Intersection of $\{2^a 3^b 5^c 7^d\}$ and its translates

Question

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)

Answer 1

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).

Answer 2

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

Answer 3

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]