Here is my question :

Are there monic polynomials with degree $\geq 5$ such that they have the same real all non zero roots and coefficients ?

Mathematically, prove or disprove the existence of $n \geq 5$ such that $$\exists (z_1,\ldots, z_n) \in \left(\mathbb R-\{0\}\right)^n, (X-z_1)...(X-z_n)=X^n+\sum_{i=1}^{n}z_iX^{n-i}$$

State of the problem: There's no such polynomial for $n \geq 6$ (see answer below). It remains to prove/disprove the $n=4,5$ cases.

Here are all such real polynomials with degree $\leq 3$:



$X^3+\alpha X^2 + \beta X + \gamma$ where $\alpha$ is the real root of $2X^3+2X^2-1$ (which determines $\gamma$ and $\beta$)

There remains complex degree 3 polynomials, as in Barry's answer.


As pointed out by Jyrki Lahtonen, if $P$ is a satisfactory polynomial, then so is $XP$. For example, The family of polynomials $X^n(X-1)(X+2)$ works.

It seems therefore more interesting to look only for polynomials with non zero coefficients,and specifically those with real coefficients (they're scarcer)

This subject has been discussed here Coefficients of a polynomial also are the roots of the polynomial? but does not deal with the existence of such polynomials with real coefficients and degree $\geq 5$.

  • 2
    $\begingroup$ To add to your existing list: $x=x+0$ is one. There are no other degree 1 polynomials with your conditions, since $x-a$ has root $a$, and clearly $-a\neq a$ unless $a=0$. $\endgroup$
    – Hayden
    May 11, 2014 at 13:37
  • $\begingroup$ You have $5$ equations in $5$ unknowns, starting with $a+b+c+d+e=-a$ and ending with $abcde=-e$ (for the polynomial $x^5+ax^4+bx^3+cx^2+dx+e$). The $a$ and $e$ can be eliminated fairly quickly (splitting into two cases according to $e=0$ or not), leaving $3$ equations in $3$ unknowns. A Grobner basis algorithm might help. $\endgroup$ May 11, 2014 at 13:49
  • 5
    $\begingroup$ If $p(x)$ is such a polynomial, then it seems to me that $xp(x)$ is one also. The multiplicity of $0$ as a coefficient as well as a root goes up by one. I guess you want to disallow zero as coefficient. Otherwise $x^5+x^4-2x^3$ would work, too (zero is a triple root and a triple coefficient). Alternatively you may want to disallow repeated roots. A cool question, though (+1). $\endgroup$ May 11, 2014 at 13:58
  • $\begingroup$ @BarryCipra Thanks for your comments. I added the extra requirement that $0$ be not a root/coefficient. $\endgroup$ May 11, 2014 at 14:07
  • $\begingroup$ @GabrielR., maybe you want to change the question itself now to ask for examples with degree $\ge3$. $\endgroup$ May 11, 2014 at 14:19

7 Answers 7


I think I got the proof that no such real polynomial with degree $ \geq 6$ exists.

Let $n \geq 6$

Suppose for contradiction that $z_1,\ldots,z_n \in \mathbb R-\{0\}^n$ are such that $(X-z_1)...(X-z_n)=X^n+\sum_{k=1}^{n-1}z_iX^{n-i}$

Then three useful identities appear $$\sum_{k=1}^{n}z_k=-z_1 \; \; \; \;(1)$$

$$\sum_{\large1\leq i<j \leq n}z_iz_j=z_2 \; \; \; \;(2)$$

$$\prod_{k=1}^n z_k=(-1)^n z_n \; \; \; \;(3)$$

Since $$(\sum_{k=1}^{n}z_k)^2=\sum_{k=1}^{n}z_k^2+2\sum_{\large1\leq i<j \leq n}z_iz_j$$it follows that$$z_1^2=2z_2+\sum_{k=1}^{n}z_k^2$$

Hence $$0< \sum_{k=2}^{n}z_k^2=-2z_2 \; \; \; \;(4)$$ and $$0<\sum_{k=3}^{n}z_k^2=1-(z_2+1)^2 \; \; \; \;(5) $$

