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I am new to number theory, I was wondering if the following questions have been studied before.

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).

It's easy to see that if a polynomial $P(x)$ has a (real or complex) root with modulus $1$, then for any $Q(x) \in \mathbb Z[x]$, the product $P(x)Q(x)$ does not have a big coefficient.

I was wondering if the converse of this is true:

Question: Suppose $P(x)$ does not have a root with modulus $1$. Does this implies that there exists $Q(x) \in \mathbb Z[x]$ such that $P(x)Q(x)$ has a big coefficient?

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

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I think the answer is yes.

We factor $P(x)$ as $$a_i\prod_{i = 1}^n (x - r_i), r_i \in \mathbb{C}.$$ Then it is a familiar fact that $$R_N(x) = \prod_{i = 1}^n (x^N - r_i^N)$$ is a polynomial in $\mathbb{Q}[x]$ that is divisible by $P(x)$. We now show that $R_N(x)$ has a big coefficient for all sufficiently large $N$.

WLOG, assume $r_1, \cdots, r_k$ have modulus greater than $1$, and $r_{k + 1}, \cdots, r_n$ has modulus less than $1$. Let $$\alpha = \max\left(\frac{1}{|r_1|}, \cdots, \frac{1}{|r_k|}, |r_{k + 1}|, \cdots |r_n|\right) < 1.$$ If we expand $R_N(x)$ as $$R_N(x) = \sum_{k = 0}^n b_{kN} x^{kN},$$ then we have $$\frac{b_{n - k}}{r_1^N \cdots r_k^N} = (-1)^k + \sum_{\{i_1, \cdots, i_k\} \neq \{1,2,\cdots, k\}} \frac{(-1)^k r_{i_1}^N \cdots r_{i_k}^N}{r_1^N \cdots r_k^N}.$$ Now we notice that $$\left\vert\frac{(-1)^k r_{i_1}^N \cdots r_{i_k}^N}{r_1^N \cdots r_k^N}\right\vert \leq \alpha^N.$$ So by the triangle inequality, we have $$|b_{n - k}| \geq |r_1^N \cdots r_k^N| \cdot \left(1 - \binom{n}{k} \alpha^N\right).$$ Similarly, we can check that for any $j \neq n - k$, we have $$|b_j| \leq \binom{n}{j} |r_1^N \cdots r_k^N| \cdot \alpha^N.$$ So we conclude that $$\frac{|b_{n - k}|}{|b_0| + \cdots + |b_N|} \geq \frac{1 - \binom{n}{k} \alpha^N}{1 + 2^n \alpha^N}$$ which actually converges to $1$ as $N \to \infty$. Thus, for sufficiently large $N$, $R_N$ has a big coefficient.

Finally, we can set $Q_N(x) = \frac{R_N(x)}{P(x)}$. This is a polynomial in $\mathbb{Q}[x]$, and we can clear the denominator to get a polynomial in $\mathbb{Z}[x]$, as scalar multiplication don't break the big coefficient.

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  • $\begingroup$ I haven't looke over your proof super well, but I am cofused how the final part works: if I understand correctly, you have to show that the left hand side is strictly bigger than 1 for some $N$. You showed that for any $\epsilon > 0$, there's an $N$ for which the left hand side is bigger than $1 - \epsilon$. Those are not the same statements. You have, I think, not shown that there is an $N$ for which LHS $\geq 1$, nor the stronger statement that LHS is strictly bigger than 1. $\endgroup$
    – kabel abel
    Commented Nov 7, 2023 at 14:50
  • $\begingroup$ @kabelabel I think I only need to show that LHS is at least $1/2$ $\endgroup$
    – abacaba
    Commented Nov 7, 2023 at 17:22
  • $\begingroup$ @abacaba OH okay, I though the denominator excluded $|b_{n-k}|$ for some reason. $\endgroup$
    – kabel abel
    Commented Nov 7, 2023 at 17:24

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