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For all solutions of n and m that satisfy.

  1. Jan 5, 2010 #1

    For some positive integers m and n.
    Last edited: Jan 5, 2010
  2. jcsd
  3. Jan 5, 2010 #2
  4. Jan 5, 2010 #3
    A nice way to solve this is to factor:
    [tex]3\times 2^m = (n-1)(n+1)[/tex]
    Now clearly one factor on the right is a power of 2 since the factor 3 can only occur in one of them. Split it into cases depending on which it is and you should get the answers.
  5. Jan 5, 2010 #4
    of (n-1)(n+1) one factor has to be a power of 2 and the other 2*3=6. so the two possibiltites are 4*6 and 6*8.

    So all answers are 3*2^3+1=5^2 and 3*2^4+1=7^2.
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