An interesting property of the roots

Question

Let $x^2+x+1=0$ be a quadratic equation having two roots $x_1$ and $x_2$.

How would you prove that $(x_1)^2=x_2$ and $(x_2)^2=x_1$?

P.S. I could show $(x_1)^2=x_2$ and $(x_2)^2=x_1$ by finding the roots explicitly then squaring and comparing. But that's not a proper approach.

Answer 1

Note that we have $$x_1^2+x_1+1=x_2^2+x_2+1 $$

That is $$ (x_1-x_2)(x_1+x_2 +1) =0$$

Since $$x_1 \ne x_2$$ we get $$x_1+x_2 +1=0$$

Compare with $$x_1^2+x_1+1=x_2^2+x_2+1=0 $$

You get $$x_1=x_2^2$$ and $$x_2 = x_1 ^2 $$

Answer 2

There are a few ways. Notice that $\, x^3 - 1 = (x - 1)(x^2 + x + 1). \,$ If $\,x_1\,$ is a root of the equation $\, E_1\!:\, x^2 + x + 1 = 0\,$ then this implies $\, x_1^3 = 1 \,$ and $\,x_1\,$ is a root of the equation $\, E_2\!:\, x^3 -1 = 0.\,$ Define $\, x_2 := x_1^2. \,$ Now $\, x_2^3 = (x_1^2)^3 = (x_1^3)^2 = 1^2 =1 \,$ which implies that $\, x_2 \,$ is also a root of equation $\, E_2, \,$ Since $\,x_1 \ne 1, \,$ $\, x_2 \ne 1 \,$ and $\, x_1 \ne x_2, \,$ then $\,x_2\,$ is the other root of the equation $\, E_1.\,$