# Cyclotomic polynomial

In mathematics the nth cyclotomic polynomial, for any positive integer n, is the unique irreducible polynomial with integer coefficients that is a divisor of $x^{n}-1$ and is not a divisor of $x^{k}-1$ for any k < n. Its roots are all nth primitive roots of unity $e^{2i\pi {\frac {k}{n}}}$ , where k runs over the positive integers not greater than n and coprime to n. In other words, the nth cyclotomic polynomial is equal to

$\Phi _{n}(x)=\prod _{\stackrel {1\leq k\leq n}{\gcd(k,n)=1}}\left(x-e^{2i\pi {\frac {k}{n}}}\right).$ It may also be defined as the monic polynomial with integer coefficients that is the minimal polynomial over the field of the rational numbers of any primitive nth-root of unity ($e^{2i\pi /n}$ is an example of such a root).

An important relation linking cyclotomic polynomials and primitive roots of unity is

$\prod _{d\mid n}\Phi _{d}(x)=x^{n}-1,$ showing that x is a root of $x^{n}-1$ if and only if it is a dth primitive root of unity for some d that divides n.

## Examples

If n is a prime number, then

$\Phi _{n}(x)=1+x+x^{2}+\cdots +x^{n-1}=\sum _{k=0}^{n-1}x^{k}.$

If n = 2p where p is an odd prime number, then

$\Phi _{2p}(x)=1-x+x^{2}-\cdots +x^{p-1}=\sum _{k=0}^{p-1}(-x)^{k}.$

For n up to 30, the cyclotomic polynomials are:

{\begin{aligned}\Phi _{1}(x)&=x-1\\\Phi _{2}(x)&=x+1\\\Phi _{3}(x)&=x^{2}+x+1\\\Phi _{4}(x)&=x^{2}+1\\\Phi _{5}(x)&=x^{4}+x^{3}+x^{2}+x+1\\\Phi _{6}(x)&=x^{2}-x+1\\\Phi _{7}(x)&=x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\\\Phi _{8}(x)&=x^{4}+1\\\Phi _{9}(x)&=x^{6}+x^{3}+1\\\Phi _{10}(x)&=x^{4}-x^{3}+x^{2}-x+1\\\Phi _{11}(x)&=x^{10}+x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\\\Phi _{12}(x)&=x^{4}-x^{2}+1\\\Phi _{13}(x)&=x^{12}+x^{11}+x^{10}+x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\\\Phi _{14}(x)&=x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1\\\Phi _{15}(x)&=x^{8}-x^{7}+x^{5}-x^{4}+x^{3}-x+1\\\Phi _{16}(x)&=x^{8}+1\\\Phi _{17}(x)&=x^{16}+x^{15}+x^{14}+x^{13}+x^{12}+x^{11}+x^{10}+x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\\\Phi _{18}(x)&=x^{6}-x^{3}+1\\\Phi _{19}(x)&=x^{18}+x^{17}+x^{16}+x^{15}+x^{14}+x^{13}+x^{12}+x^{11}+x^{10}+x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\\\Phi _{20}(x)&=x^{8}-x^{6}+x^{4}-x^{2}+1\\\Phi _{21}(x)&=x^{12}-x^{11}+x^{9}-x^{8}+x^{6}-x^{4}+x^{3}-x+1\\\Phi _{22}(x)&=x^{10}-x^{9}+x^{8}-x^{7}+x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1\\\Phi _{23}(x)&=x^{22}+x^{21}+x^{20}+x^{19}+x^{18}+x^{17}+x^{16}+x^{15}+x^{14}+x^{13}+x^{12}+x^{11}\\&\qquad \quad +x^{10}+x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\\\Phi _{24}(x)&=x^{8}-x^{4}+1\\\Phi _{25}(x)&=x^{20}+x^{15}+x^{10}+x^{5}+1\\\Phi _{26}(x)&=x^{12}-x^{11}+x^{10}-x^{9}+x^{8}-x^{7}+x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+1\\\Phi _{27}(x)&=x^{18}+x^{9}+1\\\Phi _{28}(x)&=x^{12}-x^{10}+x^{8}-x^{6}+x^{4}-x^{2}+1\\\Phi _{29}(x)&=x^{28}+x^{27}+x^{26}+x^{25}+x^{24}+x^{23}+x^{22}+x^{21}+x^{20}+x^{19}+x^{18}+x^{17}+x^{16}+x^{15}+x^{14}\\&\qquad \quad +x^{13}+x^{12}+x^{11}+x^{10}+x^{9}+x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1\\\Phi _{30}(x)&=x^{8}+x^{7}-x^{5}-x^{4}-x^{3}+x+1.\end{aligned}}

