Solve the system of equations below, listing all unordered triplets of integers:
x^2 −(y+z+yz)x+ (y+z)yz = 0
y^2 −(z+x+zx)y+ (z+x)zx = 0
z^2 −(x+y+xy)z+ (x+y)xy = 0
(In reply to
my short solution by Ady TZIDON)
there are 3 missing solutions, namely when two values are 1 and the third is 1
(1,1,1), (1,1,1), and (1,1,1)
this corresponds to x=yz, y=xz, and z=xy
using the first two we get
x=xz^2
if x=0 then we have y=0 and z=0 which we already have
if x!=0 then we have
z^2=1
case 1:
z=1
then x=y and xy=1 giving x^2=1 thus x=1 or x=1
x=1 we already have but x=1 gives us (1,1,1)
case 2:
z=1
then x=y and xy=1 giving x^2=1 x^2=1 thus x=1 or x=1
this gives us the other 2 solutions (1,1,1) and (1,1,1)

Posted by Daniel
on 20161027 05:29:47 