[**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.)