Hard problem, discrete math.

Question

If x+y+z=3 , x²+y²+z²=4, and x³+y³+z³=5 what is x, y, and z? I found that none of them can be equal or greater than 2.

Answer 1

Indeed, the system has the following solutions

(1,  1-1/√2,  1+1/√2)

(1-1/√2, 1,  1+1/√2)

(1-1/√2,  1+1/√2, 1)

You are right,  they are between 1-1/√2 and 1+1/√2.