If n = 2p where p is a prime number other than 2, then
For n up to 30, the cyclotomic polynomials are:[2]
The case of the 105th cyclotomic polynomial is interesting because 105 is the least positive integer that is the product of three distinct odd prime numbers (3×5×7) and this polynomial is the first one that has a coefficient other than 1, 0, or −1:[3]
The cyclotomic polynomials are monic polynomials with integer coefficients that are irreducible over the field of the rational numbers. Except for n equal to 1 or 2, they are palindromes of even degree.
The degree of , or in other words the number of nth primitive roots of unity, is , where is Euler's totient function.
The fact that is an irreducible polynomial of degree in the ring is a nontrivial result due to Gauss.[4] Depending on the chosen definition, it is either the value of the degree or the irreducibility which is a nontrivial result. The case of prime n is easier to prove than the general case, thanks to Eisenstein's criterion.
A fundamental relation involving cyclotomic polynomials is
which means that each n-th root of unity is a primitive d-th root of unity for a unique d dividing n.
This provides a recursive formula for the cyclotomic polynomial , which may be computed by dividing by the cyclotomic polynomials for the proper divisors d dividing n, strarting from :
These formulas may be applied repeatedly to get a simple expression for any cyclotomic polynomial in terms of a cyclotomic polynomial of square free index: If q is the product of the prime divisors of n (its radical), then[5]
This allows formulas to be given for the nth cyclotomic polynomial when n has at most one odd prime factor: If p is an odd prime number, and h and k are positive integers, then
For other values of n, the computation of the nth cyclotomic polynomial is similarly reduced to that of where q is the product of the distinct odd prime divisors of n. To deal with this case, one has that, for p prime and not dividing n,[6]
The problem of bounding the magnitude of the coefficients of the cyclotomic polynomials has been the object of a number of research papers.[7]
If n has at most two distinct odd prime factors, then Migotti showed that the coefficients of are all in the set {1, −1, 0}.[8]
The first cyclotomic polynomial for a product of three different odd prime factors is it has a coefficient −2 (see above). The converse is not true: only has coefficients in {1, −1, 0}.
If n is a product of more different odd prime factors, the coefficients may increase to very high values. E.g., has coefficients running from −22 to 23; also , the smallest n with 6 different odd primes, has coefficients of magnitude up to 532.
Let A(n) denote the maximum absolute value of the coefficients of . It is known that for any positive k, the number of n up to x with A(n) > nk is at least c(k)⋅x for a positive c(k) depending on k and x sufficiently large. In the opposite direction, for any function ψ(n) tending to infinity with n we have A(n) bounded above by nψ(n) for almost all n.[9]
A combination of theorems of Bateman and Vaughan states that[7]: 10 on the one hand, for every , we have
for all sufficiently large positive integers , and on the other hand, we have
for infinitely many positive integers . This implies in particular that univariate polynomials (concretely for infinitely many positive integers ) can have factors (like ) whose coefficients are superpolynomially larger than the original coefficients. This is not too far from the general Landau-Mignotte bound.
for certain polynomials An(z) and Bn(z) with integer coefficients, An(z) of degree φ(n)/2, and Bn(z) of degree φ(n)/2 − 2. Furthermore, An(z) is palindromic when its degree is even; if its degree is odd it is antipalindromic. Similarly, Bn(z) is palindromic unless n is composite and n ≡ 3 (mod 4), in which case it is antipalindromic.
The Sister Beiter conjecture is concerned with the maximal size (in absolute value) of coefficients of ternary cyclotomic polynomials where are three odd primes.[12]
Cyclotomic polynomials over a finite field and over the p-adic integers
Over a finite field with a prime number p of elements, for any integer n that is not a multiple of p, the cyclotomic polynomial factorizes into irreducible polynomials of degree d, where is Euler's totient function and d is the multiplicative order of p modulo n. In particular, is irreducible if and only ifp is a primitive root modulo n, that is, p does not divide n, and its multiplicative order modulo n is , the degree of .[13]
These results are also true over the p-adic integers, since Hensel's lemma allows lifting a factorization over the field with p elements to a factorization over the p-adic integers.
If x takes any real value, then for every n ≥ 3 (this follows from the fact that the roots of a cyclotomic polynomial are all non-real, for n ≥ 3).
For studying the values that a cyclotomic polynomial may take when x is given an integer value, it suffices to consider only the case n ≥ 3, as the cases n = 1 and n = 2 are trivial (one has and ).
The values that a cyclotomic polynomial may take for other integer values of x is strongly related with the multiplicative order modulo a prime number.
More precisely, given a prime number p and an integer b coprime with p, the multiplicative order of b modulo p, is the smallest positive integer n such that p is a divisor of For b > 1, the multiplicative order of b modulo p is also the shortest period of the representation of 1/p in the numeral baseb (see Unique prime; this explains the notation choice).
The definition of the multiplicative order implies that, if n is the multiplicative order of b modulo p, then p is a divisor of The converse is not true, but one has the following.
