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Let $S$ be the set of positive integers of the form $2^a3^b 5^c 7^d$. I need information about the cardinality of the intersection of $S$ and its translates. In particular, is $S \cap (S+t)$ infinite for every integer $t$? For some values of $t$?

Photo of the some of the solutions for $t=1$. Only integer solutions count, however. This still lacks proof it has infinitely many solutions over the set of integers. The graph looks similar for the solutions to 10, 100, and pretty much any number, which leads me to believe there are infinite solutions. A proof for this, though, would be invaluable.

Another way this can be worded is if the equation $2^A 3^B 5^C 7^D - 2^a 3^b 5^c 7^d = 1 \{A,B,C,D,a,b,c,d \in \mathbb{Z} \}$ has an upper bound. This would determine if it has infinite solutions, and may be able to be generalized for all of $n>0$ instead of 1. This function has way too many variables to be graphed so an algebraic way to determine if a function has bounds would be best.

A very special similar problem would be to determine whether there are infinitely many pairs of numbers differing by $1$ whose only prime factors are $2$ and $3$. (The answer is "no" for pure powers of $2$ and $3$: distance between powers of 2 and powers of 3)

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    $\begingroup$ The intersection is finite. See here arxiv.org/pdf/1311.3743.pdf- namely the third version of the subspace theorem. In your case the group is generated by $2,3,5,7$ and $K=\mathbb{C}$. Note that if you consider your equation as you wrote it does not seem it applies. But note that either $A=0$ or $a=0$; so wlog assume $a=0$. Then you can write $2^A3^B5^C7^D-2^a3^b5^c7^d=2\cdot 2^{A-1}3^B5^C7^D-2^a3^b5^c7^d$ and you are done. $\endgroup$ – Vlad Matei Dec 12 '18 at 16:20
  • $\begingroup$ I tried clicking the link and it said paper not found, can you resend it or check it @VladMatei ? $\endgroup$ – Ryan Shesler Dec 12 '18 at 16:25
  • $\begingroup$ @Ryan A small typo: Remove the hyphen at the end of that link, and it works. $\endgroup$ – Andrés E. Caicedo Dec 12 '18 at 16:27
  • $\begingroup$ @VladMatei does this apply to all natural numbers instead of 1? And is there a way for me to find the last instance in which this works? $\endgroup$ – Ryan Shesler Dec 12 '18 at 16:47
  • $\begingroup$ S is an infinite set - does the third version of the subspace theorem still hold because from what I understand you are saying that $\Gamma$ is the set S, and $z_i \in \Gamma$, but this does not apply when there is no $z_n$ @VladMatei $\endgroup$ – Ryan Shesler Dec 12 '18 at 17:05
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This is an example of an $S$-unit equation. For ones of a shape similar to this, the solutions can be found rather easily using bounds for linear forms in logarithms and lattice basis reduction. By way of example, Theorem 5.5 of de Weger's thesis (from 1989) explicitly determines the $605$ relatively prime solutions to the inequality $$ 0 < x-y < \sqrt{y} $$ with $x$ and $y$ with prime factors entirely in $\{ 2, 3, 5, 7, 11, 13 \}$. From this result, at a glance it seems that the last example for the problem you're interested in corresponds to $4375-4374=1$.

If you replace the $1$ by an arbitrary constant $k$, you still can solve the problem effectively (though the number of solutions may grow somewhat).

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  • $\begingroup$ This is absolutely fantastic, thank you. I have a quick question, just out of curiosity - if I take the limit as $k \to \infty$, is the number of solutions always finite, even at infinity? $\endgroup$ – Ryan Shesler Dec 12 '18 at 17:37
  • $\begingroup$ For fixed $k$, the number can be bounded in terms of the number of prime divisors of $k$. Conjecturally, if $k$ is relatively prime to $210$, the number should be absolutely bounded. $\endgroup$ – Mike Bennett Dec 12 '18 at 17:47
  • $\begingroup$ Right, so does this mean that as k increases and so does the number of prime divisors, the bound increases without bound (literary paradox intended but irrelevant) @MikeBennett $\endgroup$ – Ryan Shesler Dec 12 '18 at 18:26
  • $\begingroup$ Well, the bound we can prove increases (or so I believe), but the conjectured bound does not. $\endgroup$ – Mike Bennett Dec 12 '18 at 20:30
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Long before anyone spoke of S-unit equations, there was Stormer's Theorem of 1897: given any finite set of primes, there are only finitely many pairs of consecutive numbers divisible only by those primes. Moreover, Stormer gave a method for finding all the solutions, using Pell equations. Lehmer gave practical improvements in 1964. Many details, references, and links to OEIS and elsewhere, at https://en.wikipedia.org/wiki/Størmer%27s_theorem

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As Mike Bennett said, this is an example of an $S$-unit equation, although you are asking for solution to $u-v=1$ in $\mathbb Z_S^*\cap \mathbb Z$. More generally, one drops the requirement that $u$ and $v$ be in $\mathbb Z$, so one allows the primes in $S$ to also appear in the denominators of $u$ and $v$. More generally, as you ask, one can look for solutions to $u-v=k$, where if we allow $u$ and $v$ be be general $S$-units, then it's enough to consider the case that $k$ is relatively prime to the primes in $S$.

It is a theorem of Evertse that for any $a,b,c\in\mathbb Q^*$ and any finite set of primes $S=\{p_1,\ldots,p_r\}$, the equation $au+bv=c$ has at most $3\cdot7^{2|S|+3}$ solutions with $u,v\in\mathbb{Z}_S^*$, i.e., where $u$ and $v$ are rational numbers composed entirely of powers of the primes in $S$. See J.-H. Evertse [Invent. Math. 75 (1984), no. 3, 561–584; MR0735341].

There are also results saying that if you take $S=\{2,3,\ldots,p_r\}$ to be the first $r$ primes, there will be a value of $k$ so that $u+v=k$ has at least something like $C^{r^{1/3}}$ solutions in $S$-units, where $C>1$ is an absolute constant. (I may not have this quite correct.) This is in the paper Erdös, P.; Stewart, C. L.; Tijdeman, R. [Some Diophantine equations with many solutions. Compositio Math. 66 (1988), no. 1, 37–56. MR0937987]

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  • $\begingroup$ So does this mean that the number of solutions to the intersection of the sets is always the same constant? @JoeSilverman $\endgroup$ – Ryan Shesler Dec 13 '18 at 1:15
  • $\begingroup$ I don't know what you mean by the "intersection of the sets"? The number of solutions to $ax+by=c$ with $x,y\in\mathbb Z_S^*$ will vary as you change $a,b,c$ or if you vary $S$. $\endgroup$ – Joe Silverman Dec 13 '18 at 1:19
  • $\begingroup$ sorry, "the equation $au+bv=c$ has at most $3⋅7^{2|S|+3}$ solutions" threw me off. I meant does this mean that for any value of c, even as it approaches infinity, yields at most the same number of solutions? @JoeSilverman $\endgroup$ – Ryan Shesler Dec 13 '18 at 1:21
  • $\begingroup$ Yes, Evertse's bound for the number of solutions is independent of $a,b,c$. So as you vary $c$, for example, the upper bound for the number of solutions stays the same. Of course, the actual number of solutions is likely the vary. There's also a result that says that for most equivalence classes of triples $(a,b,c)$, the equation has at most 2 solutions in $S$-units, but of course the set of $(a,b,c)$ yielding more than 2 solutions will change as you make the set $S$ larger. $\endgroup$ – Joe Silverman Dec 13 '18 at 3:10

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