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Call two sets, $A$ and $B$ close iff there exists a finite $k$ such that there are infinitely many pairs of elements $(a,b)$ with $a\in A$ and $b\in B$ where $|a-b|\le k$. If two sets are not close, call them far. If such a $k$ exists, call the smallest such one the radius of closeness of those sets.

Call two numbers, $c$ and $d$ close iff the sets $\{c^n|n\in \mathbb{N}\}$ and $\{d^n|n\in \mathbb{N}\}$ are close and far otherwise and their radius of closeness the same as that of those sets. Call two numbers trivially close iff one is a rational power of the other or both of them have an absolute value less than or equal to 1. Call two numbers trivially far iff one of the absolute values is greater than one, and the other not. If two numbers are close but not trivially close, we shall call them non-trivially close.

I have several questions:
- Are $2$ and $3$ close or far? If they are close, what is their radius of closeness?
- Are $e$ and $\pi$ close or far? If they are close, what is their radius of closeness?
- Are there any pairs of integer/rational/real/complex numbers which are non-trivially close?
- Is there an algorithm to determine if two numbers are close? If so, what is it?
- Is there a smallest radius of closeness for integer/rational/real/complex numbers? If so, what is it, and what are some pairs of numbers that have that radius of closeness? If not, what is the limit of the radius of closeness, and what pairs of numbers approach that limit?

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  • $\begingroup$ I think your definition of "trivially close" should be extended to cover the case where one of the numbers is a (positive) rational power of the other, e.g., $\sqrt2$ and $\root3\of2$. $\endgroup$
    – Gerry Myerson
    Commented Dec 13, 2013 at 21:30
  • $\begingroup$ Yes. Thanks. I edited the question. $\endgroup$
    – Dylan Pizzo
    Commented Dec 13, 2013 at 21:40
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    $\begingroup$ I suspect it should be possible to construct two real numbers that are nontrivially close, recursively: for simplicity fix one of them to equal $2$. Given a tiny interval such that any real number in that interval has $m$ powers that are within $k$ of powers of $2$, look far far out until a large power of that interval contains another power of $2$, and then reduce the tiny interval even further so that the large power of the interval is all within $k$ of the power of $2$. Now we have $m+1$ such close calls, and recurse. $\endgroup$
    – Greg Martin
    Commented Dec 13, 2013 at 21:47
  • $\begingroup$ @GregMartin Why would this method not simply converge to one of the rational powers of two in that inteval? $\endgroup$
    – Dylan Pizzo
    Commented Dec 14, 2013 at 13:28
  • $\begingroup$ @Dpiz: excellent point! As stated, that is a real danger. But there are only countably many rational powers of $2$; so after the $n$th recursive step, we could diminish the interval if necessary to make sure it doesn't include the $n$th rational power of $2$ (in some fixed enumeration). $\endgroup$
    – Greg Martin
    Commented Dec 14, 2013 at 20:05

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For any fixed integers $a$ and $b$ that are multiplicatively independent and for any integer $k$, there are only finitely many pairs of positive integers $m$ and $n$ such that $$ a^n - b^m = k. $$ To see this, write $n=3N+i$ and $m=3M+j$ with $i,j$ between $0$ and $2$. Then $(a^N,b^M)$ is a solution to the Thue equation $$ a^iX^3 - b^jY^3 = k. $$ Each such equation has only finitely many solutions (as proven by Thue), and you're only considering 9 such equations.

Various people have written papers on upper bounds for the number of solutions to $AX^3+BY^3=k$ and $|AX^3+BY^3|\le k$ as a function of $k$.

Something similar should work in the ring of integers of an imaginary quadratic field, where there is only one archimedean absolute value to worry about.

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