If n > 0 is a positive integer and b > 1 is an integer, then (see below for a proof)
where
k is a non-negative integer, always equal to 0 when b is even. (In fact, if n is neither 1 nor 2, then k is either 0 or 1. Besides, if n is not a power of 2, then k is always equal to 0)
g is 1 or the largest odd prime factor of n.
h is odd, coprime with n, and its prime factors are exactly the odd primes p such that n is the multiplicative order of b modulo p.
This implies that, if p is an odd prime divisor of then either n is a divisor of p − 1 or p is a divisor of n. In the latter case, does not divide
It follows from above factorization that the odd prime factors of
are exactly the odd primes p such that n is the multiplicative order of b modulo p. This fraction may be even only when b is odd. In this case, the multiplicative order of b modulo 2 is always 1.
There are many pairs (n, b) with b > 1 such that is prime. In fact, Bunyakovsky conjecture implies that, for every n, there are infinitely many b > 1 such that is prime. See OEIS: A085398 for the list of the smallest b > 1 such that is prime (the smallest b > 1 such that is prime is about , where is Euler–Mascheroni constant, and is Euler's totient function). See also OEIS: A206864 for the list of the smallest primes of the form with n > 2 and b > 1, and, more generally, OEIS: A206942, for the smallest positive integers of this form.
Proofs
Values of If is a prime power, then
If n is not a prime power, let we have and P is the product of the for k dividing n and different of 1. If p is a prime divisor of multiplicity m in n, then divide P(x), and their values at 1 are m factors equal to p of As m is the multiplicity of p in n, p cannot divide the value at 1 of the other factors of Thus there is no prime that divides
Ifnis the multiplicative order ofbmodulop, then By definition, If then p would divide another factor of and would thus divide showing that, if there would be the case, n would not be the multiplicative order of b modulo p.
The other prime divisors ofare divisors ofn. Let p be a prime divisor of such that n is not be the multiplicative order of b modulo p. If k is the multiplicative order of b modulo p, then p divides both and The resultant of and may be written where P and Q are polynomials. Thus p divides this resultant. As k divides n, and the resultant of two polynomials divides the discriminant of any common multiple of these polynomials, p divides also the discriminant of Thus p divides n.
gandhare coprime. In other words, if p is a prime common divisor of n and then n is not the multiplicative order of b modulo p. By Fermat's little theorem, the multiplicative order of b is a divisor of p − 1, and thus smaller than n.
gis square-free. In other words, if p is a prime common divisor of n and then does not divide Let n = pm. It suffices to prove that does not divide S(b) for some polynomial S(x), which is a multiple of We take
The multiplicative order of b modulo p divides gcd(n, p − 1), which is a divisor of m = n/p. Thus c = bm − 1 is a multiple of p. Now,
As p is prime and greater than 2, all the terms but the first one are multiples of This proves that
Suppose is a finite list of primes congruent to modulo Let and consider . Let be a prime factor of (to see that decompose it into linear factors and note that 1 is the closest root of unity to ). Since we know that is a new prime not in the list. We will show that
Let be the order of modulo Since we have . Thus . We will show that .
Assume for contradiction that . Since
we have
for some . Then is a double root of
Thus must be a root of the derivative so
But and therefore This is a contradiction so . The order of which is , must divide . Thus
The constant-coefficient linear recurrences which are periodic are precisely the power series coefficients of rational functions whose denominators are products of cyclotomic polynomials.
In the theory of combinatorial generating functions, the denominator of a rational function determines a linear recurrence for its power series coefficients. For example, the Fibonacci sequence has generating function
and equating coefficients on both sides of gives for .
Any rational function whose denominator is a divisor of has a recursive sequence of coefficients which is periodic with period at most n. For example,
has coefficients defined by the recurrence for , starting from . But , so we may write
which means for , and the sequence has period 6 with initial values given by the coefficients of the numerator.
^ abSanna, Carlo (2021), "A Survey on Coefficients of Cyclotomic Polynomials", arXiv:2111.04034 [math.NT]
^Isaacs, Martin (2009), Algebra: A Graduate Course, AMS Bookstore, p. 310, ISBN978-0-8218-4799-2
^Maier, Helmut (2008), "Anatomy of integers and cyclotomic polynomials", in De Koninck, Jean-Marie; Granville, Andrew; Luca, Florian (eds.), Anatomy of integers. Based on the CRM workshop, Montreal, Canada, March 13-17, 2006, CRM Proceedings and Lecture Notes, vol. 46, Providence, RI: American Mathematical Society, pp. 89–95, ISBN978-0-8218-4406-9, Zbl1186.11010
Gauss's book Disquisitiones Arithmeticae [Arithmetical Investigations] has been translated from Latin into French, German, and English. The German edition includes all of his papers on number theory: all the proofs of quadratic reciprocity, the determination of the sign of the Gauss sum, the investigations into biquadratic reciprocity, and unpublished notes.