, 2023

The set of equations given in the problem statement is Flensburgian precisely when $n$ is even or $n = 1$.

To see that it is not Flensburgian when $n \ge 3$ is odd, notice that if $(a, b, c)$ satisfies the set of equations then so does $(-a, -b, -c)$. Hence, if there exists a single solution to the set of equations where all the variables are different then the set of equations cannot be Flensburgian. This is in fact the case, e.g., consider $(a, b, c) = \left(\frac{1}{2}, \frac{2^{n-1}-1}{2^n}, \left(\frac{2^{n-1}-1}{2^{2n}}\right)^{\frac{1}{n+1}}\right)$.

If $n = 1$ from the first equation we know $b = 0$. But then from the second equation it follows $c = 0$. Hence, there cannot be any solution with pairwise different variables $a, b, c$. Therefore, the set of equations is Flensburgian for $n = 1$.

The rest of the solution is dedicated to prove that the set of equations is indeed Flensburgian when $n$ is even.

The first equation yields $b = a - a^n \le a$, since $a^n \ge 0$ when $n$ is even. The inequality is strict whenever $a \ne 0$ and the case $a = 0$ implies $b = 0$, i.e. $a = b$, which we can disregard. Substituting the relation $b = a - a^n$ into the second equation yields
$$0 = c^{n+1} + (a - a^n)^2 - a(a - a^n) = c^{n+1} + a^{2n} - a^{n+1}, \text{ i.e.\ } \\ c^{n+1} = a^{n+1} - a^{2n} < a^{n+1}$$
since we can disregard $a = 0$ and $2n$ is even. Since $n + 1$ is odd, the polynomial $x^{n+1}$ is strictly increasing, implying that $c < a$. Hence, when $n$ is even, all solutions of the set of equations where $a, b, c$ are pairwise different satisfy $a > b$ and $a > c$.