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 $\geq 5$.
 A: 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.
A: 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.
A: 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 $1954$.
A: 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,
$a^4 -7a^7 + 2a^8 -10a^9 + 19a^{10} + 10a^{11} + 45a^{12} -51a^{13}  -38a^{14}  -51a^{15} + 38a^{16} + 11a^{17} + 115a^{18}  -2a^{19} -14a^{20}  -136a^{21} + 20a^{23} + 96a^{24}  -32a^{25}  -32a^{27} + 16a^{28}=0$
A: 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.
