Determine all integer solutions (x,y,z) for the system of equations

x²z + y²z + 4xy = 40,

x² + y² + xyz = 20

LET

X^2+Y^2=W AND XY=T

WE GET: WZ+4T=40 (1)

AND W+TZ=20 ( 2) multiply by z and

SUBTRACT (1)-(2)*Z ====> T=20/(2+Z)

CHECK FOR Z = 0,2,3,8 AND 18 to get integer T and then evaluate W. ONLY Z=3 AND Z=18 QUALIFY.

The rest follows easily.T=4 ===>x=y +2 or -2

T=1 x=y +1 or -1

remark:

There is much shorter way to solve- by reasoning that symmetry implies x=y (writing x instead of y and vice versa leaves the set of equations unchanged...)

So ( 2x^2)*z+4*x^2=40

the other equation becomes redundant, - both imply

x^2=20/(z+2 ) and we throw away the non-integer solutions

ady

*Edited on ***April 1, 2006, 4:05 am**