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obvious???

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  • leavemsg1
    Hi, Is it entirely obvious that 2^(p-1)+3^(p-2) -1 is always divisible by 2 and 3 where p is prime??? Bill
    Message 1 of 3 , Jan 16, 2007
      Hi,

      Is it entirely obvious that 2^(p-1)+3^(p-2) -1 is always divisible by 2
      and 3 where p is prime???

      Bill
    • 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 2 of 3 , Jan 16, 2007
        --- 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


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      • 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 3 of 3 , Jan 16, 2007
          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
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