The case of the 105th cyclotomic polynomial is interesting because 105 is the lowest 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:

{\begin{aligned}\Phi _{105}(x)&=x^{48}+x^{47}+x^{46}-x^{43}-x^{42}-2x^{41}-x^{40}-x^{39}+x^{36}+x^{35}+x^{34}+x^{33}+x^{32}+x^{31}-x^{28}-x^{26}\\&\qquad \quad -x^{24}-x^{22}-x^{20}+x^{17}+x^{16}+x^{15}+x^{14}+x^{13}+x^{12}-x^{9}-x^{8}-2x^{7}-x^{6}-x^{5}+x^{2}+x+1.\end{aligned}}

## Properties

### Fundamental tools

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 palindromics of even degree.

The degree of $\Phi _{n}$ , or in other words the number of nth primitive roots of unity, is $\varphi (n)$ , where $\varphi$  is Euler's totient function.

The fact that $\Phi _{n}$  is an irreducible polynomial of degree $\varphi (n)$  in the ring $\mathbb {Z} [x]$  is a nontrivial result due to Gauss. 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

$x^{n}-1=\prod _{1\leqslant k\leqslant n}\left(x-e^{2i\pi {\frac {k}{n}}}\right)=\prod _{d\mid n}\prod _{1\leqslant k\leqslant n \atop \gcd(k,n)=d}\left(x-e^{2i\pi {\frac {k}{n}}}\right)=\prod _{d\mid n}\Phi _{\frac {n}{d}}(x)=\prod _{d\mid n}\Phi _{d}(x).$

which means that each n-th root of unity is a primitive d-th root of unity for a unique d dividing n.

The Möbius inversion formula allows the expression of $\Phi _{n}(x)$  as an explicit rational fraction:

$\Phi _{n}(x)=\prod _{d\mid n}(x^{d}-1)^{\mu \left({\frac {n}{d}}\right)},$

where $\mu$  is the Möbius function.

The Fourier transform of functions of the greatest common divisor together with the Möbius inversion formula gives:

$\Phi _{n}(x)=\prod _{k=1}^{n}\left(x^{\gcd(k,n)}-1\right)^{\cos \left({\frac {2\pi k}{n}}\right)}.$

The cyclotomic polynomial $\Phi _{n}(x)$  may be computed by (exactly) dividing $x^{n}-1$  by the cyclotomic polynomials of the proper divisors of n previously computed recursively by the same method:

$\Phi _{n}(x)={\frac {x^{n}-1}{\prod _{\stackrel {d|n}{{}_{d

(Recall that $\Phi _{1}(x)=x-1$ .)

This formula allows computation of $\Phi _{n}(x)$  on a computer for any n, as soon as integer factorization and division of polynomials are available. Many computer algebra systems have a built in function to compute the cyclotomic polynomials. For example, this function is called by typing cyclotomic_polynomial(n,x) in SageMath, numtheory[cyclotomic](n,x); in Maple, and Cyclotomic[n,x] in Mathematica.

