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Math Help - minimal power and congruence

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    12

    minimal power and congruence

    Hi,

    ------------------------------------
    Problem:
    Let a and b be integers.
    Let p be an odd prime divisor of a^16 + 1. (i.e. a^16 == -1 mod p)
    Show that if e is the smallest positive integer such that b^e == 1 mod p,
    then e=32.
    -------------------------------------

    This is kind of a weird problem I know. Just using the fact that p is an odd prime divisor of a^16 + 1 I showed that p == 1 mod 32. That might not be useful, but the number 32 is in there so it might be relevant.

    Also, I know that b^x == 1 mod p if and only if e divides x, with x>0.

    Another thing I know is that if b^x == 1 == b^y mod p, with x and y positive, then a^d == 1 mod p where d = gcd(x,y).

    I'm not sure what is relevant and what is not, I've been messing with it for a while. Does anyone have any hints? If you need the problem clarified more let me know. Thanks.
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  2. #2
    Senior Member
    Joined
    Apr 2009
    From
    Atlanta, GA
    Posts
    408

    Clarity

    Just to clarify, your conjecture is for natural number a and odd prime p satisfying a^{16}\equiv 1(\bmod p), there are no solutions to b^e\equiv -1 (\bmod p) for e<32. Is this correct?
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