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Math Help - Wondering if a difference of 1 in a relation affects my proof

  1. #1
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    Wondering if a difference of 1 in a relation affects my proof

    Let k,n be integers such that 1 <= k <= n <= p-1 where p is prime.
    Assume (a,p) = 1.
    Show that ak \equiv an mod p iff  k = n.

    Proof:
    ak \equiv an mod p
    k \equiv n mod p (since (a,p) = 1)

    Then I used a Lemma that states:
    Let  i and j be integers with 1 <= i <= j <= m. If i \equiv j mod m, then i = j.

    Where I use i = k, j = n, and m = p. So that gives k = n. My question is, am I allowed to use the theorem if the equivalences are 1 off?:
    1 <= k <= n <= p-1 Proof
    1 <= i <= j <= m Lemma

    Thanks
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  2. #2
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    Quote Originally Posted by Th3sandm4n View Post
    Proof:
    ak \equiv an mod p
    k \equiv n mod p (since (a,p) = 1)
    Once you know k\equiv n(\bmod p) it would follow that k=n. This is because k\leq n and so 0\geq 0 n-k<p. The only way for n-k to be divisible by p is for n-k = 0\implies n=k.

    This is Mine 12,3th Post!!!
    Last edited by ThePerfectHacker; March 29th 2009 at 08:57 PM.
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  3. #3
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    yeah I started my proof trying to get it to show  n-k = 0 but then saw the Lemma and thought that would show it better. Thank you for the clearer explanation though, and congrats on the posts, that is a lot of helping!
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