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2^m = 3^n +5

  1. Sep 24, 2004 #1
    Hello,


    2^m = 3^n + 5, given m,n >=0

    ... Find all possible m,n... Any ideas?
     
  2. jcsd
  3. Sep 24, 2004 #2

    Hurkyl

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    Try looking at it in varios moduli?
     
  4. Sep 25, 2004 #3
    what does that mean lol pls explain.
     
  5. Sep 25, 2004 #4

    Hurkyl

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    For example, if you reduce the equation mod 2^m, and there are no values of n such that 3^n + 5 = 0 (mod 2^m), then you've found an upper limit for m.

    I don't know if this will work...
     
  6. Sep 25, 2004 #5

    Tide

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    Try rewriting your equation as
    [tex]2^m - 2 = 3^n + 3[/tex]
    That might provide some insight.
     
  7. Sep 26, 2004 #6
    Well, two small solutions are: 2^3 = 3+5; 2^5=3^3+5.
     
  8. Sep 26, 2004 #7
    Try looking up Stroeker and Tijdeman's theorem (a.k.a. the solution to Pillai's conjecture).
     
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