Consider the case $m=n=4.$ If we take $t_j=\cos(\frac{2j\pi}{9})+I\sin(\frac{2j\pi}{9})$ then $\sum_{j=0}^{8}t_j^q= 1 + \sum_{j=1}^{4} \left(t_j^{q} + {t_j^*}^{q}\right)=0$ for general systems of equations there are also $1 \le q \le 8.$ This is because the particular features set of this problem. Taking values $t_j=\exp{\frac{2j\pi I}{2n+1}}=\cos(\frac{2j\pi }{2n+1})+I\sin (\frac{2j\pi }{2n+1})$ gives, together with t_j^q$is just the nine$1$and conjugation9$th roots of unity (or in two cases, the $2n+1$st third roots of unity . That should worktaken three times). To get a real solution (which Admittedly, this is also complex) not exactly what you wanted.
If you take just the real parts for $t_j=t_j^*=\cos(\frac{2j\pi }{2n+1})$j=1,2,3,4$you get •$t_1=t_1^*=.766044443118979$•$t_2=t_2^*=.173648177666934$•$t_3=t_3^*=-.500000000000$•$t_4=t_4^*=-0.939692620785905$It can be seen that this makes$1 + \sum_{j=1}^4 \left(t_j^{q} + {t_j^*}^{q}\right)=0$correct for$q=1,3,5,7.$1 Aside from algorithms for general systems of equations there are also the particular features of this problem. Taking$t_j=\exp{\frac{2j\pi I}{2n+1}}=\cos(\frac{2j\pi }{2n+1})+I\sin (\frac{2j\pi }{2n+1})$gives, together with$1$and conjugation, the$2n+1$st roots of unity. That should work. To get a real solution (which is also complex) take$t_j=t_j^*=\cos(\frac{2j\pi }{2n+1})\$