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K(m^2-n^2) = 2(m^3+n^3-lm)

  1. Dec 31, 2009 #1
    Any pointers and/or help with proving the following would be appreciated

    For every prime number n, there exist positive integers (k,l,m) such that

    k(m2-n2)=2(m3+n3-lm)

    some examples (there could be several/many solutions for a given n)

    {n,k,l,m}
    {2, 14, 8, 2}
    {3, 59, 264, 1}
    {5, 53, 192, 3}
    {7, 71, 264, 5},{7, 239, 6080, 1}
    {11, 163, 1856, 5}

    If the statement is true for prime n's, it can be shown to hold for composite n's as well.

    Thanks, Mathador
     
  2. jcsd
  3. Dec 31, 2009 #2
    Re: k(m^2-n^2)==2(m^3+n^3-lm)

    Perhaps you can think of it as a linear problem in k and l. Given prime n see whether a specific m can be found such that a solution to the linear problem is positive.

    As you may be aware, a trivial solution is (n,k, 2n2, n) k=1,2,3,...
     
  4. Jan 1, 2010 #3
    Re: k(m^2-n^2)==2(m^3+n^3-lm)

    Also trival are solutions where m = 1
     
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