### Easy cases for computation

As noted above, if n is a prime number, then

$\Phi _{n}(x)=1+x+x^{2}+\cdots +x^{n-1}=\sum _{i=0}^{n-1}x^{i}.$

If n is an odd integer greater than one, then

$\Phi _{2n}(x)=\Phi _{n}(-x).$

In particular, if n = 2p is twice an odd prime, then (as noted above)

$\Phi _{n}(x)=1-x+x^{2}-\cdots +x^{p-1}=\sum _{i=0}^{p-1}(-x)^{i}.$

If n = pm is a prime power (where p is prime), then

$\Phi _{n}(x)=\Phi _{p}(x^{p^{m-1}})=\sum _{i=0}^{p-1}x^{ip^{m-1}}.$

More generally, if n = pmr with r relatively prime to p, then

$\Phi _{n}(x)=\Phi _{pr}(x^{p^{m-1}}).$

These formulas may be applied repeatedly to get a simple expression for any cyclotomic polynomial $\Phi _{n}(x)$  in term of a cyclotomic polynomial of square free index: If q is the product of the prime divisors of n (its radical), then

$\Phi _{n}(x)=\Phi _{q}(x^{n/q}).$

This allows to give formulas 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:

$\Phi _{2^{h}}(x)=x^{2^{h-1}}+1$
$\Phi _{p^{k}}(x)=\sum _{i=0}^{p-1}x^{ip^{k-1}}$
$\Phi _{2^{h}p^{k}}(x)=\sum _{i=0}^{p-1}(-1)^{i}x^{i2^{h-1}p^{k-1}}$

For the other values of n, the computation of the nth cyclotomic polynomial is similarly reduced to that of $\Phi _{q}(x),$  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,

$\Phi _{np}(x)=\Phi _{n}(x^{p})/\Phi _{n}(x).$

### Integers appearing as coefficients

The problem of bounding the magnitude of the coefficients of the cyclotomic polynomials has been the object of a number of research papers.

If n has at most two distinct odd prime factors, then Migotti showed that the coefficients of $\Phi _{n}$  are all in the set {1, −1, 0}.

The first cyclotomic polynomial for a product of three different odd prime factors is $\Phi _{105}(x);$  it has a coefficient −2 (see its expression above). The converse is not true: $\Phi _{231}(x)=\Phi _{3\times 7\times 11}(x)$  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., $\Phi _{15015}(x)=\Phi _{3\times 5\times 7\times 11\times 13}(x)$  has coefficients running from −22 to 23, $\Phi _{255255}(x)=\Phi _{3\times 5\times 7\times 11\times 13\times 17}(x)$ , the smallest n with 6 different odd primes, has coefficients up to ±532.

Let A(n) denote the maximum absolute value of the coefficients of Φn. 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.

### Gauss's formula

Let n be odd, square-free, and greater than 3. Then:

$4\Phi _{n}(z)=A_{n}^{2}(z)-(-1)^{\frac {n-1}{2}}nz^{2}B_{n}^{2}(z)$

where both An(z) and Bn(z) have integer coefficients, An(z) has degree φ(n)/2, and Bn(z) has 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 ≡ 3 (mod 4), in which case it is antipalindromic.

The first few cases are

{\begin{aligned}4\Phi _{5}(z)&=4(z^{4}+z^{3}+z^{2}+z+1)\\&=(2z^{2}+z+2)^{2}-5z^{2}\\[6pt]4\Phi _{7}(z)&=4(z^{6}+z^{5}+z^{4}+z^{3}+z^{2}+z+1)\\&=(2z^{3}+z^{2}-z-2)^{2}+7z^{2}(z+1)^{2}\\[6pt]4\Phi _{11}(z)&=4(z^{10}+z^{9}+z^{8}+z^{7}+z^{6}+z^{5}+z^{4}+z^{3}+z^{2}+z+1)\\&=(2z^{5}+z^{4}-2z^{3}+2z^{2}-z-2)^{2}+11z^{2}(z^{3}+1)^{2}\end{aligned}}

### Lucas's formula

Let n be odd, square-free and greater than 3. Then

$\Phi _{n}(z)=U_{n}^{2}(z)-(-1)^{\frac {n-1}{2}}nzV_{n}^{2}(z)$

where both Un(z) and Vn(z) have integer coefficients, Un(z) has degree φ(n)/2, and Vn(z) has degree φ(n)/2 − 1. This can also be written

