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Math Help - is a≡b still valid??

  1. #1
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    is a≡b still valid??

    Show that if a^k≡ b^k (mod m) and a^(k+1) ≡ b^(k+1) (mod m), where a,b,k, and m are integers with k > 0, m > 0 such that (a,m) = 1, then a ≡ b (mod m). If the condition (a,m) is dropped, is the conclusion that a ≡ b still valid?

    just throwing out my brain storm ideas, but is this an induction problem, or do i have to set k=1 or something else?..
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  2. #2
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    Quote Originally Posted by alice8675309 View Post
    Show that if a^k≡ b^k (mod m) and a^(k+1) ≡ b^(k+1) (mod m), where a,b,k, and m are integers with k > 0, m > 0 such that (a,m) = 1, then a ≡ b (mod m). If the condition (a,m) is dropped, is the conclusion that a ≡ b still valid?

    just throwing out my brain storm ideas, but is this an induction problem, or do i have to set k=1 or something else?..
    Hint: a has a multiplicative modular inverse mod m iff gcd(a,m) = 1.

    When the condition (a,m) = 1 is dropped, the statement is false. Consider a=2, b=0, m=4, k=2.
    Last edited by undefined; October 13th 2010 at 07:49 PM.
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