Baltic Way 2023 Shortlist

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 $$\begin{gathered} 0=c^{n+1}+(a-a^n{)}^2-a(a-a^n)=c^{n+1}+a^{2n}-a^{n+1},\text{ i.e. } \\ {c}^{n+1}=a^{n+1}-a^{2n}<a^{n+1} \end{gathered}$$ 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 $a>c$.