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

Re: [PrimeNumbers] obvious???

Expand Messages
  • Phil Carmody
    ... Humans and mail are slow. Computers and Pari/GP are fast. ? for(i=2,12,print(i : 2^(i-1)%6 3^(i-2)%6 (2^(i-1)+3^(i-2))%6)) 2 : 2 1 3 3 : 4 3 1 4 : 2
    Message 1 of 3 , Jan 16, 2007
    • 0 Attachment
      --- leavemsg1 <leavemsg1@...> wrote:
      > Is it entirely obvious that 2^(p-1)+3^(p-2) -1 is always divisible by 2
      > and 3 where p is prime???

      Humans and mail are slow. Computers and Pari/GP are fast.

      ? for(i=2,12,print(i" : "2^(i-1)%6" "3^(i-2)%6" "(2^(i-1)+3^(i-2))%6))
      2 : 2 1 3
      3 : 4 3 1
      4 : 2 3 5
      5 : 4 3 1
      6 : 2 3 5
      7 : 4 3 1
      8 : 2 3 5
      9 : 4 3 1
      10 : 2 3 5
      11 : 4 3 1
      12 : 2 3 5

      So it's not just true when p's an (odd) prime, but true whenever p's any odd
      number.

      Phil


      () ASCII ribbon campaign () Hopeless ribbon campaign
      /\ against HTML mail /\ against gratuitous bloodshed

      [stolen with permission from Daniel B. Cristofani]



      ____________________________________________________________________________________
      Cheap talk?
      Check out Yahoo! Messenger's low PC-to-Phone call rates.
      http://voice.yahoo.com
    • Joshua Zucker
      ... Yes. 2^whatever is even, 3^whatever is odd, so 2^x + 3^y - 1 is even. 2^even = 1 (mod 3), 3^whatever = 0 (mod 3), so 2^even + 3^whatever - 1 is divisible
      Message 2 of 3 , Jan 16, 2007
      • 0 Attachment
        On 1/16/07, leavemsg1 <leavemsg1@...> wrote:
        > Is it entirely obvious that 2^(p-1)+3^(p-2) -1 is always divisible by 2
        > and 3 where p is prime???

        Yes.

        2^whatever is even, 3^whatever is odd, so 2^x + 3^y - 1 is even.

        2^even = 1 (mod 3), 3^whatever = 0 (mod 3), so
        2^even + 3^whatever - 1 is divisible by 3.

        So if p is odd (not necessarily prime, and not 2), then the
        divisibility will happen as you have said.

        --Joshua ucker
      Your message has been successfully submitted and would be delivered to recipients shortly.