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Taiwan IMO Selection Camp, 2023

Algebra Difficulty 5.7 AIME, harder Prove it Taiwan

Let $m$ be a positive integer, and suppose the real numbers $a_1, a_2, \dots, a_m$ satisfy
$\frac{1}{m} \sum_{i=1}^{m} a_i = 1,$
$\frac{1}{m} \sum_{i=1}^{m} a_i^2 = 11,$
$\frac{1}{m} \sum_{i=1}^{m} a_i^3 = 1,$
$\frac{1}{m} \sum_{i=1}^{m} a_i^4 = 131.$
Prove that $m$ must be a multiple of 7.

Solution

Consider the polynomial $f(x) = (x-3)^2(x+4)^2 = x^4 + 2x^3 - 23x^2 - 24x + 144$ , then we have
$0 \le \frac{1}{m} \sum_{i=1}^{m} f(a_i) = 131 + 2 - 253 - 24 + 144 = 0,$
hence each $a_i$ must be either 3 or -4. Suppose among them there are $p$ copies of 3 and $q$ copies of -4, then we have
$\begin{aligned} \frac{3p - 4q}{m} &= \frac{1}{m} \sum_{i=1}^{m} a_i = 1, \\ \frac{9p + 16q}{m} &= \frac{1}{m} \sum_{i=1}^{m} a_i^2 = 11. \end{aligned}$
From this we must have $p/m = 5/7$ and $q/m = 2/7$ . This shows that $m$ must be a multiple of 7. Q.E.D.

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.