Polynomials such that roots=coefficients

Question

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.

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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.

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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 $\geq 5$.

Answer 1

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 $1954$.

Answer 2

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,

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

Answer 3

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.

Answer 4

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}

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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.

Answer 5

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 $$\sum_{k=1}^{n}z_k=-z_1 \; \; \; \;(1)$$

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

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

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

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

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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 $(7)$,

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

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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.

Answer 6

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

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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]

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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.

Answer 7

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*}

2. 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.