Find $n \geq 4$, natural number

Question

Find $n \geq 4$, natural number such that for every distinct complex numbers $a,b,c$ different of $0$ which satisfy $(a-b)^n + (b-c)^n + (c-a)^n =0$ implies that $a, b, c$ are the vertices of an equilateral triangle. I tried using Newton's identities. I guess $n$ can be $ 4,5,7$. Any idea using Newton's identities?

Answer 1

Partial answer. I believe $n=2,4,5,7$ are the only such $n.$

Given distinct values $a,b,c,$ let $x=a-b,y=b-c,z=c-a.$ We have that $x+y+z=0$ and $xyz\neq 0.$ Letting $$s_1=x+y+z=0, s_2=xy+yz+xz, s_3=xyz,$ we have the general rule that:

$$x^n+y^n+z^n=\sum_{i+2j+3k=n} (-1)^j\frac{n}{i+j+k}\binom{i+j+k}{i,j,k}s_1^is_2^js_3^k$$

But since $s_1=0,$ this turns into:

$$x^n+y^n+z^n=\sum_{2j+3k=n} (-1)^j\frac{n}{j+k}\binom{j+k}{k}s_2^js_3^k\tag{1}$$

Since $a,b,c$ are distinct, we have $s_3\neq 0.$

The cases $n=2,4,5,7$ there is only one term in $(1)$, and it has $j>0.$ (The case $n=3$ has one term, but it is a multiple of $s_3\neq 0.$) In the case $n=2,4,5,7,$ then, if $x^n+y^n+z^n=0$ and $s_1=x+y+z=0$ and $s_3=xyz\neq 0$ you have that $s_2=0.$

But when $s_2=0$ we get that $p(\alpha)=(\alpha-x)(\alpha-y)(\alpha-z)=\alpha^3-s_3,$ so the roots all have $|x|=|y|=|z|$ and the triangle $a,b,c$ is $1.$

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From here, we'll normalize $a,b,c$ so that $x+y+z=0$ and $s_3=xyz=1.$

For other $n,$ (1) has multiple terms.

For example, when $n=8,$ we have $$x^6+y^6+z^6=c_1s_2^4+c_2 s_2s_3^2=s_2(c_1s_2^3+c_2)$$ This means that we can solve $s_2^3=-c_2/c_1$ to get the polynomial $p(\alpha)=\alpha^3-s_2\alpha - 1$ to get three roots $x,y,z$ which satisfies $x+y+z=1,xyz=1,x^8+y^8+z^8=0$ but the roots are not all of the same absolute value.

I think we can show this in general - when $(1)$ has multiple terms, we have that we can solve $x^n+y^n+z^n=0$ with some $s_2\neq 0.$