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


3 Answers 3


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.

  • 2
    $\begingroup$ Thank you for your answer; that $|z|^2=\frac{n}{n-2}$ is a simple but nice observation. $\endgroup$ 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 $n=3$.

  • $\begingroup$ 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? $\endgroup$ 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. $\endgroup$ May 19, 2023 at 17:53
  • $\begingroup$ You only need to check that this quadratic guy is irreducible. $\endgroup$ 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 $p_n$. $\endgroup$ 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. :-) $\endgroup$ 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 $n<n|z|^{n-2}<3$. $\quad\Box$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.