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2^n-1 is prime

  1. Nov 18, 2009 #1

    Char. Limit

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    Gold Member

    First, remember that the OP is new to pure number theory.

    We all know that not every number that follows the formula 2n-1 is prime. My question is, without using trial and error, how would you prove or disprove this statement?

    "All numbers that obey the formula 2n-1, when n is a real integer number greater than 1, are prime."
     
  2. jcsd
  3. Nov 18, 2009 #2

    Borek

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    Staff: Mentor

    Re: 2^n-1

    Find counterexample.
     
  4. Nov 18, 2009 #3
    Re: 2^n-1

    Another possibility is noting that, for an even n = 2h, we have that 2^n-1 = (2^h+1)(2^h-1), which makes the original number composite for h > 1.
     
  5. Nov 18, 2009 #4
    Last edited by a moderator: Apr 24, 2017
  6. Nov 18, 2009 #5
    Re: 2^n-1

    For every even n, 2^n-1 is divisible by 3.
     
  7. Nov 23, 2009 #6
    Re: 2^n-1

    It doesn't have to be even. If ab is composite, then [tex]\frac{2^{ab}-1}{2^a-1}=1+2^a+2^{2a} +++2^{ab-a}[/tex]
     
    Last edited: Nov 23, 2009
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