Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

[Edit: I've since realized that my question is confused: in particular, the minimum value of k that you need to sum from increases with the largest exponent under consideration so that the sum over all exponents consists of terms that do go to zero exponentially. I've left the question up for people who have the question that I did in the future. See also Terence Tao's post titled The probabilistic heuristic justification of the ABC conjecture which is relevant.]

The Fermat-Catalan conjecture states that there are only finitely many sex-tuples $(a, b, c, d, e, f)$ of positive integers such that

(i) $\gcd(a, b, c) =1 $,

(ii) $a^d + b^e = c^f$,

(iii) $\frac{1}{d} + \frac{1}{e} + \frac{1}{f} > 1$.

If one fixes each of the exponents, one can make a heuristic argument for the finiteness of the solutions to the resulting equation as follows.

Let $S_d$, $S_e$ and $S_f$ be random subsets of the positive integers such that the number of $x \leq n$ in $S_\alpha$ is the floor of $n^\frac{1}{\alpha}$.

We can give an upper bound for the expected number of triples $x_d, x_e, x_f$ such that $x_i \in S_i$ and $x_d + x_e = x_f$ by giving an upper bound for the expected number of solutions $E_k$ in the interval $(10^{k-1}, 10^{k}]$ and then by computing $\sum_{k=1}^{\infty} E_k$. To bound $E_k$, we bound the $x_i$ by $10^{\frac{k}{i}}$. The expected number of solutions is bounded above by the product of the density of eligible pairs $(x_d, x_e)$ with sum in the interval with the density of $x_f$ in the interval all multiplied by the cardinality of $(10^{k-1}, 10^{k}]$. This is bounded above by

$\frac{10^{k(1/d + 1/e)}}{10^k - 10^{k -1}} \cdot \frac{10^{k/f}}{10^{k} - 10^{k-1}} \cdot (10^k - 10^{k-1}) = C \cdot 10^{k (1/d + 1/e + 1/f - 1)}$

Where $C$ is independent of $d, e, f, k$. Summing over $k$ gives a finite quantity of

$Q_{d, e, f} = \frac{10C}{1 - 10^{\alpha}}$

provided that $\alpha < 0$ where $\alpha = 1/d + 1/e + 1/f - 1$.

That's all fine and good, but what confuses me is the fact that the finiteness is expected even if one is allowed to vary the exponents. Say we restrict ourselves to $d = e = 4$ and allow $f$ to vary. Then the expected number of solutions is bounded above by

$\sum_{d = 3}^{\infty} Q_{4, 4, d}$

But the terms of this sum don't tend toward zero, so the sum diverges.

Either I've made an error, gave too much away in one of the approximations that I made in connection with random sets of suitable densities, or the Fermat-Catalan conjecture is getting at something more refined than what density arguments predict. Which of these is the case? If it's the last of the three, is there a more refined heuristic that predicts the truth of the conjecture? (I'm aware that it follows from the abc-conjecture, but an appeal to that without explanation would assume the conclusion.)

share|improve this question
    
This is far from my area, so forgive me if I am simply very confused. If $\frac1f > 1 - \frac1d - \frac1e$, then $f$ ranges over only a finite list. Well, we could have $d = e = 2$, in which case the list is infinite... –  Theo Johnson-Freyd Sep 24 '12 at 2:00
    
Hi Theo, Thanks, I had the inequality reversed. –  Jonah Sinick Sep 24 '12 at 2:04
    
That reversion was incorrect; look at the wikipedia on the subject. This i also somehow obvious: the "greater than 1"-version fixes a very low upper bound on d,e,f. But the catalan-fermat-conjecture is surely concerned with a lower bound on that variables... –  Gottfried Helms Sep 24 '12 at 7:17

1 Answer 1

Warning: I don't actually know anything about this subject.

It seems to me that for fixed $d,e,f$ we can get a much better upper bound than you've given. Namely: Among the first $N$ positive integers, I expect roughly $N^{1/d}$ d'th powers and $N^{1/e}$ e'th powers, hence $N^{1/d+1/e}$ numbers of the form $A^d+B^e$. And I expect roughly $N^{1/f}$ f'th powers, hence $N^{1/f}$ of the form $C^f$. So the probability that a given number is of both forms is $${N^{1/d+1/e}\over N}\cdot{N^{1/f}\over N}$$ For $1/d + 1/e+ 1/f < 1$, this is less than $1/N$. Thus the expected number of solutions among the first $N$ numbers is less than $(1/N)\cdot N$, i.e. less than 1.

So for a given $d,e,f$, the expected number of solutions is not just finite; it's less than 1. This suggests that for a given $d,e,f$ there should be no solutions at all.

The question now is why we expect only finitely many solutions when $d,e,f$ are allowed to vary. I don't immediately see how we can get there from here, but since we do have a heuristic that tells us to expect zero solutions for any given $d,e,f$, perhaps there's another heuristic that tells us this heuristic should almost never fail.

share|improve this answer
    
Thanks for writing a response. I can't immediately see a specific error in what you've written, but nevertheless feel uncomfortable the argument insofar as if 1/d + 1/e + 1/f is less than 1, then not only is the probability less than 1/N, it's 1/N^x where x is greater than 1, so that upon multiplication by N it's 1/N^(x-1), which decreases as N increases. But the claim that the expected number of solutions among the first N natural numbers decreases as N grows is absurd. –  Jonah Sinick Sep 24 '12 at 5:22
    
As for the finiteness as d, e & f vary, the proof uses the fact that the first i'th power greater than 1 grows exponentially with i. –  Jonah Sinick Sep 24 '12 at 5:22

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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