Determine all integer solutions (x,y,z) for the system of equations
x²z + y²z + 4xy = 40,
x² + y² + xyz = 20
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Submitted by Bractals
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Rating: 2.5000 (6 votes)
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Solution:
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Subtract 2 times the second equation from the first to obtain
x2z + y2z + 4xy -
2x2 - 2y2 - 2xyz = 0
Factoring this equation we get
(z - 2)(x - y)2 = 0
Therefore, z = 2 or x = y.
If z = 2, then the second equation gives (x + y)2 = 20. Since 20 is not a perfect square, no integer solutions are possible.
If x = y, then the second equation gives x2(z + 2) = 20. Since the only perfect squares that divide 20 are 1 and 4, x = 1, -1, 2, and -2 with values for z of 18, 18, 3, and 3 respectively.
Therefore, the integer solutions for (x,y,z) are (1,1,18), (-1,-1,18), (2,2,3), and (-2,-2,3). |
