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Edit: I'm beginning to suspect I either misread the sources or perhaps something wasn't stated, but it's my guess now that there is no nontrivial way of showing the ABC conjecture implies Fermat's Last Theorem in full.

Original:

If we assume the ABC conjecture, then one can prove with relative ease that $x^n+y^n=z^n$ possibly has solutions in positive integers only for $n< n_0$, where $n_0$ is some finite number. This is the "Asymptotic Fermat's Last Theorem."

My question is, does this somehow imply with further calculations that the full Fermat's Last Theorem follows? Indeed I've read that ABC conjecture implies the full FLT and I thought it would follow from the asymptotic case, again bootstrapping the ABC conjecture.

The proof for the asymptotic case goes as follows:

Recall that the ABC conjecture states that for any $\epsilon>0$, there exists an $N(\epsilon)>0$ such that for all nonnegative $a,b,c$ relatively prime with $a+b=c$, one has $c\leq N(\epsilon)\mbox{rad}(abc)^{1+\epsilon}$.

Let $x^n+y^n=z^n$, where $x,y,z$ are relatively prime, so that $\mbox{rad}(x^ny^nz^n)=\mbox{rad}(xyz)\leq xyz\leq z^3$. Supposing $n\geq 3$, we get that $z\geq 3$, so invoking the ABC conjecture with $\epsilon=1$ and $K:=\max(1,N(1))$, we get

$$z^n\leq N(\epsilon)\mbox{rad}(x^ny^nz^n)<Kz^6$$

so

$$n<6+\frac{\log K}{\log z}\leq 6+\frac{\log K}{\log 3}=:n_0$$

and we are done.

Is there a way to push this toward the full proof of FLT? I feel like we would need to consider the reformulation of the ABC conjecture, that for any $\epsilon>0$, the set of exceptions $c>\mbox{rad}(abc)$ is finite, so that there is some $m$ such that $c\leq (abc)^{m}$ (for every $a+b=c$) and therefore FLT holds for every integer that is at least $3m$. Is seems impossible to suggest that $m=1$ since there are plenty of counterexamples of triples $(a,b,c)$ for which this is false. Perhaps one would need to show no counterexamples exist when $a,b,c$ satisfy $a^3+b^3=c^3$?

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If you believe the $abc$ conjecture is true with $N(1) = 1$ (that is, $c < {\rm rad}(abc)^2$, and no counterexample to that is known) then the abc conjecture implies FLT for exponents above 6. Your argument shows that, taking $N(1) = 1$. –  KCd Oct 11 '12 at 2:23
    
@KCd: This is true. Would there be a way to then to show that $a^3+b^3=c^3$ implies $c<rad(abc)^2$? –  Alex R. Oct 11 '12 at 3:52
2  
Alex: You have your request backwards. You must have meant to ask if the bound on c implies FLT for exponent 3. Going the other way around is completely unrealistic, since then a known theorem (FLT for exponent 3) would imply the abc conjecture with epsilon = 1. In any event, off the top of my head I don't see how to get FLT for exponent 3 from that special instance of abc, but I also don't consider it too bad, because FLT for exponent 3 is very classical. Look, if you had abc then you get FLT for exponents above 6, so just be happy and appeal to known theorems for the lower exponents. –  KCd Oct 12 '12 at 1:35

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