Determine all integer solutions (x,y,z) for the system of equations
x²z + y²z + 4xy = 40,
x² + y² + xyz = 20
(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 20060330 14:41:07 