# Constant difference with finite prime divisors

Is there a finite set $P$ of primes such that for any integer $k\geq 1$, there exists an infinite list $a_1,a_2,\dots$ of positive integers such that $a_i$ and $a_i+k$ has only prime divisors in $P$ for all $i$?

If $|P|=1$ then this set cannot satisfy the condition because the difference between consecutive terms in the list $1,a,a^2,\dots$ is growing. But if $|P|\geq 2$ it is possible that this can work.

• Of course the set of every prime itself satisfies your condition. I think such $P$ should be infinite, but it is too hard to show it explicitly. – didgogns Mar 30 '17 at 13:50
• I doubt it but I'm only guessing. It might be possible to show that for a given set $P,$ the existence of $k$ and $a_1,a_2,...$ for finite a $P$ implies existence of a smaller $k'$ and $a'_1,a'_2,$... for some finite $P'$ (or for $P$) which would refute it. .... Interestingly the case $k=1, |P|=2$ may have something to do with the question of whether $A^x=B^y+1$ has a solution in integers $A,B,x,y$ all greater than $1$, other than $3^2=2^3+1,$ which is currently an unsolved problem – DanielWainfleet Mar 30 '17 at 18:54
• Do you not mean "an infinite set $\{a_1,a_2,a_3,...\}$ such that (etc.)..."? – DanielWainfleet Mar 30 '17 at 19:02
• You're all correct, sorry, I slightly changed the formulation before posting, hence the errors. Corrected. – pi66 Mar 31 '17 at 2:24

For $k=1$ you require infinite many pairs $(a,a+1)$ such that both $a$ and $a+1$ contain only prime factors in $P$.
It is well known that this is impossible for every finite set $P$ because of Stormer's theorem (See here : https://en.wikipedia.org/wiki/St%C3%B8rmer%27s_theorem ) stating that only finite many pairs exist.
Maybe, for larger $k$ this is possible.