Complex numbers, polynomials

Question

Let $a$ be complex number such that $a^5 + a + 1 = 0$. What are possible values of $a^2(a - 1)$?

I have tried to find $a$. Is there any way to find it?

Answer 1

$$ \begin{align} a^5+a+1&=(a^2+a+1)(a^3-a^2+1)\\ &=(a^2+a+1)(a^2(a-1)+1) \end{align}$$

If $a$ vanishes the second factor then $a^2(a-1)=-1$.

If $a$ vanishes the first factor, then divide $a^2(a-1)$ by $a^2+a+1$. We get $$a^3-a^2 = (a-2) × (a^2+a+1)+a+2$$

Therefore $a^2(a-1)=a+2$. We know that the roots of $a^2+a+1$ are $\frac{-1\pm\sqrt{1-4}}{2}$. Then only need to add $2$ to them.

In conclusion, the possible values are

$$-1, \frac{3\pm\sqrt{-3}}{2}.$$

Answer 2

Groebner Basis is a general method which (among other things) allows variable elimination. In the original system of equations we want to exclude variable a (aka "x") so we obtain equation over a^2(a−1) (aka "y"), hopefully of small polynomial degree. Derivation is entirely mechanical; one just copied and pastes CAS command

 GroebnerBasis[{x^5+x+1=0, x^2(x−1)=y} ,{x,y},{x}]

into WolframAlfa or some such and witnesses

 {3 - 2 y^2 + y^3}