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Pillai's conjecture -- that the gap between (nontrivial) powers is unbounded below -- is still open (it would be a consequence of the $abc$ conjecture, were that proven). But I wonder what the right order of magnitude for it is, even though a proof seems far off.

Suppose $n=|a^x-b^y|$ for integers $a,b,x,y$ with $a,b\ge1$ and $x,y\ge2.$ What is the (conjectural) maximal order of $a^x$?

$17 = 378661^2 - 5234^3$ so we should not expect it to be too tiny. But I don't even know if the right order should be polynomial, exponential, doubly-exponential, etc. Does anyone have insight here?

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up vote 3 down vote accepted

Perfect Powers: Pillai's works and their developments is related to your question.

p.9: Conjecture 3.1. For any $\epsilon > 0$, there exists a constant $\kappa(\epsilon) > 0$ such that, for any positive integers $(a, b, x, y)$, with $x \ge 2, y \ge 2$ and $a^x \ne b^y$ , $$|a^x - b^y | \ge \kappa(\epsilon) \max (a^x , b^y)^{ 1-(1/x)-(1/y)-\epsilon}$$

p.7 In 1986, J. Turk [126] gave an effective estimate from below for $|x^n - y^m |$, which was improved by B. Brindza, J.-H. Evertse and K. Gyory in 1991 and further refined by Y. Bugeaud in 1996:

Let $x$ be a positive integer and $y, n, m$ be integers which are $\ge 2$. Assume $x^n \ne y^m$ . Then $|x^n - y^m | \ge m^{2/(5n)} n^{-5} 2^{-6-42/n}.$

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Hmm. I don't know the exponents a priori, so the first result is $a^x\ll n^{6+\varepsilon}$ in my case (if I'm not mistaken). The second doesn't seem to apply at all since my $x$ is not fixed. –  Charles Dec 11 '12 at 15:35
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Hall conjectured the existence of a positive constant $C$ such that if $y^2\ne x^3$ then $$|y^2-x^3|\gt C\sqrt{|x|}$$ This has not been disproved, but is considered unlikely to be true. Nowadays one often calls Hall's Conjecture the weaker statement that for any positive $\epsilon$ there is a constant $c(\epsilon)$ such that if $y^2\ne x^3$ then $$|y^2-x^3|\gt C(\epsilon)x^{{1\over2}-\epsilon}$$ See this link.

See also Noam Elkies' page, http://www.math.harvard.edu/~elkies/hall.html

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So the weaker version gives $a^x\ll n^{2+\varepsilon}$. But this applies only to the common case of exponents 2 and 3; is this the worst case? –  Charles Dec 11 '12 at 15:47
    
Since everything is conjectural, I'm not sure that "worst case" has a well-defined meaning. In any event, all I know is what I wrote. –  Gerry Myerson Dec 11 '12 at 22:07
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