I have just completed 2 days of our National Olympiad. This year's problems are not difficult, yet new and strange to many of the students. Mentioned below will be problem 5 out of 7, the first problem of Day 2.

To anyone who wonders why it is VMO 2021, we always do national olympiad for a year before the previous year ends, so that the result will be published in early 2021


Given that $P(x)=a_{21}x^{21}+a_{20}x^{20}+a_{19}x^{19}+ \dotsb + a_1x+a_0$ is a polynomial with real coefficients such that $a_i \in [1011,2021]$ for all $ i \in \{ 0,1,2,\dotsb,21\}$ and $c \in \mathbb{R}$ such that $ | a_{k+2}-a_{k} | \le c$ for all $k \in \{ 0,1,2, \dotsb,19 \}$ Given that $P(x)$ has at least one integer root.

  1. Prove that $P(x)$ has exactly one integer root
  2. Prove the following inequality: $\displaystyle\sum_{k=0}^{10} |a_{2k+1}-a_{2k}|^2 \le 440c^2$

My works

  1. For part 1, it is easy to see that since $a_i$ are all positive, then if $P(x)$ has a root, that root must be negative. From the condition $a_i \in [1011,2021]$ it is also easy to see that $ \frac{a_i}{a_j} <2$ for all $0 \le i,j \le 21$, thus if $x$ is an integer root of $P(x)$ then $x \ge (-2) $. Thus $x = (-1)$ is the only integer root of $P(x)$.

  2. For part 2, some of my teammates say that it could be solved using Jensen inequality, but none of us succeeded. I noticed that $ \displaystyle\sum_{k=0}^{10} a_{2k+1} =\displaystyle\sum_{k=0}^{10} a_{2k}$, but how do I finish the problem?

Any help is appreciated.

  • $\begingroup$ Are you able to find a (close to) equality case, say for $ c = 1$? I suspect that we have a much tighter bound of $ 60c$. $\endgroup$
    – Calvin Lin
    Dec 26, 2020 at 15:09
  • $\begingroup$ Well I doubt that 440 could not be the best coefficient to use but I still think that the power of $c$ in the inequality must be at least 2, which means it's not that tight as $60c$ (unless you can give the solution to it). Moreover, just to remind, we can not compare $60c$ to $440c^2$ as $c$ runs through the real axis. $\endgroup$ Dec 26, 2020 at 15:15
  • $\begingroup$ Right, after realizing that $c$ is real, I believe I have a counter example. Can you check? $\endgroup$
    – Calvin Lin
    Dec 26, 2020 at 15:17
  • $\begingroup$ Sorry I posted an edit to the inequality. My memory was bad $\endgroup$ Dec 26, 2020 at 15:19
  • 1
    $\begingroup$ With your edit, 440 is the best coefficient, and I have stated an equality case.IIRC there is a pretty nice approach for this problem (without the polynomial skin). $\endgroup$
    – Calvin Lin
    Dec 26, 2020 at 15:20

1 Answer 1


(This is before the edit.)

The inequality is not true.

Take $ c = 0.1$ (Any $ 0 < c < \frac{3}{22} $ will work.)

Take $ a_{2k} = 1011 + kc $, $a_{2k+1} = 1011 + (10-k) c $.
This satisfies the conditions as 1) $P(-1) = 0$ is an integer root, 2) The difference between terms is exactly $c$.

However, $ \sum |a_{2k+1} - a_{2k}| = 10c + 8c + \ldots + 0c + 2c + \ldots + 10c = 60c > 440 c^2$.

My suspicion is that $ \sum |a_{2k+1} - a_{2k}| \leq 60c$, though I don't know how to prove it. The equality case is as above.

Claim: If

  • $\sum_{i=0}^{21} (-1)^i a_i = 0$
  • $|a_{k+2} - a_k | \leq |c| $

Then $ \sum_{k=0}^{10} |a_{k+1} - a_k |^2 \leq 440 c^2$.

Equality case is when $ a_{2k} = a + kc $, $a_{2k+1} = a + (10-k) c $.

  • $\begingroup$ yes, that was my bad, i posted the wrong equality. It must be $\sum |a_{2k+1}-a_{2k}|^2$. Can you please resolve it? $\endgroup$ Dec 26, 2020 at 15:19

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