$(4)$ and $(5)$ imply $$\; \; \; \;-2<z_2<0 \; \; \; \;(6)$$

thus $(6)$ and $(4)$ imply $$0<\sum_{k=2}^{n}z_k^2 < 4 \; \; \; \; (7)$$

Also $(6)$ and $(5)$ imply $$0<\sum_{k=3}^{n}z_k^2 \leq 1 \; \; \; \; (8)$$

By AM-GM, $$\left(|z_3|^2\ldots|z_{n-1}|^2 \right)^{1/(n-3)} \leq \frac{1}{n-3}\sum_{k=3}^{n-1}z_k^2 \leq \frac{1}{n-3}\sum_{k=3}^{n}z_k^2$$

Hence $$|z_3|^2\ldots|z_{n-1}|^2 \leq \left(\frac{1}{n-3}\sum_{k=3}^{n}z_k^2\right)^{n-3} $$

Squaring, $$|z_3|\ldots|z_{n-1}| \leq \left(\frac{1}{n-3}\sum_{k=3}^{n}z_k^2\right)^{\large \frac{n-3}{2}} \leq_{ \large (8)} \dfrac{1}{{(n-3)}^{(n-3)/2}} \; \; \; \; (9)$$

By triangle inequality $(1)$, and Cauchy-Schwarz

$$2|z_1| \leq \sum_{k=2}^{n}|z_k| \leq \sqrt{n-1} \sqrt{\sum_{k=2}^{n}z_k^2} $$

Hence by $(7)$,

$$|z_1| \leq \sqrt{n-1} \; \; \; \; (10)$$

Rewriting $(6)$ as $$|z_2|\lt2 \; \; \; \; (11) $$

Recalling $(3)$ (with $z_n$ cancelled from both sides) and putting together $(9)$, $(10)$ and $(11)$, we have

$$1=|z_1||z_2||z_3|\cdots|z_{n-1}| < \dfrac{ 2\sqrt{n-1}}{{(n-3)}^{(n-3)/2}}$$

This inequality fails for $n\geq 6$.


I can't prove anything for $n=5$ so maybe the conjecture doesn't hold.

  • 7
    $\begingroup$ This is so brilliant. What was the thought process here? $\endgroup$
    – math_lover
    Mar 27, 2016 at 23:04

The OP's edited problem (disallowing $0$ as root/coefficient) is worth looking at for polynomials of degree $3$, where the pertinent equations are

$$\begin{align} a&=-(a+b+c)\\ b&=ab+bc+ca\\ c&=-abc \end{align}$$

The assumption $abc\not=0$ turns the third equation into $a=-1/b$, which turns the first equation into $c=(2-b^2)/b$, and these, if I've done the algebra correctly, turn the second equation into


The root $b=-1$ gives $a=1$ and $c=-1$, corresponding to


The cubic has one real root at $b\approx-1.76929235424$ and two complex roots. Each of these will give a polynomial, so there are $4$ examples in all of cubic equations with nonzero root/coefficients.

Historical note: Googling on the number $1.76929235424$ leads to an earlier appearance of the cubic case about $12$ years ago at the Math Forum @ Drexel. The discussion there dates it back to at least $1954$.


Using the Magma CAS online calculator at http://magma.maths.usyd.edu.au/calc/

with the commands

I:=[2*a+b+c+d, a*(b+c+d)+b*(c+d)+c*d-b, a*b*(c+d)+c*d*(a+b)+c, a*b*c-1];

results in the components

$d=0, c =-1 , b=-1, a = 1$, which is one of the excluded solutions because of $d=0$,

$-d= - b^2 - b + 1, -c = b^2 + 2⋅b + 1, a = 1)$ for the roots of $0=b^3 + 2⋅b^2 + b + 1$, one of them $b=-1.754877666246692760049508...$,

