 -> 7 is non sequitur?)

But I doubt you can call that a circulus vitiosus.

If you do not assume that Fermat's theorem is true, then our proof is false by non sequitur.

... an enthymeme ... false by petitio principii ...