no solution yet? ok giving one.........pls corect me if i'm wrong 1st multiply the second equation bothside by $x_1$......we get ..... $ x_1^2 - \prod x_i =x_1$ i.e. $ x_1^2 = x_1 + 1$ ........ $(i)$ 2nd multiply the third equation bothside by $x_1x_2$......we get ..... $ x_1^2x_2^2 - \prod x_i =x_1x_2$ i.e. $ x_1^2x_2^2 = x_1x_2 +1$ ............$(ii)$ $\vdots$ lastly we get.......... $ (x_1x_2 \cdots x_{n-1})^2 = x_1x_2 \cdots x_{n-1} +1$ now consider equation $(ii)$ ,,,,,,,$ x_1^2x_2^2 = x_1x_2 +1$ = $ (x_1 + 1)x_2^2 = x_1x_2 +1$ [from $(i)$] =$ x_1x_2^2 + x_2^2 = x_1x_2 +1$ = $x_1x_2(x_2 -1) + (x_2 -1)(x_2 +1) =0$ = $(x_2 -1)( x_1x_2 + x_2 +1) =0$................$(a)$ similarly from $(iii)$ we get ................ $(x_3 -1)( x_1x_2x_3 + x_3 +1)=0$...............$(b)$ $\cdots$ lastly we get...... $ (x_n -1)( x_1x_2x_3 \cdots x_{n-1} + x_{n-1} +1)=0$ note that........... in the above equations...... $x_i \not= 1 \forall i$ [otherwise they won't satisfy the given system of equations] so from the equations (a) , (b) $\cdots$ we get 1) $(x_1x_2 + x_2 +1) =0 $ 2)$( x_1x_2x_3 + x_3 +1)=0$ $\vdots$ remeber from (i) we get ....... $ x_1^2 = x_1 + 1$ .....after solving the quadratic we get ... $x_1 = \frac{1\pm \sqrt{5}}{2}$ now putting the value of $x_1$ in equation (1) ..we get easily the value of $x_2$ ............after getting $x_1 , x_2$ we can find the product of them and then after putting that into equation (3) we can get $x_3$ $\cdots$ and so on......

jasperE3 wrote: Find all $n$-tuples of reals $x_1,x_2,\ldots,x_n$ satisfying the system: $$\begin{cases}x_1x_2\cdots x_n=1\\x_1-x_2x_3\cdots x_n=1\\x_1x_2-x_3x_4\cdots x_n=1\\\vdots\\x_1x_2\cdots x_{n-1}-x_n=1\end{cases}$$ I think it would make more sense if they ask for the number of solution