and 14 further non-real solutions in \begin{align} -d &= \tfrac{26000}{3301}⋅a^{13} + \tfrac{59904}{3301}⋅a^{12} + \tfrac{68448}{3301}⋅a^{11} + \tfrac{34192}{3301}⋅a^{10} - \tfrac{11712}{3301}⋅a^9 - \tfrac{31404}{3301}⋅a^8 - \tfrac{30192}{3301}⋅a^7 - \tfrac{24184}{3301}⋅a^6 - \tfrac{1658}{3301}⋅a^5 + \tfrac{16090}{3301}⋅a^4 + \tfrac{2391}{3301}⋅a^3 - \tfrac{4009}{3301}⋅a^2 + \tfrac{2426}{3301}⋅a - \tfrac{1118}{3301},\\ -c&= - \tfrac{43888}{3301}⋅a^{13} - \tfrac{139568}{3301}⋅a^{12} - \tfrac{217792}{3301}⋅a^{11} - \tfrac{192080}{3301}⋅a^{10} - \tfrac{76144}{3301}⋅a^9 + \tfrac{38644}{3301}⋅a^8 + \tfrac{92900}{3301}⋅a^7 + \tfrac{103568}{3301}⋅a^6 + \tfrac{63854}{3301}⋅a^5 - \tfrac{1016}{3301}⋅a^4 - \tfrac{21835}{3301}⋅a^3 - \tfrac{11692}{3301}⋅a^2 - \tfrac{5125}{3301}⋅a - \tfrac{4662}{3301},\\ -b&= \tfrac{17888}{3301}⋅a^{13} + \tfrac{79664}{3301}⋅a^{12} + \tfrac{149344}{3301}⋅a^{11} + \tfrac{157888}{3301}⋅a^{10} + \tfrac{87856}{3301}⋅a^9 - \tfrac{7240}{3301}⋅a^8 - \tfrac{62708}{3301}⋅a^7 - \tfrac{79384}{3301}⋅a^6 - \tfrac{62196}{3301}⋅a^5 - \tfrac{15074}{3301}⋅a^4 + \tfrac{19444}{3301}⋅a^3 + \tfrac{15701}{3301}⋅a^2 + \tfrac{9301}{3301}⋅a + \tfrac{5780}{3301},\\ 0&=a^{14} + 3⋅a^{13} + 5⋅a^{12} + 5⋅a^{11} + 3⋅a^{10} + \tfrac{1}{4}⋅a^9 - \tfrac{7}{4}⋅a^8 - \tfrac{11}{4}⋅a^7 - \tfrac{17}{8}⋅a^6 - \tfrac{3}{4}⋅a^5 + \tfrac{3}{16}⋅a^4 + \tfrac{3}{8}⋅a^3 + \tfrac{3}{8}⋅a^2 + \tfrac{3}{16}⋅a + \tfrac{1}{16} \end{align}

The next case can be generated as

I:=[ElementarySymmetricPolynomial(P,6-k)-(-1)^k*P.k : k in [1..5]];

//D[7] has a=1, generator is b with polynomial in position 4
//D[8] is parametrized by a with polynomial in position 5

where the first 6 componentsideals only have solutions where one or more components are $0$, and the last two ideals have no real solution, one has degree 18, the other has degree 78 and very large coefficients.

  • $\begingroup$ Does it mean the conjecture holds for $n=5$ ? $\endgroup$ May 11, 2014 at 23:02
  • 1
    $\begingroup$ I'd say so. The determination that the solutions are non-real was done by computing the roots with 80 digits precision. The imaginary parts all have a size greater 0.01, which should be stable enough against perturbation from the numerical to the exact coefficients. $\endgroup$ May 11, 2014 at 23:25
  • $\begingroup$ It seems worth noting that googling on −1.75487766624669 leads to a problem regarding the Mandelbrot set at math.stackexchange.com/questions/743727/… $\endgroup$ May 12, 2014 at 0:02

I slapped this together very quickly but using Mathematica, we write

F[n_] := Union[#[[1]] == #[[2]] & /@ Transpose[{r /@ Range[n], 
 Reverse[Most[CoefficientList[Product[x - r[k], {k, 1, n}], x]]]}], 
 r[#] != 0 & /@ Range[n]]

Reduce[F[4], {r[1], r[2], r[3], r[4]}, Integers]

And the output is False, so there are no nontrivial integer solutions for monic degree 4 polynomials. There do exist solutions for general complex roots. It is not difficult to modify the above commands to obtain them.


Here, we shall use methods similar to those in the answer by Gabriel Romon to rule out the case of degree $5$ polynomials. We start by sharpening some of the bounds from that answer and then we introduce the idea that evaluating a polynomial at the value $1$ is the same as summing its coefficients. Enjoy!

Let $x^r+a_1x^{r-1}+a_2x^{r-2}+\dots+a_r$ be a polynomial with nonzero real coefficients such that $a_1,\dots,a_r$ are all roots (counting multiplicity) so that $$(x-a_1)\cdots(x-a_r) = x^r+a_1x^{r-1}+a_2x^{r-2}+\dots+a_r$$ (Note: We shall never consider the case $r\le 2$ below.) Equating coefficents, we have $a_1 = -\sum_{i=1}^ra_i$, as well as $a_2=\sum_{1\le i<j\le r}a_ia_j$, and $a_r=(-1)^r\prod_{i=1}^ra_i$.

