Edit to clarify and rephrase:

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 $2^A 3^B 5^C 7^D - 2^a 3^b 5^c 7^d = 1 \{A,B,C,D,a,b,c,d \in {Z} \}$ 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 $3$: https://mathoverflow.net/questions/116840/distance-between-powers-of-2-and-powers-of-3)

The original question. I am leaving this here since it has an upvoted answer.

I was browsing some set theory theorems on infinite sets and I couldn't find one that could help me with my specific case. This is a question I'm not positive has an answer, but for an infinite set with random integers $$\{ n_1, n_2, n_3, ...\} \cap \{n_1 + a, n_2 + a, n_3+a...\}$$ where a is an integer, what circumstances would yield a finite set as supposed to an infinite set. Alternatively, how could I prove that this set would be infinite instead of saying that it is 'probably' infinite. The distribution is all numbers greater than 7 who don't have any prime factors strictly greater than 7, and aren't prime. Some of the terms are $$\{80, 81, 84, 96, 100, 105, 108 ... \}$$ To clarify, these are values down the line. The first term is 8, the second 9, the third 10, etc. An answer for all of a is what I'm looking for. I do know that each progressive n is larger than the last. I'm sorry if this question is too vague, I don't have much mathematical experience. Thank you for anything you can provide.

  • $\begingroup$ how is $n_i$ distributed? $\endgroup$ – Dando18 Dec 10 '18 at 0:19
  • 1
    $\begingroup$ The notion of "random integer" is subtle, and you don't specify what you mean by it here. Once you do, the fact that these integers are "random" means that the most you can hope for is an answer that says "probably infinite". In order to help you we need much more precise information about your "specific case". Where does it come from? How do you choose your "random integers"? Please edit the question to tell us - don't use comments. $\endgroup$ – Ethan Bolker Dec 10 '18 at 0:22
  • $\begingroup$ What does "basically" mean in your edit? Why isn't $4$ in the list? Is this your only example, or are there others? Do you need an answer for some $a$ or for every $a$? $\endgroup$ – Ethan Bolker Dec 10 '18 at 1:05
  • $\begingroup$ @EthanBolker I edited some more. Sorry for the lack of detail, I'm new to stacks. $\endgroup$ – Ryan Shesler Dec 10 '18 at 1:24
  • $\begingroup$ The question is now clear - I think you are looking at the sequence of numbers of the form $2^a3^b5^c$. I don't know if the intersection of this sequence with each of its translates is finite, but I suspect so. If I have the question right you might want to do a major edit and ask it this way from the start, including in the title. Leave the current version at the end, so as not to invalidate @YiFan 's answer. $\endgroup$ – Ethan Bolker Dec 10 '18 at 1:44

There's no easy way that works for every such sequence, and for some sequences answering this question can be a research-level problem. Consider for example the sequence of all prime numbers, and take $a=2$. Finding whether this intersection is finite or infinite is equivalent to answering the twin prime conjecture which is still an open problem.

  • $\begingroup$ he said "random", not "arbitrary" $\endgroup$ – mathworker21 Dec 10 '18 at 0:22
  • $\begingroup$ @mathworker21 the OP will have to clarify the statement then. The way I interpreted it is "taking any such sequence, is there always an easy way of finding the intersection?" If he meant it in a statistical sense then he needs to state it better (especially including things like how he wants to define a random distribution of integers). $\endgroup$ – YiFan Dec 10 '18 at 0:23

Very partial answer.

You are looking for integer solutions to $$ 2^A3^B5^C7^D = 2^a3^b5^c7^d = t . $$ If $t$ is relatively prime to $2\times 3 \times 5 \times 7 = 210$ then at least one of each pair $X,x$ of exponents must be $0$.

In particular that's the case when $t=1$. That's a very strong restriction: each of the four primes appears in at most one of the terms.

When $C= c = D = d = 0$ it's known that there are only finitely many solutions.

When one of the terms and $t$ share a prime factor so must the other, so you can divide through by it and reduce to the case when $t$ is relatively prime to both terms.

I suspect further analysis along these lines will show there are always only finitely many solutions.

  • $\begingroup$ I'm looking for integer solutions to $2^A 3^B 5^C 7^D - 2^a 3^b 5^c 7^d = t$. $\endgroup$ – Ryan Shesler Dec 11 '18 at 22:13
  • $\begingroup$ I think... I'm not entirely sure but that was what I meant; integer values for A B C D that produce one more value than a b c d, and in general $t$ more than a b c d $\endgroup$ – Ryan Shesler Dec 11 '18 at 22:15
  • $\begingroup$ I found this: $$\mid a^{m}-b^{n}\mid> \frac{\max(a^{m}, b^{n})}{(e\max(m, n))^{C_1}}$$ where $C_1$ is a computable number strictly greater than 0. Make $a,b=2$ and $m={A + B\log_2(3)+C\log_2(5)+D\log_2(7)}$, and $n={a + b\log_2(3)+c\log_2(5)+d\log_2(7)}$ I don't have the mathematical knowledge to do computations like this but it seems like it could help lead to a solution. @EthanBolker $\endgroup$ – Ryan Shesler Dec 11 '18 at 22:59
  • $\begingroup$ I doubt that will help. $\endgroup$ – Ethan Bolker Dec 12 '18 at 1:11

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