# Polynomials such that roots=coefficients

Here is my question :

Are there monic polynomials with degree $$≥5\geq 5$$ such that they have the same real all non zero roots and coefficients ?

Mathematically, prove or disprove the existence of $$n≥5n \geq 5$$ such that $$∃(z1,…,zn)∈(R−{0})n,(X−z1)...(X−zn)=Xn+n∑i=1ziXn−i\exists (z_1,\ldots, z_n) \in \left(\mathbb R-\{0\}\right)^n, (X-z_1)...(X-z_n)=X^n+\sum_{i=1}^{n}z_iX^{n-i}$$

State of the problem: There’s no such polynomial for $$n≥6n \geq 6$$ (see answer below). It remains to prove/disprove the $$n=4,5n=4,5$$ cases.

Here are all such real polynomials with degree $$≤3\leq 3$$:

$$X2+X−2=(X−1)(X+2)X^2+X-2=(X-1)(X+2)$$

$$X3+X2−X−1=(X−1)(X+1)2X^3+X^2-X-1=(X-1)(X+1)^2$$

$$X3+αX2+βX+γX^3+\alpha X^2 + \beta X + \gamma$$ where $$α\alpha$$ is the real root of $$2X3+2X2−12X^3+2X^2-1$$ (which determines $$γ\gamma$$ and $$β\beta$$)

There remains complex degree 3 polynomials, as in Barry’s answer.

Edit:

As pointed out by Jyrki Lahtonen, if $$PP$$ is a satisfactory polynomial, then so is $$XPXP$$. For example, The family of polynomials $$Xn(X−1)(X+2)X^n(X-1)(X+2)$$ 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 $$≥5\geq 5$$.

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

Since it follows that

Hence and

$(4)$ and $(5)$ imply

thus $(6)$ and $(4)$ imply

Also $(6)$ and $(5)$ imply

By AM-GM,

Hence

Squaring,

By triangle inequality $(1)$, and Cauchy-Schwarz

Hence by $(7)$,

Rewriting $(6)$ as

Recalling $(3)$ (with $z_n$ cancelled from both sides) and putting together $(9)$, $(10)$ and $(11)$, we have

This inequality fails for $n\geq 6$.

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