From: Alexander A Borisov <borisov@math.psu.edu>
Subject: Re: Sums of cyclotomic polynomials.
Date: Wed, 26 Apr 2000 10:41:06 -0400
Newsgroups: sci.math.research

On Tue, 25 Apr 2000, Charles Nicol wrote:

> Let C_m(x) denote the m-th. cyclotomic polynomial.It has been observed
> that if m and n are each greater than one and C_m(x) + C_n(x) is a
> reducible polynomial,then in all cases this polynomial has a cyclotomic
> factor.For example C_7(x) + C_22(x) =(x^2+1)*(x^8-x^7+2*x^4+2).Furthermore
> the non-cyclotomic part of these polynomials always seems to be
> irreducible.
> These observations have been checked up to 150 and in a number of larger
> isolated larger cases.It would be interesting if these observations were
> always true.
> Charles Nicol.
> 
> 

 The question is really interesting. I have some doubts that
this could be true in general. I am sure it is true for almost all pairs
(n,m) in the density sense. I have no idea though how to prove it. What I
can prove is the following.

1) If deg C_n and deg C_m are different, then the non-cyclotomic part of
f_{n,m}=C_n+C_m is coprime to its reciprocal.

2) If n and m are both primes, then C_n+C_m is irreducible.



Proof of 1)
Both C_n and C_m are reciprocal, and C_n(1/x)=(1/x^{degC_n})C_n(x). The
same is true for C_m. So if x and 1/x are both roots of C_n+C_m, this
gives as a system of two linear equations on C_n(x) and C_m(x). Either
both f(x) and g(x) are zeroes, or the determinant of the system is zero.
The first case is actually impossible, because C_n and C_m are coprime.
The determinant being zero implies that x^{degC_n-DegC_m}=1, so x is a
root of unity.

Proof of 2)
Suppose n and m are primes. We can certainly assume that they
are different. The polynomial F(x)=C_n(x)+C_m(x) is in this case
(x^n+x^m-2)/(x-1). Therefore it has Mahler measure of F is at most
\sqrt{1+1+4}=\sqrt(6). If F=G*H, for some G and H, then one of the
polynomials G, H has a constant term pus or minus 2. So it has Mahler
measure at least 2. Therefor the other polynomial, say G, has Mahler
measure at most \sqrt{6}/2=\sqrt{3/2}. Because G is coprime to its
reciprocal, we can apply to it the result of Christopher Smyth,

On the product of the conjugates outside the unit circle of an algebraic
integer. Bull. London Math. Soc. 3 1971 169--175.

It implies that either F is cyclotomic or its Mahler measure is
greater than or equal to the root of the polynomial x^3-x-1. Because
\sqrt(3/2) is less than this root, G must be cyclotomic.
This already shows that the noncyclotomic part of F is
irreducible. To complete the proof of the irreducibility of F, suppose
F(x)=0 for some x, which is on the unit circle. Then x^n+x^m=2, which is
only possible if both x^nand x^m are equal to 1 (triangle equality).
Because n and m are different primes, it implies that x=1. Finally, it is
clear that x=1 is not a root of F, because all the coefficients of F are
positive.

Alexander Borisov
borisov@math.psu.edu
http://www.math.wustl.edu/~borisov