I have copied this question from StackExchange, in the hope that some experts here can provide some relevant insight. Thanks to Greg Martin for improving the question.

Given $f(x) = a_0 + a_1 x + a_2 x^2 \cdots + a_n x^n \in \mathbb Z[x]$, we say that $f(x)$ has a big coefficient $a_i$ if $|a_i| > \frac{1}{2} \sum_{k=0}^n\left|a_k\right|$ (so that $|a_i|$ is larger than the sum of the remaining coefficients' absolute values).

Suppose $P(x) \in \mathbb Z[x]$ has degree at least 3.

Are these two conditions equivalent?

  1. $P(x)$ does not have a root with modulus $1$.

  2. There exists $Q(x) \in \mathbb Z[x]$ such that $P(x)Q(x)$ has a big coefficient.

(It is clear that they are equivalent when $P(x)$ has degree $2$.)

Thank you for reading. Any relevant idea/reference would be really appreciated.

  • $\begingroup$ What does "root with modulus 1" mean? $\endgroup$
    – Max Horn
    Commented Nov 6, 2023 at 23:52
  • 5
    $\begingroup$ @MaxHorn: a $z \in \mathbb{C}$ with $P(z)=0$ and $|z|=1$. What else could it mean? $\endgroup$ Commented Nov 7, 2023 at 1:47
  • 1
    $\begingroup$ I have no idea, that's why I asked. TIL: complex modulus is a synonym for (complex) norm / absolute value. $\endgroup$
    – Max Horn
    Commented Nov 7, 2023 at 6:50
  • $\begingroup$ It is usual that you get answers on both sites if you do not wait sufficiently long. $\endgroup$ Commented Nov 7, 2023 at 13:03

1 Answer 1


A similar fact is well known: A polynomial $f(x)$ with complex coefficients divides a polynomial with positive coefficients if and only $f(x)$ has no nonnegative root. A similar strategy works here.

Surely, we may work with polynomials with real coefficients, and we will do that (approximation works).

Say that a polynomial is $N$-extremal (for $N>1$) if $|a_i|> N\sum_{j\neq i}|a_j|$ for some $i$.

Lemma. The product of two $N$-extremal polynomials is $\frac{N-1}2$-extremal.

Proof. We may assume that $$ N=|a_i|>N\sum_{j\neq i}|a_j| \quad\text{and}\quad N=|b_s|>N\sum_{t\neq s}|b_t| $$ for the coefficients of the two factors. If the $c_k$'s are the product's coefficients, then $$ |c_{i+s}|\geq |a_i||b_s|-\sum_{j\neq i}\sum_{t\neq s}|a_j||b_t|>N^2-1, $$ while $$ \sum_{k\neq i+s}|c_k|\leq |a_i|\sum_{t\neq s}|b_t|+|b_s|\sum_{j\neq i}|a_j|+\sum_{j\neq i}|a_j|\sum_{t\neq s}|b_t|< 2N+1, $$ and $\frac{N^2-1}{2N+1}>\frac{N-1}2$. $\square$

Now consider a (say, monic) polynomial $f(x)$ with no roots on the unit circle. Expand it as $$ f(x)=\prod_i(x-a_i)\prod_j(x-b_j)(x-\bar b_j), $$ where $a_i\in\mathbb R$, $b_j\in\mathbb C\setminus \mathbb R$. By the Lemma, it suffices to find multiples of each of $x-a_i$ and $(x-b_j)(x-\bar b_j)$ which are $N$-extremal for a sufficiently large $N$.

These multiples are respectively $x^n-a_i^n$ and $(x^n-b_j^n)(x^n-\bar b_j^n)$ for some sufficiently large $n$.

To summarise, the desired multiple that works has the form $$ f(x)f(\zeta x)\cdots f(\zeta^{n-1}x), $$ where $\zeta$ is some primitive $n$-th degree root of unity. It even has integer coefficients, if $f$ has integer coefficients.


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