Loading ...
Sorry, an error occurred while loading the content.

19819Re: [PrimeNumbers] Find non-integer A such that floor(A^n) is never prime?

Expand Messages
  • Jack Brennen
    Jan 16, 2009
      Ah, great -- that's just the kind of trick I was looking for.
      Thanks for the example; I'm sure it's just one example of a
      whole family of solutions.


      Robert Gerbicz wrote:
      > 2009/1/16 jbrennen <jfb@...>
      >> Just a silly little thought I had...
      >> It is almost certainly possible to find a positive non-integer A > 1
      >> such that for all integer n, floor(A^n) is not prime.
      >> Question -- does any such number A exist with a simple closed-form
      >> expression?
      >> A=5+sqrt(19) is good. To prove it use that
      >> g(n)=(5+sqrt(19))^n+(5-sqrt(19))^n is an integer sequence for that
      >> g(n)=10*g(n-1)-6*g(n-2), using this it's easy by induction that g(n)==1 mod
      >> 3 so floor(A^n)=g(n)-1 is divisible by 3 for all n>0 and isn't 3, so not
      >> prime. For n<=0 floor(A^n)=0,1 so not prime.
    • Show all 6 messages in this topic