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Integer Solutions (Posted on 2006-03-30) Difficulty: 3 of 5
Determine all integer solutions (x,y,z) for the system of equations

x²z + y²z + 4xy = 40,

x² + y² + xyz = 20

See The Solution Submitted by Bractals    
Rating: 2.5000 (6 votes)

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Some Thoughts re: I think I got it. | Comment 3 of 16 |
(In reply to I think I got it. by Jer)

Excellent, I like the method you used.  You have a typo, though, it should be (z+2)=20 in the next to last line.

Building off your work, (x^2 + y^2) must be positive, and since it times (z+2) equals 40, then doesn't it follow that (z+2) has to be positive?

This would limit z to just -1 and -2 as far as negative integers go, and I don't see how either of those would work.  If you try z = -1, then in the first equation you have -x^2 - y^2 +4xy = 40, or

4xy = 40 + x^2 + y^2,

which just doesn't seem possible.  Trying z = -2, you run into the same problem, only to a greater magnitude.


  Posted by tomarken on 2006-03-30 14:41:07
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