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

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 $\leq 3$:

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

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

$X^3+\alpha X^2 + \beta X + \gamma$ 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 $a=0$. – Hayden May 11 '14 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. – Barry Cipra May 11 '14 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). – Jyrki Lahtonen May 11 '14 at 13:58
• @BarryCipra Thanks for your comments. I added the extra requirement that $0$ be not a root/coefficient. – Gabriel Romon May 11 '14 at 14:07
• @GabrielR., maybe you want to change the question itself now to ask for examples with degree $\ge3$. – 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.

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

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

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 $n=5$ ? – Gabriel Romon May 11 '14 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. – LutzL May 11 '14 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/… – 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,

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

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