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Thread: (2^p) - 1 divides (2^kp) - 1 where p is prime and k is some integer

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    (2^p) - 1 divides (2^kp) - 1 where p is prime and k is some integer

    Doing number theory problems and it's given that this is trivial, but I'm not sure how. What's the proof behind it?

    Similar situation for (a^2) - 1 divides (a^(p-1) - 1) where a is an integer > 1 as (p-1) is given to be even. Why is this?
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    Re: (2^p) - 1 divides (2^kp) - 1 where p is prime and k is some integer

    Quote Originally Posted by patrickmanning View Post
    Doing number theory problems and it's given that this is trivial, but I'm not sure how. What's the proof behind it?
    You''re missing grouping symbols for the second exponent. When k = 0, the revised second expression would be zero, so the statement would be trivially true.
    I would recommend you rewrite it similar to the following:

    2^p - 1 divides 2^(kp) - 1, where p is prime and k is some positive integer.
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