Solve the system $(I)$ of equations $$\begin{align} 2JK+J^2+K^2+J^2K+K+JK^2+J &=0,\tag{$Ia$} \\ 2KL+K^2+L^2+K^2L+L+KL^2+K &=0,\tag{$Ib$} \\ 2LJ+L^2+J^2+L^2J+J+LJ^2+L &=0\tag{$Ic$} \end{align}$$ for $J, K, L$. Assume that $J, K, L$ are $n$th roots of unity for some $n\in\Bbb N$.
My Attempt:
By symmetry in $(I)$, we have $J=K=L$, so that, from $(Ia)$,
$$2K^2+K^2+K^2+K^3+K+K^3+K=0,$$
i.e., $2K^3+4K^2+2K=0$, which gives $K=0$ or $K=-1$. But $K\neq 0$.
Thus the solutions to $(I)$ are given by $J=K=L=-1$.
Is this the only solution?
Please help :)
Symmetry in the system does not imply that the three variables are equal. All factor, you have $$ (J+K)(1+J)(1+K) = 0, $$ $$ (K+L)(1+K)(1+L) = 0, $$ $$ (L+J)(1+L)(1+J) = 0. $$
If all $J,K,L \neq -1$ the remaining linear system says $J,K,L = 0.$
If $L = -1$ but $J,K \neq -1$ then $K = -J$ are anything, as long as $J,K \neq 1$ either. They are saying roots of unity, sure, why not.
If $L,K = -1$ then $J$ is anything at all.