Determine all integer solutions (x,y,z) for the system of equations
x²z + y²z + 4xy = 40,
x² + y² + xyz = 20
If we devide both the equations by xy we get
z*(x/y+y/x) +4 = 40/xy and
(x/y+y/x) +z = 20/xy
(40/xy4)*z = (20/xyz)*2
Simple solution is z= 2 but then x,y = 2*5^0.5
If we add & substract 2xyz in equation 1 & 2xy in equation 2 we get
z* ( x+y)^2 +xy*(42z)=40
(x+y)^2 + xy( z2)=20
from which we get
xy*( 42zz^2+2z)=4020z
xy*(2z)(2+z)=20(2z)
xy(2+z)=20
if xy is 1 then z=18
if xy is 2 then z=8 All interger solution is not possible
if xy is 4 then z=3
if xy is 5 then z=2 This also does not have all integer solution
if xy is 10 z=0 This also does not have all integer solution

Posted by salil
on 20060331 03:52:49 