Can the equation $1+z+z^2=z^n$ for natural $n$ have multiple complex roots $z$?

The question is stated in the title of this post.

It is easy to see that, if $$z$$ is a multiple root of $$p_n(z):=1+z+z^2-z^n$$, then $$(n-2)z^2+(n-1)z+n=0$$, so that we can successively express $$z^2,\dots,z^n$$ as affine functions of $$z$$, but the expressions seem hard to analyze.

Trying to use the discriminant of $$p_n$$ and letting $$d_n:=(-1)^{n (n - 1)/2} (-1)^{n - 1}\,\text{discriminant}(p_n),$$ we get $$(d_1,\dots,d_{15})=(-4, 1, 44, 279, 2300, 57425, 841436, 14201719, 442321436, 10095992037, 254419300556, 9827983382723, 304507125159644, 10182574354147897, 472932455198902268).$$

There is nothing in the OEIS on $$4, 1, 44, 279, 2300, 57425, 841436, 14201719$$ or on $$1, 44, 279, 2300, 57425, 841436, 14201719.$$

Shown here are the roots of $$p_{30}$$ and the unit circle in $$\mathbb C$$:

Suppose that $$z$$ is our multiple root, $$n\geq 4$$. Since $$z$$ is a root of equation $$(n-2)z^2+(n-1)z+n=0$$, we have $$|z|^2=\frac{n}{n-2}$$. Indeed, the roots are non-real and conjugate, so the other root is $$\overline{z}$$ and by Vieta's formula we get $$|z|^2=z\overline{z}=\frac{n}{n-2}.$$ Next, we have $$z^n=z^2+z+1$$, hence $$2z^2+z=n(z^2+z+1)=nz^n$$. Consequently, $$n|z|^n\leq 2|z|^2+|z|\leq 3|z|^2,$$ hence $$1<|z|^{n-2}\leq \frac{3}{n},$$ which is a contradiction.

• Thank you for your answer; that $|z|^2=\frac{n}{n-2}$ is a simple but nice observation. May 19, 2023 at 17:50

As I was typing essentially Alexander Kalmynin's argument, his answer popped up ...

Therefore, I now offer a different argument: The roots of $$p_n(z)$$ are algebraic integers, while the roots of $$(n-2)z^2+(n-1)z+n$$ aren't except for $$n=3$$.

• Thank you for your answer. I understand that the polynomial $q_n(z):=(n-2)z^2+(n-1)z+n$ is monic only if $n=3$ -- but don't you have to do a bit work to show that then roots of $q_n$ cannot be roots of some monic polynomial with integral coefficients? May 19, 2023 at 17:44
• @IosifPinelis No. By the Gauß Lemma, if a monic polynomial like $q_n(z)/(n-2)$ from $\mathbb Q[z]$ divides a monic polynomial from $\mathbb Z[z]$, then the coefficients are automatically integers. May 19, 2023 at 17:53
• You only need to check that this quadratic guy is irreducible. May 19, 2023 at 18:01
• @FedorPetrov Yes, either that way (which is easy enough), or by noting that the roots of $q_n$ are complex conjugate, so if $q_n$ and $p_n$ have a common root, then $q_n$ divides $p_n$. May 19, 2023 at 18:09
• @PeterMueller : Thank you for this detail. I think I learned the Gauss lemma over 50 years ago, but hardly ever used it since then. :-) May 19, 2023 at 18:20

This is a slight modification of Alexander Kalmynin's answer. For $$n\ge3$$. let $$z$$ be a multiple root of $$p_n$$. Then $$0=p_n'(z)=1+2z-nz^{n-1}$$ and, as shown by Alexander Kalmynin, $$|z|>1$$. So, $$n|z|^{n-1}\le1+2|z|<3|z|$$ and hence $$n. $$\quad\Box$$