Vietnamese Mathematical Olympiad

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 $b_0 + \dots + b_{10} = 0$. 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
$$|b_i - b_j| \le 2(i - j)c \quad \text{for all } i > j \text{ and } i, j \in \{0, 1, \dots, 10\}.$$
Therefore,
$$\sum_{k=0}^{10} b_k^2 = \sum_{k=0}^{10} (b_k - b_5)^2 + 2b_5 \sum_{k=0}^{10} b_k - 10b_5^2 \le 4c^2 \sum_{k=0}^{10} (k-5)^2 = 440c^2.$$
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