Here is my question :

Are there monic polynomials with degree ≥5 such that they have the same

realall non zeroroots and coefficients ?Mathematically, prove or disprove the existence of n≥5 such that ∃(z1,…,zn)∈(R−{0})n,(X−z1)...(X−zn)=Xn+n∑i=1ziXn−i

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

Here are all such real polynomials with degree ≤3:

X2+X−2=(X−1)(X+2)

X3+X2−X−1=(X−1)(X+1)2

X3+αX2+βX+γ where α is the real root of 2X3+2X2−1 (which determines γ and β)

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 Xn(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.

**Answer**

I think I got the proof that no such real polynomial with degree ≥6 exists.

Let n≥6

Suppose for contradiction that z1,…,zn∈R−{0}n are such that (X−z1)...(X−zn)=Xn+∑n−1k=1ziXn−i

Then three useful identities appear n∑k=1zk=−z1(1)

∑1≤i<j≤nzizj=z2(2)

n∏k=1zk=(−1)nzn(3)

Since (n∑k=1zk)2=n∑k=1z2k+2∑1≤i<j≤nzizjit follows thatz21=2z2+n∑k=1z2k

Hence 0<n∑k=2z2k=−2z2(4) and 0<n∑k=3z2k=1−(z2+1)2(5)

(4) and (5) imply −2<z2<0(6)

thus (6) and (4) imply 0<n∑k=2z2k<4(7)

Also (6) and (5) imply 0<n∑k=3z2k≤1(8)

By AM-GM, (|z3|2…|zn−1|2)1/(n−3)≤1n−3n−1∑k=3z2k≤1n−3n∑k=3z2k

Hence

|z3|2…|zn−1|2≤(1n−3n∑k=3z2k)n−3

Squaring, |z3|…|zn−1|≤(1n−3n∑k=3z2k)n−32≤(8)1(n−3)(n−3)/2(9)

By triangle inequality (1), and Cauchy-Schwarz

2|z1|≤n∑k=2|zk|≤√n−1√n∑k=2z2k

Hence by (7),

|z1|≤√n−1(10)

Rewriting (6) as |z2|<2(11)

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

1=|z1||z2||z3|⋯|zn−1|<2√n−1(n−3)(n−3)/2

This inequality fails for n≥6.

Contradiction.

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

**Attribution***Source : Link , Question Author : Gabriel Romon , Answer Author : Barry Cipra*