My Math Forum for N

 Algebra Pre-Algebra and Basic Algebra Math Forum

 February 2nd, 2014, 03:41 AM #1 Senior Member   Joined: Nov 2013 Posts: 434 Thanks: 8 for N solve in N natural number x^3-y^3-xy=61
 February 2nd, 2014, 04:03 AM #2 Senior Member     Joined: Dec 2013 From: some subspace Posts: 212 Thanks: 72 Math Focus: real analysis, vector analysis, numerical analysis, discrete mathematics Re: for N I found two solutions: x = -5, y = -6 (but negative numbers) x = 6, y = 5 But I'm not sure if there are other solutions.
 February 2nd, 2014, 10:22 AM #3 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: for N I guess not, (x, y) = (5, 6) is the only solution s.t. 1 <= x, y <= 6. For y > x > 5 (and x, y in N), y^3 - x^3 - x*y > 61, for x > y > 5, y^3 - x^3 - x*y < 61.
 February 2nd, 2014, 10:39 AM #4 Senior Member     Joined: Sep 2007 From: USA Posts: 349 Thanks: 67 Math Focus: Calculus Re: for N Nevermind.

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