Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 6.3 National olympiad Prove it

Let $n \geq 4$ be an integer. Find all positive real solutions to the following system of $2 n$ equations:
$\begin{array}{rlrl} a_{1} & =\frac{1}{a_{2 n}}+\frac{1}{a_{2}}, & a_{2} & =a_{1}+a_{3}, \\ a_{3} & =\frac{1}{a_{2}}+\frac{1}{a_{4}}, & a_{4} & =a_{3}+a_{5}, \\ a_{5} & =\frac{1}{a_{4}}+\frac{1}{a_{6}}, & a_{6} & =a_{5}+a_{7}, \\ & \vdots & \vdots \\ a_{2 n-1} & =\frac{1}{a_{2 n-2}}+\frac{1}{a_{2 n}}, & a_{2 n} & =a_{2 n-1}+a_{1} . \end{array}$

Solution

We will prove $a_{2 k}$ is a constant sequence, at which point the result is obvious. \ First approach (Andrew Gu). Apparently, with indices modulo 2n, we should have
$a_{2 k}=\frac{1}{a_{2 k-2}}+\frac{2}{a_{2 k}}+\frac{1}{a_{2 k+2}}$
for every index $k$ (this eliminates all $a_{\text{odd}}$ 's). Define
$m=\min _{k} a_{2 k} \quad \text{and} \quad M=\max _{k} a_{2 k}$
Look at the indices $i$ and $j$ achieving $m$ and $M$ to respectively get
$\begin{aligned} & m=\frac{2}{m}+\frac{1}{a_{2 i-2}}+\frac{1}{a_{2 i+2}} \geq \frac{2}{m}+\frac{1}{M}+\frac{1}{M}=\frac{2}{m}+\frac{2}{M} \\ & M=\frac{2}{M}+\frac{1}{a_{2 j-2}}+\frac{1}{a_{2 j+2}} \leq \frac{2}{M}+\frac{1}{m}+\frac{1}{m}=\frac{2}{m}+\frac{2}{M} \end{aligned}$
Together this gives $m \geq M$ , so $m=M$ . That means $a_{2 i}$ is constant as $i$ varies, solving the problem.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.