# Polynomials such that roots=coefficients

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 $$\leq 3$$:

$$X^2+X-2=(X-1)(X+2)$$

$$X^3+X^2-X-1=(X-1)(X+1)^2$$

$$X^3+\alpha X^2 + \beta X + \gamma$$ 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.

Edit:

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 $$\geq 5$$.

• 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 $a=0$. Commented May 11, 2014 at 13:37
• 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. Commented May 11, 2014 at 13:49
• 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). Commented May 11, 2014 at 13:58
• @BarryCipra Thanks for your comments. I added the extra requirement that $0$ be not a root/coefficient. Commented May 11, 2014 at 14:07
• @GabrielR., maybe you want to change the question itself now to ask for examples with degree $\ge3$. Commented May 11, 2014 at 14:19

## 7 Answers

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 $n\geq 6$.

Contradiction.

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

• This is so brilliant. What was the thought process here? Commented 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

$$(b+1)(b^3-2b+2)=0$$

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

$$X^3+X^2-X-1=(X-1)(X+1)(X+1)$$

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

P<d,c,b,a>:=PolynomialRing(Rationals(),4);
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];
PrimaryDecomposition(ideal<P|I>);


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

P<e,d,c,b,a>:=PolynomialRing(Rationals(),5);
I:=[ElementarySymmetricPolynomial(P,6-k)-(-1)^k*P.k : k in [1..5]];
D:=PrimaryDecomposition(ideal<P|I>);
D[7];
//D[8];

CC<i>:=ComplexField(80);
R<x>:=PolynomialRing(CC);
//D[7] has a=1, generator is b with polynomial in position 4
Roots(Evaluate(BasisElement(D[7],4),[x,x,x,x,x]));
//D[8] is parametrized by a with polynomial in position 5
Roots(Evaluate(BasisElement(D[8],5),[x,x,x,x,x]));


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.

• Does it mean the conjecture holds for $n=5$ ? Commented May 11, 2014 at 23:02
• 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. Commented May 11, 2014 at 23:25
• It seems worth noting that googling on −1.75487766624669 leads to a problem regarding the Mandelbrot set at math.stackexchange.com/questions/743727/… Commented May 12, 2014 at 0:02

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 $$a_2=\sum_{1\le i, and $$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. $$a_2\le-\frac2{2+\sqrt{r-2}}\cdot\left(\frac{r-3}{\sqrt{r-2}}\right)^{r-3}$$

Proof:

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.

• 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. :) Commented Dec 23, 2019 at 16:54

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.

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,

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

• By the way, I nearly confirm @Barry's derivation of $2 +2b^2 +b^3 +b^4=0$.
– user65203
Commented May 11, 2014 at 15:23
• 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?) Commented May 11, 2014 at 15:28
• @BarryCipra According to the all-mighty Wolfram, it reduces to $16 a^{17}-16 a^{15}-36a^{14}+10 a^{13}+7 a^{12}+22a^{11}+10a^{10}+27a^9-28 a^8-13a^7-11a^6+8a^5-4 a^4+9 a^3-2 a^2+a=1$ Commented May 11, 2014 at 16:29
• Which has only $1.1695676495049220889$ as real solution. Commented May 11, 2014 at 16:39
• @Barry: why rational roots ?
– user65203
Commented 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.