A website ( http://www.math.unicaen.fr/~nitaj/abc.html#Consequences ) says that the $abc$ conjecture implies that there are only finitely many solutions to the equation $x^n+y^n=z^n$ with $\gcd(x,y,z)=1$ and $n\ge 4$. This one I have proven.

Lang's Algebra (p. 196) says that the $abc$ conjecture implies that for all $n$ sufficiently large, there are no solutions to the equation $x^n+y^n=z^n$ with $x,y,z\ne 0$. This one I have not proven (is it true?).

What are the strongest known assertions about Fermat's Last Theorem that follows from the $abc$ conjecture, and how are they proven? I searched the web but people tend to just say vague things like "implies asymptotic version of FLT" and such.


1 Answer 1


There is a slight ambiguity what "the ABC conjecture" is as there are some variation. However, the most common and what you likely mean is this fomulation (or something equialent to it):

For every $\epsilon >0$, there is a $C_{\epsilon}$ such that: if $a+b=c$, with positive coprime intergers, then $$c < C_{\epsilon} \ \text{rad}(abc)^{1 +\epsilon}.$$

Now, for the equation $x^n + y^n = z^n$ this means that $$ z^n < C_{\epsilon}\ \text{rad}(x^ny^nz^n)^{1 +\epsilon}. $$ Yet $\text{rad}(x^ny^nz^n) = \text{rad}(xyz)$ so one actually has $$ z^n < C_{\epsilon} \ \text{rad}(xyz)^{1 +\epsilon}. $$ and since $z$ is the largest and since $\text{rad}(a) \le a$ this further means $$ z^n < C_{\epsilon} z^{3 + 3\epsilon}. $$ Now, take $\epsilon =1/4$, say. Then on the one hand this cannot hold for any $z>1$ for $n$ sufficiently large (so no solution for large $n$, what you ask) and also not for any $n \ge 4$ fixed for $z$ sufficiently large (so only finitely many for fixed $n$) or also only finitely many couples $(z,n)$ that fulfill this (for $n \ge 4$).

[Added:] Since the question was changed to remove the gcd condition in Langs's version, I add for completeness, that each solution (for given $n$) implies the existence of a coprime one (for this $n$) so since above establishes there are no coprime solutions for some $n$ then there are none at all. [End Added]

However, to make these things effective/explicit one would need to know something about $C_{\epsilon}$ (in dependence of $\epsilon$).

Regarding "strongest possible": I think this is about what can be said, from the conjecture in the way I stated it. If one assumes stronger conjectures where one would have an explicit dependence of $C_{\epsilon}$ on $\epsilon$ then one could give explicit bounds. But (I think) the argument essentially always passes throught the last displayed equation and checking what this yields.

  • $\begingroup$ Thanks, quid. So there are two asymptotic versions of FLT which none of my sources cared to distinguish! $\endgroup$
    – Favst
    May 17, 2013 at 17:41
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    $\begingroup$ You are welcome. Yes in some sense one can consider so to say "two limits" (in $z$ and in $n$) and thus there are in some sense different versions. Yet the finiteness of all solutions (under n> 3) contains all, as if there are only finitely many in total then there is a largest $n$ and a largest $z$ and so on. And, while I read it differently at first, thus my edit, I think the version you link to actually is meant to assert the finiteness of the set of all solutions , ie couples $(z,n)$ and thus contains Lang's. $\endgroup$
    – user9072
    May 17, 2013 at 17:49
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    $\begingroup$ If we would know that $C_1=1$ for $\epsilon =1$ (and no counterexample is known to that), then FLT would follow for exponents $n\ge 7$. $\endgroup$ May 17, 2013 at 17:52
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    $\begingroup$ @Dietrich Burde: I think even for $n \ge 6$, as the inequality is strict. $\endgroup$
    – user9072
    May 17, 2013 at 20:13
  • $\begingroup$ @DietrichBurde Please see mathoverflow.net/questions/303141 $\endgroup$ Dec 14, 2018 at 1:10

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