# Multiply an integer polynomial with another integer polynomial to get a "big" coefficient

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.

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

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.