Are $x_1,x_2,\dots,x_n$ supposed to be distinct? Because if not, then let $x_1=x_2=\cdots =x_{n-1}=x_n=1$. Then any $k$ works.

It should be that at least two of them are distinct. Can somebody post the correct formulation?

Let $k>1$ be a positive integer and $n>2018$ an odd positive integer. The non-zero rational numbers $x_1,x_2,\ldots,x_n$ are not all equal and: $$x_1+\frac{k}{x_2}=x_2+\frac{k}{x_3}=x_3+\frac{k}{x_4}=\ldots=x_{n-1}+\frac{k}{x_n}=x_n+\frac{k}{x_1}$$Find the minimum value of $k$, such that the above relations hold.

ultralako wrote: Let $k>1$ be a positive integer and $n>2018$ an odd positive integer. The non-zero rational numbers $x_1,x_2,\ldots,x_n$ are not all equal and: $$x_1+\frac{k}{x_2}=x_2+\frac{k}{x_3}=x_3+\frac{k}{x_4}=\ldots=x_{n-1}+\frac{k}{x_n}=x_n+\frac{k}{x_1}$$Find the minimum value of $k$, such that the above relations hold. Fixed it, thanks.

Results?

Remark: If all relations are true, we must have ALL numbers are equal. Proof: We can suppose that ${{x}_{1}}>{{x}_{2}}>...>{{x}_{n}}$ (1) We have ${{x}_{1}}-{{x}_{2}}=k\left( \frac{1}{{{x}_{3}}}-\frac{1}{{{x}_{2}}} \right),{{x}_{2}}-{{x}_{3}}=k\left( \frac{1}{{{x}_{4}}}-\frac{1}{{{x}_{3}}} \right),...,{{x}_{n-1}}-{{x}_{n}}=k\left( \frac{1}{{{x}_{1}}}-\frac{1}{{{x}_{n}}} \right)$ But from the last inequality we get ${{x}_{n}}>{{x}_{1}}$ (2) From (1) and (2) result that all numbers are equal. Is something wrong in my demosntration?

Hmm, not sure but I think the answer is $k=4$ when $3\mid n$ and there's no possible such $k$ for other $n$

The quantifiers in the problem statement are not clear. Does it mean: "Find the minimum value of $k$, for which there exist an integer $n>2018$ and $n$ rational numbers that satisfy the equation system." Or does it mean: "For every integer $n>2018$, find the minimum value of $k$ for which there exist $n$ rational numbers that satisfy the equation system."

For $n=2019$ and $k=2$, there exists the following solution: $x_{3k-2}=2-\sqrt{2}$ $x_{3k-1}=-1$ $x_{3k}=-2(1+\sqrt{2})$

@above Sorry but $2-\sqrt{2}\not\in \mathbb{Q}$

The equation system yields $x_i-x_{i+1} = k\cdot \frac{x_{i+1}-x_{i+2}}{x_{i+1}x_{i+2}}$. Case 1. If there is an index $i$ with $x_i=x_{i+1}$, then all $x_i$ are equal; contradiction. (Here and in the following we use $x_{n+1}=x_1$) Case 2. If there is no index $i$ with $x_i=x_{i+1}$, then we multiply all these equations, and we divide both sides by the product of all terms $x_i-x_{i+1}$. This implies $\prod_{i=1}^{n}x_i=k^{n/2}$. Case 2a. If $k=2$ or $k=3$, then $k^{n/2}$ is non-rational (as $n$ in the exponent is odd). Contradiction, as the product of rational numbers is always rational. Case 2b. If $k=4$, there exists a solution for $n=2019$: $x_{3k-2}=2$ $x_{3k-1}=-1$ $x_{3k}=-4$

Note that the essence of the above solution is basically the same with this VJIMC 2017. And I think we can further prove that there's no such $k$ when $3\nmid n$.

ThE-dArK-lOrD wrote: Note that the essence of the above solution is basically the same with this VJIMC 2017. And I think we can further prove that there's no such $k$ when $3\nmid n$. How to prove there is no such $k$ for $3\nmid n$?

The original problem states: Find: a) the product x1x2 . . . xn as a function of k and n b) the least value of k, such that there exist n, x1, x2, . . . , xn satisfying the given conditions.