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 Integer Solutions (Posted on 2006-03-30)
Determine all integer solutions (x,y,z) for the system of equations

x²z + y²z + 4xy = 40,

x² + y² + xyz = 20

 Submitted by Bractals Rating: 2.5000 (6 votes) Solution: (Hide) 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).

 Subject Author Date Alternative Methodology K Sengupta 2007-07-04 14:45:51 kiss- corrected and reinforced Ady TZIDON 2006-04-01 03:49:27 kiss- corrected and reinforced Ady TZIDON 2006-04-01 03:49:26 solutions salil 2006-03-31 21:24:14 re(2): KISS indeed Ady TZIDON 2006-03-31 17:43:33 re: KISS tomarken 2006-03-31 07:43:30 Solution goFish 2006-03-31 06:43:27 solution salil 2006-03-31 03:52:49 KISS Ady TZIDON 2006-03-31 00:20:23 re: Congrats, and a question John Reid 2006-03-30 17:05:25 Congrats, and a question tomarken 2006-03-30 16:28:17 Answer, without detailed explanation John Reid 2006-03-30 16:11:14 (partial?) solution Dej Mar 2006-03-30 16:08:22 re: I think I got it. tomarken 2006-03-30 14:41:07 Solution (not sure if it's complete) tomarken 2006-03-30 14:26:15 I think I got it. Jer 2006-03-30 14:25:56

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