Does ABC implies Fermat's last theorem?

Question

I read from the newspaper that Mochizuki's proof of the ABC conjecture implies the Fermat's last theorem. Is it true? I think it implies the proof only for large enough exponents?

Answer 1

On Wikipedia there is a proof of Fermat's Last Theorem for $n \geq 6$ given a "strong version" of the ABC conjecture, in which we assume for coprime $x, y, z$ where $x + y = z$ that $z < rad(xyz)^2$. It is as follows:

Letting $a^n + b^n = c^n$ and assuming WLOG that they are coprime, we have $$c^n < rad(a^n b^n c^n)^2 = rad((abc)^n)^2 = rad(abc)^2 \leq \; (abc)^2 < (c^3)^2 = c^6$$ And so any integer solution must have $n < 6$. Since individual proofs exist for the 3, 4, and 5 case, FLT follows.

Answer 2

From Wolfram MathWorld, Fermat's Last Theorem for sufficiently large powers is a consequence of the abc conjecture.

So, if Mochizuki's proof is valid, then it would show this.