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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?

chaos
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2 Answers2

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

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

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

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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}
Tegiri Nenashi
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    Please explain what a Groebner Basis is, and why it is useful. – nanofarad Nov 15 '14 at 02:50
  • You might as well say "Just type roots of a^5 + a + 1 into Wolfram Alpha and make no effort to understand the underlying principals." –  Nov 15 '14 at 19:08
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    One can exhibit each step of Buchberger's algorithm; I'm not sure how they do solve[{x^5+x+1=0}] – Tegiri Nenashi Nov 15 '14 at 19:21
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    Please note, that there is a computational step in the other solution as well. In order to factorize polynomial a^5 + a + 1 you have to apply Kronecker algorithm. – Tegiri Nenashi Nov 15 '14 at 19:50
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    This answer is actually the best of the two answers. This is the method to solve not only this problem, but all problems of this type. Don't mind the blindness of those that can't see it. I can only up-vote once which I did long ago. –  Nov 16 '14 at 01:36