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

  • Thread starter mathador
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  • #1
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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
 

Answers and Replies

  • #2
6
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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,...
 
  • #3
841
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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,...
Also trival are solutions where m = 1
 

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