I claim $P(x)=x^{2023}$ is the only such polynomial. Set $N=2023$ and let \[ -x_1\le \cdots \le -x_k\le 0\le x_{k+1}\le \cdots \le x_N<1 \]be the roots of $P$, where $x_i\ge 0$. In particular, using Vieta's relations, we have $\textstyle \sum_{1\le i\le k}x_i = \sum_{k+1\le i\le n}x_i$. We have $1=P(1)=\textstyle (1+x_1)\cdots(1+x_k)(1-x_{k+1})\cdots(1-x_n)$. Note that each term appearing in the product is positive. Using AM-GM, \[ 1\le \left(1 + \frac{1}{2023}\sum_{1\le i\le k}x_i - \frac{1}{2023}\sum_{k+1\le i\le n}x_i\right)^{2023}=1. \]Thus, $1+x_1=\cdots=1+x_k=1-x_{k+1}=\cdots=1-x_n$. This implies $x_i+x_j=0$ for any $i\in\{1,\dots,k\}$ and $j\in\{k+1,\dots,n\}$, so $x_i=0$ identically. Hence, $P(x)=x^{2023}$.

Let the roots be $-r_1,\ldots,-r_{2023}$. Then $r_i>-1$ for every $i$, $P(x)=(x+r_1)\ldots (x+r_{2023})$ and $r_1+\ldots+r_{2023}=0$ $1=P(1)=(1+r_1)\ldots (1+r_{2023})$. Denote by $u_i=1+r_i$ - positive. Then $u_1+\ldots+u_{2023}=2023$ and $u_1\ldots u_{2023}=1$, so all $u$ are $1$. Then $P(x)=x^{2023}$, and it works.

Write $P(x) = (x-\alpha_1)(x-\alpha_2)\cdots (x-\alpha_n)$ with $\alpha_i \in \mathbb{R}_{<1}$ By AM-GM and Vieta: \[ 1 = P(1) = \prod_{i=1}^n (1-\alpha_i) \leq \left( \frac{\sum_{i=1}^n 1 - \alpha_i}{2023} \right)^{2023} = \left( \frac{2023 - a_{2022}}{2023} \right)^{2023} = 1\]This works since $1 - \alpha_i > 0$ Equality in AM-GM holds if and only if all the terms are equal, therefore $1-\alpha_i = 1-\alpha_j \implies \alpha_i = \alpha_j$ for all $1\leq i,j \leq n$. Thus: \[ P(n) = (x-\alpha)^{2023} \implies 1 = P(1) = (1-\alpha)^n \implies \alpha = 0 \]Therefore $P(x) = x^{2023}$

Noice Problem

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