There are many problems involving relations between complex numbers and the geometrical configuration of their corresponding affixes. Here is a generalized problem of the special case $n=2:$
Find the natural numbers $n>1$ such that the following implication holds: given distinct $a,b,c \in \mathbb{C}$, with $(a-b)^n+(b-c)^n+(c-a)^n=0,$ the triangle determined by the affixes of $a,b,c$ is equilateral.
The problem is converted into: $z_1+z_2+z_3=z_1^n+z_2^n+z_3^n=0$, where $z_1:=a-b$ and so on. For $n=2$, this is a standard exercise, which implies: $|z_1|=|z_2|=|z_3|.$
I proved that, for $n\in \{2,4,5,7\}$, the implication does hold, and for $n\in \{3,6,8,10\}$, the implication fails. Of course, one can keep checking the next cases of $n$, but I am wondering whether there is a straightforward, rigurous method of finding all those $n$-s.
I was able to reduce the problem to this one: (to be explicit, I keep the $z_i$ notation and I denote $z:=z_2/z_3$, after the elimination $z_1=-z_2-z_3$)
Find the natural numbers $n>1$ such that any solution of $(-z-1)^n+z^n+1=0$, where $z\in \mathbb{C}, z\notin \{0,-1\}$, have absolute value $1$.
Why $z \neq -1?$ Because otherwise, $z_1=0$, then $a=b$. Thus, it is clear that $z_i\neq0,$ and from here we need $z \neq 0$. Also, one can notice that $z \neq 1$.
For example, if $n=3$, we get: $z^2+z=0$, with no solutions in the above domain.
For $n=7$, we get: $z(z+1)(z^2+z+1)^2=0,$ and this is a valid solution of $n$, since if we divide this polynomial to $z(z+1)$, all roots of the quotient polynomial lie on the unit circle, as wanted.
P.S. After trying some small $n$-s, I realize that it has something to to with the primitive root of order $3$ of unity, $\epsilon^3=1$, maybe this could help, as it appears as a root in the valid cases $n\in \{2,4,5,7\}$.
Best Answer
From $|z_2|=|z_3|$ and $z_2=zz_3,$ we conclude $|z|=1.$ From $|z_1|=|z_3|$ and $z_1=-z_2-z_3,$ we conclude $|z+1|=1.$ If we put this together, we get $z\in\{\epsilon,\bar\epsilon\}$ where $\epsilon$ and $\bar\epsilon$ are the solutions of $z^2+z+1=0.$
We have to check if $p(z)=(-z-1)^n+z^n+1=0$ has any other solutions but $0,-1,\epsilon,\bar\epsilon.$
$0$ and $-1$ are roots of $p$ if and only if $n$ is odd. Furthermore, it can easily be verified (using the derivative of $p$) that they are simple roots in this case.
We also know that $\epsilon$ and $\bar\epsilon$ have the same multiplicity as roots of $p,$ because the coefficients of $p$ are real. If we put all this together and scale the polynomials such that the coefficients of $z^{n-1}$ match, we get the following equations that must hold if the choice of $n$ enforces the triangle to be equilateral: \begin{eqnarray} nz(z+1)(z^2+z+1)^{(n-3)/2} &=& (z+1)^n -z^n-1 \text{ for odd } n \\ 2(z^2+z+1)^{n/2} &=& (z+1)^n +z^n + 1 \text{ for even } n \end{eqnarray} For odd $n\geq 5,$ the coefficient of $z^3$ in $nz(z+1)(z^2+z+1)^{(n-3)/2}$ after expansion is $\frac{(n+3)n(n-3)}{8}$, while the coefficient of $z^3$ in $(z+1)^n -z^n-1$ is $\binom n3.$ Those two match only if $n\in\{5,7\}.$
For even $n\geq 6,$ the coefficient of $z^3$ in $2(z^2+z+1)^{n/2}$ after expansion is $\frac{(n+8)n(n-2)}{24}$, while the coefficient of $z^3$ in $(z+1)^n +z^n+1$ is $\binom n3$ again. Those two do not match for any value of $n\geq 6.$
The values you have found are the only ones that enforce the triangle to be equilateral.