The 65th IMO China National Team Selection Test

Proof. Let $\ell \in \{1,2,\dots ,n\}$. For a complex number $\alpha \in \mathbb{C}$, consider $$\begin{array}{rl}Q(z)& =P(z)(1+\alpha z^{\ell })\\ & =\alpha a_n{z}^{n+\ell }+\cdots +\alpha a_{n-\ell +1}{z}^{n+1}+(a_n+\alpha a_{n-\ell })z^n+\cdots +(a_{\ell }+\alpha a_0)z^{\ell }\\ & +a_{\ell -1}{z}^{\ell -1}+\cdots +a_0.\end{array}$$ Take $M>2n$, then $$\begin{array}{rl}\sum _{j=1}^M{\left|Q\left(e^{i\frac{2j\pi }{M}}\right)\right|}^2& =\sum _{j=1}^M{\left|P\left(e^{i\frac{2j\pi }{M}}\right)\right|}^2\cdot {\left|\left(1+\alpha e^{i\frac{2j\pi }{M}\ell }\right)\right|}^2\\ & \le \sum _{j=1}^M{\left|\left(1+\alpha e^{i\frac{2j\pi }{M}\ell }\right)\right|}^2=M(1+|\alpha {|}^2).\end{array}$$ On the other hand, $$\sum _{j=1}^M{\left|Q\left(e^{i\frac{2j\pi }{M}}\right)\right|}^2=M\cdot \left(|\alpha a_n{|}^2+\cdots +|\alpha a_{n-\ell +1}{|}^2+\sum _{j=\ell }^n|a_{n-j+\ell }+\alpha a_{n-j}{|}^2+|a_{\ell -1}{|}^2+\cdots +|a_0{|}^2\right).$$ Combining the above two estimates, we get $$|\alpha {|}^2(|a_n{|}^2+\cdots +|a_{n-\ell +1}{|}^2)+\sum _{j=\ell }^n|a_{n-j+\ell }+\alpha a_{n-j}{|}^2+|a_{\ell -1}{|}^2+\cdots +|a_0{|}^2\le 1+|\alpha {|}^2.$$ In particular, $$|\alpha a_n{|}^2+|a_n+\alpha a_{n-\ell }{|}^2\le 1+|\alpha {|}^2.$$ Choosing $\alpha \in \mathbb{C}$ such that $|\alpha |=\frac{1}{|a_n|}$ and $\arg (a_n)=\arg (\alpha a_{n-\ell })$ in the above equation, we get $$1+{\left(|a_n|+\frac{|a_{n-\ell }|}{|a_n|}\right)}^2\le 1+\frac{1}{|a_n{|}^2}.$$ Solving this gives $|a_{n-\ell }|\le 1-|a_n{|}^2$. The proof is complete.