Thus, $a_1^2 = \sum_{i=1}^ra_i^2 + 2a_2$ so that $0 = a_2^2+2a_2+\sum_{i=3}^ra_i^2$. The discriminant of this quadratic polynomial is nonnegative and the roots of the given polynomial are nonzero, so $0<\sum_{i=3}^ra_i^2\le 1$. In particular, $a_2^2+2a_2<0$. This means $a_2\in(-2,0)$.

Having reiterated what we need from an earlier post, let's get to work.

PRP 1: If $r\ge 4$, then

  1. $|a_1|\le \frac{2+\sqrt{r-2}}2$.

  2. $\prod_{j=3}^r|a_j| \le \left(\frac{\sqrt{-a_2}\sqrt{2+a_2}}{\sqrt{r-2}}\right)^{r-2}$.

  3. if $3\le j_0\le r$ then $\prod_{3\le j\le r\\j\ne j_0}|a_j| < \left(\frac{\sqrt{-a_2}\sqrt{2+a_2}\sqrt{r-2}}{r-3}\right)^{r-3}$.

Proof: 1. Using the Cauchy-Schwarz inequality and $|a_2|\le 2$, we have \begin{equation*} |a_1| \le \frac12(|a_2|+\sum_{i=3}^r|a_i|) \le \frac{2+\sqrt{r-2}}2. \end{equation*}

2-3. We use $\sum_{j=3}^ra_j^2 = -a_2(2+a_2)$. By the Cauchy-Schwarz inequality, we have \begin{equation*} \sum_{j=3}^r|a_j| \le \sqrt{-a_2}\sqrt{2+a_2}\sqrt{r-2}. \end{equation*} Also, since $a_j\ne 0$ for $j=1,\dots,r$, we get \begin{equation*} \sum_{3\le j\le r\\j\ne j_0}|a_j| < \sqrt{-a_2}\sqrt{2+a_2}\sqrt{r-2} \end{equation*} Now, we apply the AM-GM inequality to deduce (2.), (3.).

We purposely kept instances of $a_2$ in the bounds in PRP 1. We are going to multiply by $|a_2|$ and take a maximum of a function of $|a_2|$ using the following lemma, which is proved with methods of single-variable calculus.

LEM 2: If $a,b>0$, then the function $f(x)=x^a(2-x)^b$ has its maximum on $[0,2]$ at $x=2a/(a+b)$.

Here's what we get.

PRP 3: If $r\ge 4$, then

  1. $\prod_{j=2}^r|a_j| \le \frac{r^{r/2}}{(r-1)^{r-1}}$.

  2. for $3\le j_0\le r$ we have $\prod_{2\le j\le r\\j\ne j_0}|a_j| \le \frac{(r-1)^{(r-1)/2}}{(r-3)^{(r-3)/2}(r-2)^{(r-1)/2}}$.

Proof: 1. By PRP 1.2 and LEM 2, we have \begin{align*} \prod_{j=2}^r|a_j| &\le \frac{|a_2|^{r/2}(2-|a_2|)^{(r-2)/2}}{(r-2)^{(r-2)/2}} \\ &\le \frac{(r/(r-1))^{r/2}((r-2)/(r-1))^{(r-2)/2}}{(r-2)^{(r-2)/2}} \\ &= \frac{r^{r/2}}{(r-1)^{r-1}} \end{align*}

  1. Using PRP 1.3 and including LEM 2, we derive the second inequality as follows \begin{align*} \prod_{2\le j\le r\\j\ne j_0}|a_j| &\le \frac{|a_2|^{(r-1)/2}(2-|a_2|)^{(r-3)/2}(r-2)^{(r-3)/2}}{(r-3)^{r-3}} \\ &\le \frac{((r-1)/(r-2))^{(r-1)/2}((r-3)/(r-2))^{(r-3)/2}(r-2)^{(r-3)/2}}{(r-3)^{r-3}} \\ &=\frac{(r-1)^{(r-1)/2}}{(r-2)^{(r-1)/2}(r-3)^{(r-3)/2}} \end{align*}

Next, we use the fact that $a_1a_2\dots a_{r-1}=(-1)^r$ to derive some fresh inequalities.

