Let $a_i=1011b_i$ then $1 \leq b_i<2$ Let $t$ is integer root, then $t<0$ and $b_{21}|t|^{21} \leq b_{20}|t|^{20}+...+b_1|t|+b_0$ so $|t|^{21}<2(|t|^{20}+...+|t|+1) \to |t|^{22} -|t|^{21} < 2|t|^{21}-2 \to |t|<3$ and so $t=-1$ or $t=-2$ If $t=-2$ then $0=b_0-2b_1+...-b_{21}2^{21} < 2(1+2^2+...+2^{20})-(2+2^3+...+2^{21})=0 \to $ contradiction. So $t=-1$ is root So $a_0-a_1+a_2-a_3+...+a_{20}-a_{21}=0$ Let $a_{2k+1}-a_{2k}=c_k$ We have $c_0+...+c_{10}=0$ Let $S=c_0^2+...+c_{10}^2$ $|c_{k+1}-c_k|=|a_{2k+3}-a_{2k+2}-a_{2k+1}+a_{2k}| \leq |a_{2k+3}-a_{2k+1}|+|a_{2k+2}-a_{2k}| \leq 2c$ Let $0 \leq j<i \leq 10$ then $|c_i-c_j| \leq |c_i-c_{i-1}|+|c_{i-1}-c_{i-2}|+...+|c_{j+1}-c_j| \leq 2(i-j)c$ $c_i^2+c_j^2-4(i-j)^2 c^2 \leq 2c_ic_j$ $0=(c_0+...+c_{10})^2 =c_0^2+...+c_{10}^2+2\sum_{0 \leq j<i \leq 10} c_ic_j \geq S + \sum_{0 \leq j<i \leq 10}(c_i^2+c_j^2-4(i-j)^2c^2)=11S-4\sum_{0 \leq j<i \leq 10}( i-j)^2c^2 =11S-4*1210c^2 \to S \leq 440c^2$

jacks3000 wrote: Let the polynomial $P(x)=a_{21}x^{21}+a_{20}x^{20}+\cdots +a_1x+a_0$ where $1011\leq a_i\leq 2021$ for all $i=0,1,2,...,21.$ Given that $P(x)$ has an integer root and there exists an positive real number$c$ such that $|a_{k+2}-a_k|\leq c$ for all $k=0,1,...,19.$ a) Prove that $P(x)$ has an only integer root. b) Prove that $$\sum_{k=0}^{10}(a_{2k+1}-a_{2k})^2\leq 440c^2.$$ (1) Obviously if ${a}$ is a integer root of $P(x)$, then $a<0$. For $1\leq i\neq j\leq n$, we have $\frac 12<\frac{1011}{2021}\leqslant\frac{a_i}{a_j}\leqslant\frac{2021}{1011}<2$. For $0\leq i\leq 20$, define $q_i=\frac{a_i}{a_{21}}$, then $q_i\in \left(\frac 12,2\right)$ and ${a}$ satisfies $a^{21}+q_{20}a^{20}+\cdots +q_1a+q_0=0$. Therefore $1+\frac{q_{20}}a+\cdots +\frac{q_0}{a^{21}}=0$. If $|a|\geqslant 3$, $0\geqslant 1-\frac {2}{3}+\frac{1}{2\times 3^2}-\frac{2}{3^3}+\frac{1}{2\times 3^4}+\cdots >\frac 13-\frac{1}{2\times 3^3}\left(\dfrac{1}{1-\frac 19}\right)>0$. Therefore $a\in \{-1,-2\}$. If $a=-2$, for $0\leq i\leq 10$, $a_{2i+1}(-2)^{2i+1}+a_{2i}(-2)^{2i}=2^{2i}(-2a_{2i+1}+a_{2i})<0$. So $a=-1$.$\blacksquare$