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Alex. Lupas science forum Guru Wannabe
Joined: 06 May 2005
Posts: 245

Posted: Tue Jun 07, 2005 3:51 am Post subject:
Poly. degree four = ?



Suppose that A,B,C,D are integers and let f(x)=x^4+Ax^3+Bx^2+Cx+D
having the roots z_1,z_2,z_3,z_4 with z_1=z_2=z_3=z_4=1.
Find the coefficients A,B,C,D.
[Generalization !]
Possible source: [Leopold Kronecker, solution by Ludwig Bieberbach ] 

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Virgil science forum Guru
Joined: 24 Mar 2005
Posts: 5536

Posted: Tue Jun 07, 2005 5:48 am Post subject:
Re: Poly. degree four = ?



In article <slrndaaejg.soe.timusenet@soprano.littlepossums.net>,
Timothy Little <timusenet@littlepossums.net> wrote:
Quote:  Alex. Lupas wrote:
Suppose that A,B,C,D are integers and let f(x)=x^4+Ax^3+Bx^2+Cx+D
having the roots z_1,z_2,z_3,z_4 with z_1=z_2=z_3=z_4=1.
Find the coefficients A,B,C,D.
Is there something wrong with the obvious A = B = C = 0, D = +1?
 Tim

Or even z_1 = z_2 = z_3 = z_4 = 1 so that A = C = 4, B = 6, D = 1 ? 

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Timothy Little science forum Guru Wannabe
Joined: 30 May 2005
Posts: 295

Posted: Tue Jun 07, 2005 6:10 am Post subject:
Re: Poly. degree four = ?



Alex. Lupas wrote:
Quote:  Suppose that A,B,C,D are integers and let f(x)=x^4+Ax^3+Bx^2+Cx+D
having the roots z_1,z_2,z_3,z_4 with z_1=z_2=z_3=z_4=1.
Find the coefficients A,B,C,D.

Is there something wrong with the obvious A = B = C = 0, D = +1?
 Tim 

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Alex. Lupas science forum Guru Wannabe
Joined: 06 May 2005
Posts: 245

Posted: Tue Jun 07, 2005 12:02 pm Post subject:
Re: Poly. degree four = ?



Hi Timothy /Virgil,
I appreciate that your solution is'nt complete./Alex 

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Chip Eastham science forum Guru
Joined: 01 May 2005
Posts: 412

Posted: Tue Jun 07, 2005 3:46 pm Post subject:
Re: Poly. degree four = ?



Alex. Lupas wrote:
Quote:  Suppose that A,B,C,D are integers and let f(x)=x^4+Ax^3+Bx^2+Cx+D
having the roots z_1,z_2,z_3,z_4 with z_1=z_2=z_3=z_4=1.
Find the coefficients A,B,C,D.
[Generalization !]
Possible source: [Leopold Kronecker, solution by Ludwig Bieberbach ]

Hi, Alex:
I notice that for roots on the unit circle, real implies +1,1.
Removing such a factor preserves the integrality of polynomial
coefficients, so your problem simplifies to asking what pairs of
quadratic factors will multiply to give an integral quartic:
(x^2 + ax + b)(x^2 + cx + d)
with the constraint that the quadratics also have roots on the
unit circle.
But since the conjugate pairs of roots on the unit circle are
necessarily reciprocals, b = d = 1.
This is I believe enough to force a,c to be integers & polish
off the possibilities.
regards, chip 

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Alex. Lupas science forum Guru Wannabe
Joined: 06 May 2005
Posts: 245

Posted: Wed Jun 08, 2005 12:15 am Post subject:
Re: Poly. degree four = ?



HINT:Denote by M the set of all
f(x)=f(A,B,C,D;x)=x^4+Ax^3+Bx^2+Cx+D
having the roots z_1,z_2,z_3,z_4 such that
z_1=z_2=z_3=z_4=1.
If Card(M) denotes the number of polynomials from M, then
Card(M) is finite. This follows from the fact that
A=z_1+z_2+z_3+z_4 =< 4
B=z_1z_2+z_1z_3+z_1z_4+z_2z_3+z_2z_4+z_3z_4 =< 6
C =< 6
D =< 1 .
Let q be the number of all roots of polynomials f from M.
Also q is a finite positive integer; denote by
Q={r_1,r_2,....,r_q}
the roots of all f, f in M. If r is in Q , then
each numbers from r,r^2,r^3, ....,r^{q+1} is also in Q .
Because Card(Q)=q there exists u,v, (u<v), such that r^u=r^v
that is
======
r^s=1
======
where s:=vu is a positive integer. In conclusion each
element from Q must be a root of unity, i.e. in our case
there exists positive integers s_1,s_2 ;k_1,k_2,1=<k_j=<s_j,
such that
z_1=cos(2k_1*pi/s_1) + i*sin(2k_1*pi/s_1)
z_2=cos(2k_1*pi/s_1)  i*sin(2k_1*pi/s_1)
z_3=cos(2k_2*pi/s_2) + i*sin(2k_2*pi/s_2)
z_4=cos(2k_2*pi/s_2)  i*sin(2k_2*pi/s_2)
.. ...... 

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