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Another cyclic order conjecture

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  • purushaz
    Thanks, Phil, for the previous proof on conjecture that if O(n) for p is odd, then O(n^2) is the same; otherwise, O(n^2) is half that of O (n). Eg. O(2) for 7
    Message 1 of 1 , Sep 1, 2002
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      Thanks, Phil, for the previous proof on conjecture that if O(n) for p
      is odd, then O(n^2) is the same; otherwise, O(n^2) is half that of O
      (n). Eg. O(2) for 7 is 3 and O(4) for 7 is 3.
      Then, for O(2),17 = 8, so O(4),17 is 4.

      Do you have a proof for a related conjecture: any O(N^2) is n^0,
      n^2, n^3. (mod p)..with cycle returning to 1 or p-1.

      Example: O(4), 7 is 1,2,4, 8 = 3 since 8==1 MOD 7. That's a case in
      which O(2) and O(4) are both 3. Let's try a case in which O(2) is
      even: (I'm looking a Beiler's chart 48 "Exponents to which a belongs,
      MOD p and MOD p^n) How about p=5: O(2),5 is 4, while O(4),5 is 2.

      Yup. with 1,2,4...MOD 5, O(4) is 2 since 4 is p-1 MOD 5. Let me look
      through the chart for another even:. OK for p=17, O(2) is 8 while
      it's 4 for 4. Then 1,2,4,8,16 MOD 17, order is 4 with cycle
      returning at p-1.
      So, it appears that for odds O returns to 1 while for evens it's p-
      1. Thanks for your help!. Gary W. Adamson
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