This 2014 post asks for triples $x,y,z$ such that
$$\begin{aligned}x^{k} + y^{k} + z^{k} &= 3\\x^{k+1} + y^{k+1} + z^{k+1} &= 3\\x^{k+2} + y^{k+2} + z^{k+2} &= 3\end{aligned}\tag1$$
for $k=2012$. Of course, the choice of exponent was related to the year it was posted, and the nature of the $x,y,z$ was undefined. However, I wondered about general $k$.
Let us make it challenging and require that at least one of the variables is real. Using resultants, Mathematica can resolve this system for small $k$. For example, for $k=2$,
$$\begin{aligned}x^{2} + y^{2} + z^{2} &= 3\\x^{3} + y^{3} + z^{3} &= 3\\x^{4} + y^{4} + z^{4} &= 3\end{aligned}\tag2$$
in addition to $(x,y,z) = (1,1,1)$, there is also a 9th deg eqn which solves $(2)$ using one real root and two complex conjugates.
(For $k=3$, there is a $27$ deg; for $k=4$ there is a $57$ deg; for $k=5$ there is a $102$ deg, and so on.)
Q: Is it true that, other than $(x,y,z) = (1,1,1)$, the system for $k>1$ has another triplet where one of the variables is real?
