I have frequently read and heard that given the ABC-conjecture a number of important unsolved problems of number theory can be solved (with relatively simple proofs). Among them, the celebrated Fermat's Last theorem is frequently mentioned.

So, my question is: Given that the $ABC$ conjecture is valid, can we prove that it implies Fermat's Last theorem ?

P.S.: I can understand that ABC conjecture "easily" implies the asymptotic FLT (stating that: "the equ-ation $x^n+y^n=z^n$ can have solutions in positive integers only for $n< n_0$, where $n_0$ is some finite number"). This is outlined in Lang's Algebra (p.196, 1994 edition), see also here and here.

  • 10
    $\begingroup$ No. The largeness of exponents in the asymptotic version depends crucially on what constant emerges (for a given specific epsilon) in an ABC proof. In particular, if ABC were to be proved with ineffective constants then the "largeness" would be totally mysterious, and Vesselin Dimitrov's "effective" version of Mochizuki's result (assuming the latter as a black box!) gives horrifically gigantic constants (massive exponentials, etc.). For this and other reasons, the proof of FLT via modularity methods seems likely to remain the only way to prove FLT and to understand why it is true. $\endgroup$ – nfdc23 Sep 18 '16 at 5:08
  • 1
    $\begingroup$ No, abc implies at most finitely many counterexamples to FLT, but it allows counterexamples. $\endgroup$ – joro Sep 18 '16 at 12:02
  • 3
    $\begingroup$ Actually, it isn't that inconceivable that ABC implies FLT, even though not in the obvious way - we might e.g. be able to prove that if there is a solution with exponent $n$, then there is a solution with an exponent $m>n$, in which case we would get a contradiction with ABC. I am not claiming I know how to prove FLT from ABC this way, just sharing a thought. $\endgroup$ – Wojowu Sep 18 '16 at 12:46
  • 3
    $\begingroup$ Here's one way to think about it. $ABC$ easily implies that if $n$ is sufficiently large, then the equation $X^n+Y^n=94151567435Z^n$ has no solutions in non-zero integers. But it could not imply that there are no solutions for all $n$, since in fact $(2,3,1)$ is a solution for $n=23$. So $ABC$ will, as others have said, rule out all sufficiently large $n$ (even for more general equations such as $aX^n+bY^n=cZ^n$), but then one will need some method of dealing with the remaining values of $n$. $\endgroup$ – Joe Silverman Sep 22 '16 at 0:50
  • 1
    $\begingroup$ Subsumed by mathoverflow.net/q/130980/41291 (imo) $\endgroup$ – მამუკა ჯიბლაძე Sep 22 '16 at 6:56

No, abc doesn't imply FLT.

For all exponents $n > 3$, abc implies at most finitely many counterexamples to FLT, but it allows counterexamples to FLT.

For exponent $n=3$ it allows infinitely many counterexamples and the Fermat-like equation $x^3+y^3=a z^3$ has infinitely many coprime solutions for some $a$ via the group law on the elliptic curve.

Basically finite number of abc triples of arbitrary quality don't contradict abc, while a single counterexample contradicts FLT.

Also, there is generalization of abc over number fields, widely believed to be true.

Over number fields FLT fails, e.g. $1^n+1^n=(\sqrt[n]{2})^n$.

| cite | improve this answer | |

Not the answer you're looking for? Browse other questions tagged or ask your own question.