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Consider a polynomial $P(x) = \sum_{i=1}^{n}{a_ix^{i-1}}$ in $\mathbb{C}$.

Is it true that if $\{a_i\}$ are positive and not all equal, then $P(\exp(\frac{2i\pi}{n})) \neq 0$ ?

Thanks

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2 Answers 2

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It is a partial answer.

It seems the following.

Put $\zeta_n=\exp(\frac{2i\pi}{n})$. Then $\zeta_n$ is a primitive $n$-th root of the unit. Then a minimal polynomial of $\zeta_n$ over a filed $\Bbb Q$ is the $n$-th cyclotomic polynomial $\Phi_n$, defined by

$$\Phi_n(x)=\prod (x−\zeta)$$

where $\zeta$ ranges over the primitive $n$-th roots of unity. $\Phi_n(x)$ is also given inductively by $$\Phi_n(x)=\frac{x^n−1}{\prod \Phi_d(x)}$$

where $d$ ranges over the proper divisors of $n$. Hence $\Phi_p(x)= \frac{x^p−1}{x-1}=x^{p-1}+x^{p-2}+\dots+1$ for each prime $p$. Also it is well known (see the references), that the polynomial $\Phi_n$ is irreducible over a field $\Bbb Q$.

Suppose that the number $n$ is not prime. Let $p$ be a prime divisor of $n$ and $k=n/p$. Then $$\Phi_n(x)|\frac{x^n-1}{\Phi_p(x)}=\frac{x^n-1}{x^p-1}(x-1).$$ Since the polynomial $\Phi_n(x)$ is irreducible, $\Phi_n(1)\ne 0$, so $$\Phi_n(x)|\frac{x^n-1}{x^p-1}= x^{(k-1)p}+x^{(k-2)p}+\dots+1=Q(x).$$ Then $\deg Q=n-p< n-1$ and all coefficients of the polynomial $Q$ are positive. Put $$P(x)=(x^{p-1}+2)Q(x).$$ Then all coefficients of the polynomial $P$ are positive, not all of them are equal and $$P(\zeta_n)= (\zeta_n^{p-1}+2)Q(\zeta_n)=0.$$

If the number $n$ is prime, then the polynomial $\Phi_n(x)$ has degree $n-1$ and it is irreducible over a field $\Bbb Q$. So if $P(x)\in\Bbb Q[x]$, $\deg P=n-1$ and $P(\zeta_n)=0$ then $P(x)=\lambda \Phi_n(x)$ for some constant $\lambda\in\Bbb Q$. So in this cases all coefficients of the polynomial $P(x)$ are equal.

Update. Consider the rest of cases.

If $n=1$ then $P(x)=a_1>0$ for all $x$, so $P(\zeta_n)=a_1>0$, so the claim is true.

If $n=2$ then $P(x)=a_2x+a_1$, $\zeta_n=-1$. If $P(\zeta_n)=0$ then $a_1=a_2$, so the claim is true too.

If $n=3$ then $P(x)=a_3x^2+a_2x+a_1$, $\zeta_n=\in\Bbb C\setminus\Bbb R$ and $Q(\zeta_n)=0$, where $Q(x)=x^2+x+1$. If $P(\zeta_n)=0$ then $a_3\ne 0$, so $P(\zeta_n)-a_3Q(\zeta_n)=0$, which is possible only if $P(\zeta_n)-a_3Q(\zeta_n)=0$. So all coefficients of the polynomial $P$ are equal, and the claim is true.

Now suppose that $n\ge 5$ is a prime. Let $k$ be the smallest positive integer such that $2^{k+1}>n$. Then $2^{k}<n$ (the inequality is strict, because $n$ is odd). Put $\xi=\zeta_n^{2^{k-1}}=\cos\frac{2^k\pi}{n}+i\sin\frac{2^k\pi}{n}$. Then $\xi$ is a root of an equation $(y-\xi)(y-\bar\xi)=y^2-2\cos\frac{2^k\pi}{n}y+1=0$. So $\zeta_n$ is a root of a polynomial $R(x)=x^{2^k}-2\cos\frac{2^k\pi}{n}x^{2^{k-1}}+1$. But the choice of the number $k$ implies that $\frac{\pi}{2}<\frac{2^k\pi}{n}<\pi$. So $-2\cos\frac{2^k\pi}{n}>0$ and all coefficients of the polynomial $R(x)$ are positive. Moreover, $-2\cos\frac{2^k\pi}{n}\ne 1$, because in the opposite case we have $n=3\cdot 2^{k-1},$ which is impossible, because $n\ge 5$ is a prime. So not all coefficients of the polynomial $R$ are equal and then $\zeta_n$ is a root of a polynomial $P(x)=R(x)x^{n-2^k-1}$, and not all coefficients of the polynomial $P$ are equal.

The concluding answer: The claim holds only when $1\le n\le 3$.

References

[Ge] Yimin Ge, Elementary Properties of Cyclotomic Polynomials

[Jam] G.J.O. Jameson, The cyclotomic polynomials.

[Lan] Serge Lang, Algebra, Addison-Wesley, Reading, Mass., 1965 (Russian translation, Moskow, 1968) [VIII, $\S$ 3].

[War] B. L. van der Waerden, Algebra (Russian translation), [$\S 42, 60$]

[Wei] Steven H. Weintraub, Several proofs of the irreducibility of the cyclotomic polynomials

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  • $\begingroup$ Thank you for your answer. $\endgroup$
    – John
    Commented Dec 25, 2014 at 14:19
  • $\begingroup$ @John I completed my answer. $\endgroup$ Commented Dec 25, 2014 at 18:50
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No.

$P(x) = ax^3+bx^2+ax+b$.

We get $P(exp(2i\pi/4))=0$

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