If $x_1>=x_2$ then $x_2>=x_3$ and so on but at the end we find $x_n>=x_1$ wich means there all equal wich they are not Whats wrong with my contradiction? So in b theres no k

parmenides51 wrote: Let $k > 1, n > 2018$ be positive integers, and let $n$ be odd. The nonzero rational numbers $x_1,x_2,\ldots,x_n$ are not all equal and satisfy $$x_1+\frac{k}{x_2}=x_2+\frac{k}{x_3}=x_3+\frac{k}{x_4}=\ldots=x_{n-1}+\frac{k}{x_n}=x_n+\frac{k}{x_1}$$Find: a) the product $x_1 x_2 \ldots x_n$ as a function of $k$ and $n$ b) the least value of $k$, such that there exist $n,x_1,x_2,\ldots,x_n$ satisfying the given conditions. Dunno why it seems hard last two year, it's pretty trivial. (guess I'm improving) a) \[ x_1 - x_2 = k \left( \frac{1}{x_3} - \frac{1}{x_2} \right) \Rightarrow \frac{x_1 - x_2}{x_2 - x_3} = \frac{k}{x_2 x_3} \]Multiply cyclically, we get $x_1 x_2 \dots x_n = k^{\frac{n}{2}}$. b) $k > 1$. Since $n$ is odd, $k$ must be a square. Thus, $k \ge 4$. This works. Take $n = 2019$, $(2,-1,4, 2, -1, 4, \dots)$.

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Samusasuke wrote: GorgonMathDota wrote: parmenides51 wrote: Let $k > 1, n > 2018$ be positive integers, and let $n$ be odd. The nonzero rational numbers $x_1,x_2,\ldots,x_n$ are not all equal and satisfy $$x_1+\frac{k}{x_2}=x_2+\frac{k}{x_3}=x_3+\frac{k}{x_4}=\ldots=x_{n-1}+\frac{k}{x_n}=x_n+\frac{k}{x_1}$$Find: a) the product $x_1 x_2 \ldots x_n$ as a function of $k$ and $n$ b) the least value of $k$, such that there exist $n,x_1,x_2,\ldots,x_n$ satisfying the given conditions. Dunno why it seems hard last two year, it's pretty trivial. (guess I'm improving) a) \[ x_1 - x_2 = k \left( \frac{1}{x_3} - \frac{1}{x_2} \right) \Rightarrow \frac{x_1 - x_2}{x_2 - x_3} = \frac{k}{x_2 x_3} \]Multiply cyclically, we get $x_1 x_2 \dots x_n = k^{\frac{n}{2}}$. b) $k > 1$. Since $n$ is odd, $k$ must be a square. Thus, $k \ge 4$. This works. Take $n = 2019$, $(2,-1,4, 2, -1, 4, \dots)$. Your example just doesn't work as far as I can see... Also, I proved there are no examples, tomorrow I might latex it I would really like to see your proof that there are no examples Yeah, that's a typo. Should be $(2,-1,-4,2,-1,-4, \dots)$ instead.

@above hey, that actually works! I found the mistake in my solution, but i think it's still interesting. Let's call $x_i+\frac{4}{x_{i+1}}=c$ I solved a recursion of characteristic equation $x^2-cx+1$ in the proof that no example exists. My mistake was not noticing the solution only works if this equation has two different variables (which it doesn't for $c=\pm 2$). So you example is right, and turns out I proved only that there is an example exclusively for c=+-2.

In the end I was able to find all the solutions. My pdf is too big for AOPS so here is the link to the full solution ( I encourage you to look, it's really fun I.M.O.). The TLDR is all the solutions are the following: Take any $k=l^2;c=\pm l;x_1\neq c; 3\mid n$ for some $l\in\mathbb{Z}_{>1}$. Than define the function $f(x)=\frac{k}{c-x}$, and define $x_{i+1}=f(x_i)$ . The function loops after 3 iterations, so you will always get $x_{n+1}=x_1$, and from, the very definition we get $x_i+\frac{k}{x_{i+1}}=c$

Irelang MO 2014 wrote: Three different non-zero real numbers $a,b,c$ satisfy the equations $a+\frac{2}{b}=b+\frac{2}{c}=c+\frac{2}{a}=p $, where $p$ is a real number. Prove that $abc+2p=0.$ VJIMC 2017, Category II, Problem 3 wrote: Let $n \ge 2$ be an integer. Consider the system of equations \begin{align} x_1+\frac{2}{x_2}=x_2+\frac{2}{x_3}=\dots=x_n+\frac{2}{x_1} \end{align}1. Prove that $(1)$ has infinitely many real solutions $(x_1,\dotsc,x_n)$ such that the numbers $x_1,\dotsc,x_n$ are distinct. 2. Prove that every solution of $(1)$, such that the numbers $x_1,\dotsc,x_n$ are not all equal, satisfies $\vert x_1x_2\cdots x_n\vert=2^{n/2}$.

I'l fix it, turns out I dumbly assumed in my solution that if the Euler function of a number is at most 2, the number bust be 1,2,3 or 4, forgetting that it could also be 6. That leads to the exact case where c=l