2014/021)Solve for real solutions of the following system:

2z+z^2x=x
2y+y^2z=z
2x+x^2y=y

Solution

Z cannot be +/-1 because if z = +/- 1 then 2+x = x is a contradiction. Similarly x and y cannot be -1.

So we get

x = 2z/(1-z^2)

y = 2y/(1-y^2)

z = 2x/(1-x^2)


letting z = tan a we get x = tan 2a, y = tan 4a, z = tan 8a

or tan 8a = tan a

so 8a = npi + a

or 7a = npi

so solution set

z= tan (npi/7), x = tan (2npi/7), y = tan (4npi/7)

for n = 0 to 6