# $x_i=\prod_{j \neq i}x_j$ for $i=1,2,..,6$

Find all real $$x_i^s$$ satisfying the system of equations $$x_i=\prod_{j \neq i}x_j$$ for all $$i=1,2,..,6$$ It is obvious that $$(x_1,x_2,..x_6)=(0,0,..0),(1,1,...,1),(-1,-1,-1,...-1)$$ are obvious solutions and so are $$(1,1,-1,-1,-1,-1),(-1,-1,1,1,1,1)$$

Actually we get the criteria $$x_1^2=x_2^2=...x_6^2$$ So, it seems it has finitely many solutions but the answer key claims it has infinitely many solutions. Where am I going wrong?

Define $$P:=\prod_j x_j=x_i^2$$ so $$P^6=\prod_i x_i^2=P^2$$ and $$P^2(P^4-1)=0$$, which allows finitely many values of $$P$$. Each gives at most two values for $$x_i$$, so I agree there are only finitely many choices for the $$x_i$$ with $$1\le i\le 6$$. The only sense in which there are infinitely many $$x_i$$ is if we have an infinite sequence, with no constraints on $$x_i$$ if $$i\ge 7$$. (However, I do think an answer key error is more likely here.)