Determine all integer solutions (x,y,z) for the system of equations
x²z + y²z + 4xy = 40,
x² + y² + xyz = 20
Mulitply the first equation by z and the second by 4 and subtract:
x²(z²4) + y²(z²4) = 40z80
x²(z+2)(z2) + y²(z+2)(z2) = 40(z2)
x²(z+2) + y²(z+2) = 40
(x²+y²)(z+2) = 40
As the first factor must be positivewe only need to check the positive factors of 40.
(x²+y²)=40 (z+2)=1 yields (2,6,1) which doesn't work in the original system yielding 8 and 28 respectively,
(x²+y²)=20 (z+2)=2 yields (2,4,0) which doesn't work in the original system yielding 32 and 20 respectively,
(x²+y²)=10 (z+2)=4 yields (1,3,2) which doesn't work in the original system yielding 32 and 16 respectively,
(x²+y²)=5 (z+2)=8 yields (2,1,6) which doesn't work in the original system yielding 38 and 17 respectively,
(x²+y²)=4 (z+2)=10 yields (2,0,8) which doesn't work in the original system yielding 32 and 4 respectively,
(x²+y²)=1 (z+2)=40 yields (0,1,38) which doesn't work in the original system yielding 31 and 1 respectively,
but (x²+y²)=2 (z+2)=18 yields <b>(1,1,18)</b> which works.
I think this is the only solution, but looking at the original problem it still seems that a negative value for z could also give solutions.

Posted by Jer
on 20060330 14:25:56 