VMO

a) Let $\alpha$ be an integer root of $P$. We investigate two cases. If $\alpha \ge 0$ then $P(\alpha )\ge a_0>0$. If $\alpha \le -2$ then $$P(\alpha )=\sum _{i=0}^{10}(a_{2i+1}\alpha +a_{2i}){\alpha }^{2i}\le \sum _{i=0}^{10}(-2a_{2i+1}+a_{2i}){\alpha }^{2i}<0.$$ Therefore, the only integer root of $P$ is $\alpha =-1$.

b) For each $i\in \{0,1,\dots ,10\}$, let $b_i=a_{2i+1}-a_{2i}$ then $b_0+\cdots +b_{10}=0$. From $|a_{k+2}-a_k|\le c$ for all $k\in \{0,1,\dots ,19\}$, we get that $$|b_k-b_{k+1}|=|a_{2k+1}-a_{2k+3}+a_{2k+2}-a_{2k}|\le 2c,\forall k\in \{0,1,\dots ,9\}.$$ Using triangle inequality, we obtain that $$|b_i-b_j|\le 2(i-j)c\text{for all }i>j\text{ and }i,j\in \{0,1,\dots ,10\}.$$ Therefore, $$\sum _{k=0}^{10}{b}_k^2=\sum _{k=0}^{10}(b_k-b_5{)}^2+2b_5\sum _{k=0}^{10}{b}_k-10b_5^2\le 4c^2\sum _{k=0}^{10}(k-5{)}^2=440c^2.$$ □