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Math Help - Order (mod p)

  1. #1
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    Order (mod p)

    Show that if a has order 3 (mod p), then a + 1 has order 6 (mod p).
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  2. #2
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    Quote Originally Posted by tarheelborn View Post
    Show that if a has order 3 (mod p), then a + 1 has order 6 (mod p).

    (a+1)^6=[(a+1)^3]^2=[a^3+3a^2+3a+1]^2=(2+3a(a+1))^2= 4+12a(a+1)+9a^2(a+1)^2= 4+12a^2+12a+9a^4+18a^3+9a^2= 9a+18+21a^2+12a+4=21a^2+21a+22=<br />
21(a^2+a+1)+1=0+1=1 .

    Now check that (a+1)^3=-1 and deduce what you want (remember that a^3-1=(a-1)(a^2+a+1) ... )

    Tonio

    Ps. The proof that (a+1)^k\neq 1\,\,\forall\, 1\leq k\leq 5 is pretty interesting in itself.
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  3. #3
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    If a^3 == 1 (mod p) does it imply that (a+1)^3 == -1 (mod p)? I have gotten only as far as (a+1)^3 = 2 + 3a(a+1) but I can't see that it helps me any in getting to -1.
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  4. #4
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    How do I know that (a+1) doesn't have order < 6?
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