PRP 4: If $r\ge 4$, then

  1. $\frac{(r-3)^{(r-3)/2}(r-2)^{(r-1)/2}}{(r-1)^{(r-1)/2}}\le |a_1|$.

  2. $|a_r|\le \frac{r^{r/2}}{(r-1)^{r-1}}\cdot\frac{2+\sqrt{r-2}}2$.

  3. $a_2\le-\frac2{2+\sqrt{r-2}}\cdot\left(\frac{r-3}{\sqrt{r-2}}\right)^{r-3}$


  1. From $(-1)^r=\prod_{i=1}^{r-1}|a_i|$, we derive $$|a_1| = \frac 1{\prod_{i=1}^{r-1}|a_i|}.$$ The desired inequality follows from PRP 3.2.

  2. Likewise $\prod_{i=2}^{r-1}|a_i|=1/|a_1|\ge 2/(2+\sqrt{r-2})$ by PRP 1.1. Thus, $$\frac2{2+\sqrt{r-2}}|a_r| \le \prod_{i=2}^{r}|a_i|.$$ Then, use PRP 3.1.

  3. By PRP 2.3, $$\prod_{j=3}^{r-1}|a_j| < \left(\frac{\sqrt{-a_2}\sqrt{2+a_2}\sqrt{r-2}}{r-3}\right)^{r-3} \le \left(\frac{\sqrt{r-2}}{r-3}\right)^{r-3}$$ Thus, $$\frac2{2+\sqrt{r-2}}\le \prod_{j=2}^{r-1}|a_j| \le |a_2|\left(\frac{\sqrt{r-2}}{r-3}\right)^{r-3}$$ as desired.

Case $r=5$

Now, we focus on the case $r=5$. Observe that PRP 4.1 simplifies to $|a_1|\ge 9/8$. In particular $a_1\ne 1$. So, from the equation $\prod_{i=1}^5(1-a_5) = 1+\sum_{i=1}^5a_i = 1-a_1$, we get \begin{equation*} \frac 1{(1-a_2)(1-a_5)} = (1-a_3)(1-a_4) = 1-a_3-a_4+a_3a_4 \end{equation*} Using the upper bounds on $a_2$ and $a_5$ from PRP 4, we get \begin{equation*} a_3+a_4-a_3a_4 = 1-\frac1{(1-a_2)(1-a_5)} \ge 1-\frac1{(1-0.41)(1-(-0.72))}>0 \end{equation*} We next find a lower bound for $a_3a_4$. We note that $a_1$ is positive. For if $a_1<0$, then $a_3+a_4+a_5>-a_2+9/4>2.25$. We have a contradiction, since $\sqrt3\ge |a_3|+|a_4|+|a_5|$ by Cauchy-Schwarz. So \begin{equation*} a_1 \ge 9/8=1.125. \end{equation*} From $-1=a_1a_2a_3a_4$, and using $a_1\le (2+\sqrt(3))/2$ and $-a_2<2$, we find $$a_3a_4 = \frac1{a_1(-a_2)} > \frac 1{2+\sqrt3} > 0.25$$ Hence, \begin{equation*} a_3+a_4 > a_3a_4 > 0.25. \end{equation*} Therefore, using $a_2\ge -2$ and $a_5\ge -0.41$, \begin{equation*} a_1 = \frac12(-a_2-a_3-a_4-a_5) \le \frac12(2.41-0.25) =1.08. \end{equation*} We have a contradiction.

  • $\begingroup$ The labeled propositions above apply to the case of degree $4$. In fact, one can use more-or-less the techniques above in the case $r=5$ to show that $a_1=1$ if $r=4$ by way of contradiction. From there one is left with solving low-degree polynomial equations in two variables. The computations are a little too messy for me to do by hand. :) $\endgroup$ Dec 23, 2019 at 16:54

I have tried to handle this case with this polynomial reduction applet: http://www.dr-mikes-maths.com/polynomial-reduction.html, to express the solution for a alone.

