# Show that, if $n$ is even, then the polynomial $p(x) =x^n + x^{(n-1)}+...+x+1$ does not have real roots.

This question is in a complex numbers exercises list.

Instead of a full answer, a suggestion was given to me on the solutions part of the list, which says to multiply $$p(x)$$ by $$(x-1)$$. I have done that by unfortunately do not know how does that help on proving what the question asks. In my "proof", I do that and then I try to solve $$p(x)=0$$ to find roots: $$p(x)\cdot(x-1)=(x^{n+1}+x^{n}+...+x^2+x-x^n-x^{n-1}-...-x-1)\Rightarrow$$ $$\Rightarrow p(x)\cdot(x-1)=x^{n+1}-1\Rightarrow$$ $$\Rightarrow p(x)= \frac{x^{n+1}-1}{x-1}=0 \therefore n=-1, x \ne 1$$

Could anyone please help me on this problem? How does this train of thought help? Or even, is there any other proof that even a math beginner like me can work it out (preferably using complex numbers)? Thank you for reading!

• Simply notice that when $n$ is even, there is no real number $x\ne1$ such that $x^{n+1}=1$. Aug 26, 2023 at 13:27

If $$a$$ is a zero of $$p(x)$$, then it is a zero of $$p(x)\cdot(x-1)=x^{n+1}-1$$. It follows that $$a^{n+1}=1$$. If $$a$$ was real, then necessarily $$a=\pm 1$$, though we can immediately exclude $$a=-1$$ because $$n+1$$ is odd. Also $$a=1$$ can be excludef because $$p(1)=n+1.$$ Thus, no zero $$a$$ of $$p(x)$$ can be real.

• Zuy, since $n$ is even, the equation $a^{n+1}=1$ just has one solution which is $1$, hence $a$ cannot be equal to $-1$. Aug 26, 2023 at 17:07
• @Angelo You have a point.
– Zuy
Aug 26, 2023 at 21:49
• Zuy, you too (+1), indeed I have upvoted your answer just now. Aug 26, 2023 at 21:51

As far as I am concerned, there are multiple ways to tackle this problem.

The most natural solution I found is to divide $$p(x)$$ into perfect square expressions. For example, when $$n=4$$, $$p(x)$$ can be rewritten as: $$(1/2) + (1/2*{(x+1)}^2) + (1/2*{(x+1)}^2*x^2) + (1/2*x^4)$$. Where I divide $$p(x)=x^4+x^3+x^2+x^1+1$$ into $$1/2*x^4+(1/2x^4+x^3+1/2x^2)+(1/2x^2+x+1/2)+1/2$$.

For any $$n$$, $$p(x)$$ can be rewritten as:

$$p(x) = 1/2*[1+[(1+x)^2*\sum_{i \text{ is even }}x^{i}]+x^n],$$ where $$i$$ range from $$0$$ to $$n-2$$.

Since $$1/2>0$$, $$x^{i}>0$$ for all real numbers, $$(x+1)^2>0$$ and $$x^{n}>0$$, the equation is positive. Indicating there will be no real number of solutions.

• What is $x_i$? I think, you should rewrite the expression for general even $n$, better. Aug 26, 2023 at 15:30