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 $$\exists (z_1,\ldots, z_n) \in \left(\mathbb R-\{0\}\right)^n, (X-z_1)...(X-z_n)=X^n+\sum_{k=0}^{n-1}z_iX^{n-i}$$

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 '14 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. $\endgroup$ – Barry Cipra May 11 '14 at 13:49
  • 4
    $\begingroup$ 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). $\endgroup$ – Jyrki Lahtonen May 11 '14 at 13:58
  • $\begingroup$ @BarryCipra Thanks for your comments. I added the extra requirement that $0$ be not a root/coefficient. $\endgroup$ – Gabriel Romon May 11 '14 at 14:07
  • $\begingroup$ @GabrielR., maybe you want to change the question itself now to ask for examples with degree $\ge3$. $\endgroup$ – Barry Cipra May 11 '14 at 14:19

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.

  • 3
    $\begingroup$ This is so brilliant. What was the thought process here? $\endgroup$ – Joshua Benabou Mar 27 '16 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 $n=5$ ? $\endgroup$ – Gabriel Romon May 11 '14 at 23:02
  • $\begingroup$ 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. $\endgroup$ – LutzL May 11 '14 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/… $\endgroup$ – Barry Cipra May 12 '14 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.


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 $2 +2b^2 +b^3 +b^4=0$. $\endgroup$ – Yves Daoust May 11 '14 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?) $\endgroup$ – Barry Cipra May 11 '14 at 15:28
  • $\begingroup$ @BarryCipra According to the all-mighty Wolfram, it reduces to $16 a^{17}-16 a^{15}-36a^{14}+10 a^{13}+7 a^{12}+22a^{11}+10a^{10}+27a^9-28 a^8-13a^7-11a^6+8a^5-4 a^4+9 a^3-2 a^2+a=1$ $\endgroup$ – Gabriel Romon May 11 '14 at 16:29
  • $\begingroup$ Which has only $1.1695676495049220889$ as real solution. $\endgroup$ – Gabriel Romon May 11 '14 at 16:39
  • $\begingroup$ @Barry: why rational roots ? $\endgroup$ – Yves Daoust May 12 '14 at 6:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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