$\Phi _{n}\left((-1)^{\frac {n-1}{2}}z\right)=C_{n}^{2}(z)-nzD_{n}^{2}(z).$

If n is even, square-free and greater than 2 (this forces n/2 to be odd),

$\Phi _{\frac {n}{2}}\left(-z^{2}\right)=\Phi _{2n}(z)=C_{n}^{2}(z)-nzD_{n}^{2}(z)$

where both Cn(z) and Dn(z) have integer coefficients, Cn(z) has degree φ(n), and Dn(z) has degree φ(n) − 1. Cn(z) and Dn(z) are both palindromic.

The first few cases are:

{\begin{aligned}\Phi _{3}(-z)&=\Phi _{6}(z)=z^{2}-z+1\\&=(z+1)^{2}-3z\\[6pt]\Phi _{5}(z)&=z^{4}+z^{3}+z^{2}+z+1\\&=(z^{2}+3z+1)^{2}-5z(z+1)^{2}\\[6pt]\Phi _{6/2}(-z^{2})&=\Phi _{12}(z)=z^{4}-z^{2}+1\\&=(z^{2}+3z+1)^{2}-6z(z+1)^{2}\end{aligned}}

## Cyclotomic polynomials over a finite field and over 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 $\Phi _{n}$  factorizes into ${\frac {\varphi (n)}{d}}$  irreducible polynomials of degree d, where $\varphi (n)$  is Euler's totient function, and d is the multiplicative order of p modulo n. In particular, $\Phi _{n}$  is irreducible if and only if p is a primitive root modulo n, that is, p does not divide n, and its multiplicative order modulo n is $\varphi (n),$  the degree of $\Phi _{n}$ .[citation needed]

These results are also true over the p-adic integers, since Hensel's lemma allows lifting a factorization over the field with p of elements to a factorization over the p-adic integers.

## Polynomial values

If x takes any real value, then $\Phi _{n}(x)>0$  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 $\Phi _{1}(x)=x-1$  and $\Phi _{2}(x)=x+1$ ).

For n ≥ 2, one has

$\Phi _{n}(0)=1,$
$\Phi _{n}(1)=1$  if n is not a prime power,
$\Phi _{n}(1)=p$  if $n=p^{k}$  is a prime power with k ≥ 1.

The values that a cyclotomic polynomial $\Phi _{n}(x)$  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 $x^{n}-1.$  For b > 1, the multiplicative order of b modulo p is also the shortest period of the representation of 1/p in the numeral base b (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 $\Phi _{n}(b).$  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)

$\Phi _{n}(b)=2^{k}gh,$

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 $\Phi _{n}(b),$  then either n is a divisor of p − 1 or p is a divisor of n. In the latter case $p^{2}$  does not divides $\Phi _{n}(b).$

Zsigmondy's theorem implies that the only cases where b > 1 and h = 1 are

{\begin{aligned}\Phi _{1}(2)&=1\\\Phi _{2}\left(2^{k}-1\right)&=2^{k}&&k>0\\\Phi _{6}(2)&=3\end{aligned}}

It follows from above factorization that the odd prime factors of

${\frac {\Phi _{n}(b)}{\gcd(n,\Phi _{n}(b))}}$

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 $\Phi _{n}(b)$  is prime. In fact, Bunyakovsky conjecture implies that, for every n, there are infinitely many b > 1 such that $\Phi _{n}(b)$  is prime. See for the list of the smallest b > 1 such that $\Phi _{n}(b)$  is prime (the smallest b > 1 such that $\Phi _{n}(b)$  is prime is about $\lambda \cdot \varphi (n)$ , where $\lambda$  is Euler–Mascheroni constant, and $\varphi$  is Euler's totient function). See also for the list of the smallest primes of the form $\Phi _{n}(b)$  with n > 2 and b > 1, and, more generally, , for the smallest positive integers of this form.

