Show that if $ \sigma $ is a solution to $ x^2 + x + 1 = 0 $ then the following equality occurs.

Question

Show that if $ \sigma $ is a solution to the equation $ x^2 + x + 1 = 0 $ then the following equality occurs:

$$ (a +b\sigma + c\sigma^2)(a + b\sigma^2 + c\sigma) \geq 0 $$

I looked at the solution in my textbook and it says I should multiply the parentheses and take into account that $ \sigma + \sigma^2 + 1 = 0 $. I tried factoring the rest but I just can't seem to manage to solve it?

Maybe I messed up at multiplying the parentheses? Here's what I got:

$$ a^2 + ab\sigma^2 + ac\sigma + ab\sigma + b^2\sigma^3 + bc\sigma^2 + ac\sigma^2 + bc\sigma^4 + c^2\sigma^3 $$

Answer 1

First note that $ \sigma ^3=1$. So the expression can be rewritten as \begin{eqnarray*} a^2+b^2+c^2+(ab+bc+ca)(\sigma+\sigma^2)=a^2+b^2+c^2-(ab+bc+ca) \\ =\frac{1}{2}((a-b)^2+(b-c)^2+(c-a)^2) \end{eqnarray*} which is clearly non-negative.

Answer 2

Hint: look at the $ab$ terms, where you have $ab\sigma^2+ab\sigma=ab(\sigma^2+\sigma)$: now what can you do with your existing hint to simplify this part of your expression?

That should get you a start.

Answer 3

Hint Rearranging the defining equation gives $\require{cancel} \sigma^2 = -\sigma - 1$. So, for example, the second term, $a b \sigma^2$ becomes $-a b \sigma - a b$.

Alternatively, we can avoid tedious computation by showing that $$(\sigma^2)^2 + (\sigma^2) + 1 = 0 ,$$ which implies that $\sigma^2$ is also a root of $x^2 + x + 1$, and hence that $\sigma^2 = \bar \sigma$. Here (provided $a, b, c$ are all real---the claim isn't true for general complex parameters) we can rewrite the inequality as $$z \bar z \geq 0 , \qquad \textrm{where} \qquad z := a + b \sigma + c \sigma^2 .$$

Answer 4

I suppose $a,b,c$ are real numbers. Now observe $\sigma$ is a non-real cubic root of unity, so the other non-real root is $\bar \sigma=\sigma^2$. Thus \begin{align} (a +b\sigma + c\sigma^2)(a + b\sigma^2 + c\sigma)&=(a +b\sigma + c\bar\sigma)(a + b\bar\sigma + c\sigma)=(a +b\sigma + c\bar\sigma)(\overline{a + b\sigma + c\bar\sigma})\\ &=\bigl|a +b\sigma + c\bar\sigma\bigr|^2. \end{align}