For the second degree, I got

$1 + a=0$

For the third degree,

$-1 -a -2a^2 -2a^4=0$

And for the fourth,

$a^4 -7a^7 + 2a^8 -10a^9 + 19a^{10} + 10a^{11} + 45a^{12} -51a^{13} -38a^{14} -51a^{15} + 38a^{16} + 11a^{17} + 115a^{18} -2a^{19} -14a^{20} -136a^{21} + 20a^{23} + 96a^{24} -32a^{25} -32a^{27} + 16a^{28}=0$

  • $\begingroup$ By the way, I nearly confirm @Barry's derivation of $2 +2b^2 +b^3 +b^4=0$. $\endgroup$
    – user65203
    May 11, 2014 at 15:23
  • $\begingroup$ The nonzero assumption takes last polynomial down to something of degree $24$. Have you tested it for rational roots? (Also, how did it get written with two signs in the "$+-51a^{15}$" term?) $\endgroup$ May 11, 2014 at 15:28
  • $\begingroup$ @BarryCipra According to the all-mighty Wolfram, it reduces to $16 a^{17}-16 a^{15}-36a^{14}+10 a^{13}+7 a^{12}+22a^{11}+10a^{10}+27a^9-28 a^8-13a^7-11a^6+8a^5-4 a^4+9 a^3-2 a^2+a=1$ $\endgroup$ May 11, 2014 at 16:29
  • $\begingroup$ Which has only $1.1695676495049220889$ as real solution. $\endgroup$ May 11, 2014 at 16:39
  • $\begingroup$ @Barry: why rational roots ? $\endgroup$
    – user65203
    May 12, 2014 at 6:16

I have confirmed your solutions for $n\le 3$ and solved $n=4,5$. (Using Mathematica.)

Solving for some degree $n\in\mathbb N$

Let $c_i\in \mathbb R,i=0,1,\dots,n$ be coefficients of polynomial $P_n(X)$, where $c_n=1$ multiplies $X^n$.

Expand the degree $n$ polynomial to express its coefficients as combinations of its roots $z_i$.

$$\begin{align} P_n(X) &=\sum_{i=0}^nc_iX^i\\ &=\prod_{i=1}^{n}(X-z_i) \\ &=\left((-1)^n\sum_{j=1}^1\prod_{k=1}^nz_k\right)X^0 +\left((-1)^{n-1}\sum_{j=1}^n\prod_{k=1}^n\frac{z_k}{z_j}\right)X^1 +\dots +X^n\end{align}$$

To solve the problem, we need to solve the following system of equations:

$$ c_{i-1}=z_i $$

For $i=1,\dots,n$, where $z_i\ne0$.

This can be solved with Mathematica, using Reduce[]. The function e[n] solves it for given $n$:

p[n_] := Fold[Times, Table[(x - x[i]), {i, 1, n}]]
e[n_] := Reduce[Fold[And, Table[(x[i] != 0 && x[i] == (CoefficientList[p[n], x])[[i]]), {i, 1, n}]], Variables[p[n]], Reals]

Obtained solutions

Let $Z=(z_1,z_2,\dots,z_{n})$ represent a solution.

Here are the solutions for $n\in\{1,2,3,4,5\}$ generated by above Mathematica code:

$(n=1):$ No solutions.

$(n=2):$ One solution:

  • $Z_1=(-2,1)$ giving $(X+2)(X-1)=-2 + X+ X^2$.

$(n=3):$ Two solutions:

  • $Z_1=(-1,-1,1)$ giving $(X+1)(X+1)(X-1)=-1 - X + X^2 + X^3$.

  • $Z_2=(z_1,z_2,z_3)$ whose components are real roots of the following polynomials: $$\begin{align}p_1(x)&=-2 + 4x - 2x^2 + x^3, & z_1= 0.6388969194\dots\\ p_2(x)&=2-2x+x^3, & z_2=-1.769292354\dots \\ p_3(x)&=-1+2x^2+2x^3, & z_3=0.5651977173\dots \end{align}$$ This agrees with Barry Cipra's answer.

$(n=4):$ One solution:

  • $Z_1=(z_1,z_2,z_3,z_4)$ whose components are real roots of the following polynomials: $$\begin{align}p_1(x)&=-1+2x+3x^2+x^3, & z_1= 0.3247179572\dots\\ p_2(x)&=1+2x+x^2+x^3, & z_2=-0.5698402909\dots \\ p_3(x)&=1+x+2x^2+x^3, & z_3=-1.754877666\dots \\ p_4(x)&=-1+x, & z_4=1 \end{align}$$ This agrees with Lutz Lehmann's answer.

$(n=5):$ No solutions. Additionally, it seems another answer obtained this result without Mathematica (without computer assistance).

This resolves all $n\le 5$. For $n\gt 5$, there are no solutions according to Gabriel Romon's answer.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .