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Math Help - Congruence relation

  1. #1
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    Joined
    Oct 2011
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    Congruence relation

    Hi, also need help with this one, been trying this for hours..

    Find all integers that are k>=3 (bigger or equal to 3)

    a) 11 ≡ k^2, mod k
    b) 7 ≡ k, mod k^2

    Thanks!
    Last edited by lolcat; October 2nd 2011 at 03:08 PM.
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  2. #2
    Member
    Joined
    Oct 2010
    From
    Mumbai, India
    Posts
    198

    Re: Congruence relation

    try to see the meaning of the congruence

    11\equiv k^2(\mbox{mod}\;k) means

    \exists n\in \mathbb{Z}(11-k^2=kn)

    \Rightarrow n\in \mathbb{Z}\;\mbox{and}\;11-k^2=kn

    \Rightarrow n=\frac{11-k^2}{k}=\frac{11}{k}-k

    since k is already an integer >= 3 , and n must be an integer. the only way n could be integer is if \frac{11}{k} is also an integer. there is only one value of
    k for which this happens........

    apply similar reasoning to b)
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