We essentially need to find a monic polynomial $P(x)$ such that all coefficients of $P^2(x) - \alpha^2$ are bounded by $\eta$, where $\alpha, \eta$ are given positive numbers. Let $n$ be a positive integer, $A$ be a set contained in $\{1, \dots, 2n\}$ and $\epsilon$ be a positive constant to be determined. Take \[ P(x) = x^{2n+1} + \epsilon \cdot \sum_{i \in A} x^{i} - \alpha. \]Then for $1 \leq m < 2n+1$, the coefficient of $x^{m}$ in $P^2(x) - \alpha^2$ is \[ -2 \alpha \epsilon \cdot \mathbf{1}_{\{m \in A\}} + \epsilon^2 \cdot \sum_{i, j \in A, i+j=m} 1. \]Here $\mathbf{1}_{\{m \in A\}}$ represents the indicator function for $m$ belonging to $A$. For $2n+1 < m \leq 4n+1$, the coefficient of $x^m$ is \[ 2 \epsilon \cdot \mathbf{1}_{\{m-2n-1 \in A\}} + \epsilon^2 \cdot \sum_{i, j \in A, i+j=m} 1. \] Finally, for $m = 2n+1$, the coefficient of $x^m$ is \[ - 2 \alpha + \epsilon^2 \cdot \sum_{i, j \in A, i+j=2n+1} 1. \] Our goal is to make all these coefficients bounded by $\eta$ in absolute value. For $2 \leq m \leq 4n$, let $V(m) = \sum_{i, j \in A, i+j = m} 1$ be the counting function for $A + A$. Suppose we can find $n$ and $A$ such that $V(2n+1)$ is large while all other $V(m)$ are much smaller. Then, by choosing $\epsilon$ so that $\epsilon^2 V(2n+1) = 2\alpha$, we see that the coefficients $-2 \alpha \epsilon \cdot \mathbf{1}_{\{m \in A\}} + \epsilon^2 V(m)$ (for $m < 2n+1$) and $2 \epsilon \cdot \mathbf{1}_{\{m-2n-1 \in A\}} + \epsilon^2 V(m)$ (for $m > 2n+1$) are both small, completing the proof. It remains to find such $n$ and $A$. Let $B \subset \{1, \dots, n\}$ and let $A = B \cup (\{2n+1\} - B)$. Then obviously $V(2n+1) = 2 \lvert B \rvert$. On the other hand, suppose $B$ has the property that its elements have pairwise distinct differences, then it can be shown that $V(m) \leq 4, \forall m \neq 2n+1$. Indeed, without loss assume $m < 2n+1$. Then $m$ can be represented as the sum of two elements in $A$ only if $m = i + j$ for $i, j \in B$, or $m = i + 2n+1 - j$ for $i, j \in B$. Either possibility gives rise to at most two solutions $(i, j)$, due to pairwise distinct differences. Hence, we just need to take $n$ sufficiently large and then choose a large subset $B$ having distinct differences. It is well known that such $B$ can have as many as $n^{\frac{1}{2}}$ elements, but even a greedy construction gives $n^{\frac{1}{3}}$ elements, which is more than sufficient for our argument.

This problem is proposed by Dusan Djukic.