MS_Kekas wrote: Nonzero real numbers $x_1, x_2, \ldots, x_n$ satisfy the following condition: $$x_1 - \frac{1}{x_2} = x_2 - \frac{1}{x_3} = \ldots = x_{n-1} - \frac{1}{x_n} = x_n - \frac{1}{x_1}$$ Determine all $n$ for which $x_1, x_2, \ldots, x_n$ have to be equal. Let us look for cases where all $x_i$ are not equal. WLOG (since cyclic) consider $x_2\ne x_1$. Let $u>0$ such that the common value in the equalities is $u-\frac 1u$ Let $S_n=u^{-n}-(-u)^n$ It is easy to establish (or simply check) that $x_k=\frac{S_{k-2}x_1+S_{k-1}}{S_{k-1}x_1+S_k}$ $x_{n+1}=x_1$ is $S_{n-1}x_1+S_{n}=S_{n}x_1^2+(S_{n+1}-S_{n-1})x_1+S_n=0$ Which is $S_n(x_1^2+(\frac 1u-u)x_1-1)=0$ Which is $S_n(x_1-x_2)(x_1-u+\frac 1u)=0$ But $x_1-u+\frac 1u\ne 0$ (in order $x_2$ be defined) and $x_1-x_2\ne 0$ So $S_n=0$, which is $u^{2n}=(-1)^n$ and so $n$ even and $u=1$ and $x_{k+1}=\frac 1{x_k}$, which indeed fits, whatever is $x_1\notin\{-1,0,1\}$ Hence the answer $\boxed{\text{Any odd }n}$