Polynomials such that roots=coefficients

Here is my question :

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

Mathematically, prove or disprove the existence of n5 such that (z1,,zn)(R{0})n,(Xz1)...(Xzn)=Xn+ni=1ziXni

State of the problem: There’s no such polynomial for n6 (see answer below). It remains to prove/disprove the n=4,5 cases.


Here are all such real polynomials with degree 3:

X2+X2=(X1)(X+2)

X3+X2X1=(X1)(X+1)2

X3+αX2+βX+γ where α is the real root of 2X3+2X21 (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(X1)(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 n6

Suppose for contradiction that z1,,znR{0}n are such that (Xz1)...(Xzn)=Xn+n1k=1ziXni

Then three useful identities appear nk=1zk=z1(1)

1i<jnzizj=z2(2)

nk=1zk=(1)nzn(3)


Since (nk=1zk)2=nk=1z2k+21i<jnzizjit follows thatz21=2z2+nk=1z2k

Hence 0<nk=2z2k=2z2(4) and 0<nk=3z2k=1(z2+1)2(5)

(4) and (5) imply 2<z2<0(6)

thus (6) and (4) imply 0<nk=2z2k<4(7)

Also (6) and (5) imply 0<nk=3z2k1(8)


By AM-GM, (|z3|2|zn1|2)1/(n3)1n3n1k=3z2k1n3nk=3z2k

Hence
|z3|2|zn1|2(1n3nk=3z2k)n3

Squaring, |z3||zn1|(1n3nk=3z2k)n32(8)1(n3)(n3)/2(9)


By triangle inequality (1), and Cauchy-Schwarz

2|z1|nk=2|zk|n1nk=2z2k

Hence by (7),

|z1|n1(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||zn1|<2n1(n3)(n3)/2

This inequality fails for n6.

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

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