1

I’m looking for the minimum distance between any two members of the geometric progressions 2, 4, 8,… and 3,9,27,…

It seems like the pair of numbers which has the minimum distance is (2,3). Can you help me find a proof?

Also what if one of the sequences starts later? For example if the sequence starts at 8,16,32,… or higher.

David G. Stork
  • 29,774
user499369
  • 11
  • 3
    It can be shown (I can’t do it though) that the only perfect powers that are separated by 1 are 8 and 9. – insipidintegrator Sep 05 '22 at 04:19
  • 3
    Apparently not an easy topic: https://en.wikipedia.org/wiki/Catalan%27s_conjecture – Mike O'Connor Sep 05 '22 at 04:40
  • Isn't this related to the Hall-conjecture ? It is conjectured that every difference occurs only for finite many pairs of perfect powers (only the case $1$ has been proven , there is only the pair $(8/9)$ ) , for those particular powers probably more can be said. – Peter Sep 05 '22 at 17:51

1 Answers1

-4

The numbers in the series are all natural numbers, and that means that the distance between two must be an integer. The distance between $2$ and $3$ is $1$. So, for the answer to be anything else, we need the distance between them to be $0$, so:$$x^2=x^3$$This is only true when $x=0$ or $x=1$, neither of which are a term in either of the series. So, the minimum distance is $1$, which occurs at $2$ and $3$. Unless there is another pair with distance $1$, it is $2$ and $3$.

ArthD21
  • 382
  • 1
    "This is only true when $x=0$ or $x=1$". Oh? What about $2^3$ and $3^2$? – David G. Stork Sep 05 '22 at 04:29
  • 1
    To the OP: one doesn't know the bases are the same. One base could be a and the other b. That would just mean $a^2=b^3$ which can occur as David's remark shows. Of course it cannot happen if a,b must be 2,3 but you need to show that, perhaps use unique factorization. – coffeemath Sep 05 '22 at 04:41