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Math Help - 53|2^k-2^n

  1. #1
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    53|2^k-2^n

    Find all n,k\in\mathbb{N} for which 53|2^k-2^n.
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  2. #2
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    Quote Originally Posted by james_bond View Post
    Find all n,k\in\mathbb{N} for which 53|2^k-2^n.


    Choose k>n\Longrightarrow k=n+m\,,\,m>0\Longrightarrow 53\mid (2^{n+m}-2^n)= 2^n(2^m-1)\iff

    \iff 53\mid (2^m-1)\iff 2^m=1\!\!\pmod{53} , and since 2 is not a primitive

    root modulo 53 (look here Primitive root modulo n - Wikipedia, the free encyclopedia) ,

    the above is true only when...

    Tonio

    Edit: Either the above link is wrong or else I didn't understand it. Anyway, 2 indeed is

    a primitive root modulo 53, and thus 2^m=1\iff m=k-n=0\!\!\pmod {52}
    Last edited by tonio; April 3rd 2011 at 02:11 PM.
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  3. #3
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    Quote Originally Posted by james_bond View Post
    Find all n,k\in\mathbb{N} for which 53|2^k-2^n.
    We have k\geq n. Write 2^k-2^n=2^n(2^{k-n}-1). Since (53,2^n)=1, 53 must divide 2^{k-n}-1, i.e 2^{k-n}\equiv1\pmod{53}.

    Actually 2 is a primitive root of 53 ; hence the congruence holds if and only if k-n\equiv0\pmod{52}.

    Now I think this is the answer:
    For any fixed integer t\geq0 , the solutions are given by any pair k, n such that k-n=52t.
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