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Math Help - What about b?

  1. #1
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    What about b?

    If (2^a -1) divides (2^{a(b-1)} + 2^{a(b-2)} + ... + 2^2a + 2^a + 1)
    for a, b 2^a - 1 and 2^b -1 being odd primes
    What does it tell us about b?
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  2. #2
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    Quote Originally Posted by helgamauer View Post
    If (2^a -1) divides (2^{a(b-1)} + 2^{a(b-2)} + ... + 2^2a + 2^a + 1)
    for a, b 2^a - 1 and 2^b -1 being odd primes
    What does it tell us about b?
    It would imply that b=2^a-1.

    Note that (2^a)^k\equiv1\pmod{2^a-1} for any positive integer k . Therefore, (2^a)^{b-1}+(2^a)^{b-2}+\cdots+(2^a)^2+(2^a)+1\equiv1+1+\cdots+1=b\equi  v0\pmod{2^a-1}.

    Thus 2^a-1(>1) divides b , and since b is prime we must have b=2^a-1.

    By the way, 2^{a(b-1)}+2^{a(b-2)}+\cdots+2^{2a}+2^a+1=(2^a)^{b-1}+(2^a)^{b-2}+\cdots+(2^a)^2+(2^a)+1.
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