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Take Second Degree, Solve For Real (Posted on 2007-04-27) Difficulty: 2 of 5
Determine all possible real pairs (m,n) satisfying the following system of equations:

mn2 = 15m2+ 17mn + 15n2

m2n = 20m2 + 3n2

See The Solution Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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Solution Solution | Comment 2 of 4 |

Rewriting the equations:
  mn(n-17) = 15(m^2 + n^2)        (1)
  m^2*(n-20) = 3n^2               (2)
Eq. (2) implies n != 20 and n != 17.
Multiplying (1) by n-20 gives
  mn(n-17)(n-20) = 15(m^2^[n-20] + n^2*[n-20])
                 or
  mn(n-17)(n-20) = 15(3n^2 + n^2*[n-20])
                 or
  mn(n-20) = 15n^2                (3)    
Squaring eq. (3) gives
  n^2*m^2*(n-20)(n-20) = 15^2*n^4
                 or
  n^2*3n^2*(n-20) = 15^2*n^4
                 or
  n^4*(n-95) = 0
Which implies n = 0 or n = 95.
Eq. (2) implies (m,n) = (0,0), (-19,95), and
(19,95) are solutions. But, eq. (1) implies
(-19,95) is not a solution. Therefore,
         (m,n) = (0,0) and (19,95)
            

  Posted by Bractals on 2007-04-27 12:05:06
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