Solve the system of equations below, listing all un-ordered 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)
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Posted by Daniel
on 2016-10-27 05:29:47 |
re: my short solution
