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!

  • 6
    $\begingroup$ Simply notice that when $n$ is even, there is no real number $x\ne1$ such that $x^{n+1}=1$. $\endgroup$ Aug 26, 2023 at 13:27

2 Answers 2


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.

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

  • 1
    $\begingroup$ What is $x_i$? I think, you should rewrite the expression for general even $n$, better. $\endgroup$
    – Bob Dobbs
    Aug 26, 2023 at 15:30

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