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    Let m = ((4^p) - 1))/3, where p is a prime number exceeding 3. Prove that (2^(m-1))-1 is divisible by m.

    This problem has had me stumped for over one and a half hour, more than enough time to solve this problem.

    Any hints on this problem are welcome.

    Thank
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    (Original post by EmptyMathsBox)
    Let m = ((4^p) - 1))/3, where p is a prime number exceeding 3. Prove that (2^(m-1))-1 is divisible by m.
    I know next to nothing about number theory, but I think that you may need to apply Fermat's Little theorem here. I think that you need to show that m and p are relatively prime (i.e that p does not divide m) to get it to work.
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    Showing that m and p are relatively prime is not that hard 4^p is congruent to 4 (mod p) (little theorem) so (4^p)-1 is congruent to 3 (mod p) so m = ((4^p)-1)/3 is congruent to 1 (mod p) (p is not 3 and is prime so shares no factors with 3 so we can divide by 3) so is relatively prime with p.

    However I do not see how this helps me solve the problem except for knowing that m-1 is divisible by p.

    Please post any other hints if you have them.
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    (Original post by EmptyMathsBox)
    However I do not see how this helps me solve the problem except for knowing that m-1 is divisible by p.
    Sorry, it doesn't. I only glanced at the question and misread it (and I meant to write "show that 2 and m are relatively prime", anyway). In fact, my idea doesn't make sense at all since m is not prime.

    Please post any other hints if you have them.
    Sadly, that's about the limit of my number theory. You'll have to wait for someone who knows this stuff to respond.
 
 
 
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