Sums of cyclotomic polynomials.
Charles Nicol
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.
Alexander A Borisov
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
bor...@math.psu.edu
http://www.math.wustl.edu/~borisov
David Mosher
Wed 26 Apr 2000 10:41 EDT, in message
<news:Pine.GSO.4.10.100042...@stokes.math.psu.edu>,
Alexander A Borisov <mailto:bor...@math.psu.edu> wrote,
> 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.
[...] [C_r denotes the rth cyclotomic polynomial.]
> 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).
[...]
> 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^n and x^m are equal to 1 (triangle equality). Because n and m are
> different primes, it implies that x=1.
[...]
Above (2) is an immediate consequence of this elementary observation:
An integer polynomial with no roots in the closed unit disk, whose value
at 0 is plus-or-minus a prime, is irreducible in Z[x].
Indeed, such a polynomial is not the product of two non-unit *complex*
polynomials each of whose constant term is an integer and each of whose
leading coefficient has modulus at least 1.
With, as usual, \Omega denoting the arithmetic function that counts prime
factors with multiplicity, extended to negative integers and to Z[x] (prime
factors of the content should also be counted), this may be recast somewhat
more precisely:
For non-null f \in Z[x],
\Omega(f) <= min{\Omega(f(n)): n \in Z & dist(n,roots(f))>1}.
[With appropriate definition of \Omega (a little care is needed for class
numbers > 1), this also holds for rings of integers, and polynomials over
them, of imaginary quadratic number fields.]
To apply the observation to F above (in the case that m and n are distinct),
note that, as above, (numerator of F)(x)=0 for |x|<=1 (not just for |x|=1!)
forces x=1, but F(1)=m+n, so F is zero-free in the closed unit disk. Since
F(0)=2, F is irreducible.
In the same way, a sum of polynomials (x^{n_k}-1)/(x^d-1), with
d=gcd(...,n_k,...), is zero-free in the closed unit disk. More can be said
along these lines, but most relevant here is this:
A sum of N (not necessarily distinct) cyclotomic polynomials, each of
prime index (or each of index twice-an-odd-prime), has at most \Omega(N)
(non-unit) factors in Z[x]; in particular, if N is prime, such a sum is
irreducible.
Experimenting with small composite N and low degrees, using Pari/GP:
The only example of a reducible sum of 4 prime-index cyclotomic
polynomials of degree at most 58 (i.e., index<=59) is
(C_3 + 3C_2)(x) = (x+2)^2.
There are no reducible sums of 6 prime-index cyclotomic polynomials of
degree at most 40.
The reducible sums of 8 prime-index cyclotomic polynomials of degree at
most 28 are given by these 15 octuples of indices. Each has two
non-constant factors.
[ 3, 3, 2, 2, 2, 2, 2, 2]
[ 5, 5, 5, 3, 2, 2, 2, 2]
[ 7, 3, 3, 3, 2, 2, 2, 2]
[ 7, 5, 3, 3, 3, 3, 2, 2]
[ 7, 5, 3, 3, 3, 3, 3, 3]
[ 7, 5, 5, 3, 3, 3, 3, 3]
[ 7, 5, 5, 5, 5, 3, 3, 3]
[11, 7, 5, 5, 5, 5, 2, 2]
[11, 7, 7, 5, 5, 5, 5, 3]
[11, 7, 7, 7, 2, 2, 2, 2]
[11, 7, 7, 7, 7, 5, 3, 3]
[17, 13, 13, 13, 13, 5, 3, 3]
[23, 13, 11, 11, 11, 11, 2, 2]
[23, 13, 13, 11, 11, 11, 11, 3]
[23, 19, 19, 19, 19, 5, 3, 3]
--
David Mosher <dmo...@blackhole.nyx.net>
(The ultimate effect of Hawking radiation should be applied to the address.)