February 2nd, 2014, 03:41 AM  #1 
Senior Member Joined: Nov 2013 Posts: 434 Thanks: 8  for N
solve in N natural number x^3y^3xy=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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