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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 $\mathbb C$:

enter image description here

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3 Answers 3

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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.

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    $\begingroup$ Thank you for your answer; that $|z|^2=\frac{n}{n-2}$ is a simple but nice observation. $\endgroup$
    – Iosif Pinelis
    Commented May 19, 2023 at 17:50
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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 $n=3$.

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  • $\begingroup$ 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? $\endgroup$
    – Iosif Pinelis
    Commented May 19, 2023 at 17:44
  • $\begingroup$ @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. $\endgroup$
    – Peter Mueller
    Commented May 19, 2023 at 17:53
  • $\begingroup$ You only need to check that this quadratic guy is irreducible. $\endgroup$
    – Fedor Petrov
    Commented May 19, 2023 at 18:01
  • $\begingroup$ @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 $p_n$. $\endgroup$
    – Peter Mueller
    Commented May 19, 2023 at 18:09
  • $\begingroup$ @PeterMueller : Thank you for this detail. I think I learned the Gauss lemma over 50 years ago, but hardly ever used it since then. :-) $\endgroup$
    – Iosif Pinelis
    Commented May 19, 2023 at 18:20
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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 $n<n|z|^{n-2}<3$. $\quad\Box$

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