Proofs
• Values of $\Phi _{n}(1).$  If $n=p^{k+1}$  is a prime power, then
$\Phi _{n}(x)=1+x^{p^{k}}+x^{p^{2k}}+\cdots +x^{(p-1)p^{k}}\qquad {\text{and}}\qquad \Phi _{n}(1)=p.$
If n is not a prime power, let $P(x)=1+x+\cdots +x^{n-1},$  we have $P(1)=n,$  and P is the product of the $\Phi _{k}(x)$  for k dividing n and different of 1. If p is a prime divisor of multiplicity m in n, then $\Phi _{p}(x),\Phi _{p^{2}}(x),\cdots ,\Phi _{p^{m}}(x)$  divide P(x), and their values at 1 are m factors equal to p of $n=P(1).$  As m is the multiplicity of p in n, p cannot divide the value at 1 of the other factors of $P(x).$  Thus there is no prime that divides $\Phi _{n}(1).$
• If n is the multiplicative order of b modulo p, then $p\mid \Phi _{n}(b).$  By definition, $p\mid b^{n}-1.$  If $p\nmid \Phi _{n}(b),$  then p would divide another factor $\Phi _{k}(b)$  of $b^{n}-1,$  and would thus divide $b^{k}-1,$  showing that, if there would be the case, n would not be the multiplicative order of b modulo p.
• The other prime divisors of $\Phi _{n}(b)$  are divisors of n. Let p be a prime divisor of $\Phi _{n}(b)$  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 $\Phi _{n}(b)$  and $\Phi _{k}(b).$  The resultant of $\Phi _{n}(x)$  and $\Phi _{k}(x)$  may be written $P\Phi _{k}+Q\Phi _{n},$  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 $n^{n}$  of $x^{n}-1.$  Thus p divides n.
• g and h are coprime. In other words, if p is a prime common divisor of n and $\Phi _{n}(b),$  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.
• g is square-free. In other words, if p is a prime common divisor of n and $\Phi _{n}(b),$  then $p^{2}$  does not divide $\Phi _{n}(b).$  Let n = pm. It suffices to prove that $p^{2}$  does not divides S(b) for some polynomial S(x), which is a multiple of $\Phi _{n}(x).$  We take
$S(x)={\frac {x^{n}-1}{x^{m}-1}}=1+x^{m}+x^{2m}+\cdots +x^{(p-1)m}.$
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,
$S(b)={\frac {(1+c)^{p}-1}{c}}=p+{\binom {p}{2}}c+\cdots +{\binom {p}{p}}c^{p-1}.$
As p is prime and greater than 2, all the terms but the first one are multiples of $p^{2}.$  This proves that $p^{2}\nmid \Phi _{n}(b).$

## Applications

Using $\Phi _{n}$ , one can give an elementary proof for the infinitude of primes congruent to 1 modulo n, which is a special case of Dirichlet's theorem on arithmetic progressions.

Proof

Suppose $p_{1},p_{2},\ldots ,p_{k}$  are a finite list of primes congruent to $1$  modulo $n.$  Let $N=np_{1}p_{2}\cdots p_{k}$  and consider $\Phi _{n}(N)$ . Let $q$  be a prime factor of $\Phi _{n}(N)$  (to see that $\Phi _{n}(N)\neq \pm 1$  decompose it into linear factors and note that 1 is the closest root of unity to $N$ ). Since $\Phi _{n}(x)\equiv \pm 1{\pmod {x}},$  we know that $q$  is a new prime not in the list. We will show that $q\equiv 1{\pmod {n}}.$

Let $m$  be the order of $N$  modulo $q.$  Since $\Phi _{n}(N)\mid N^{n}-1$  we have $N^{n}-1\equiv 0{\pmod {q}}$ . Thus $m\mid n$ . We will show that $m=n$ .

Assume for contradiction that $m . Since

$N^{m}-1\equiv \prod _{d\mid m}\Phi _{d}(N)\equiv 0{\pmod {q}}$

we have

$\Phi _{d}(N)\equiv 0{\pmod {q}},$

for some $d . Then $N$  is a double root of

$\prod _{d\mid n}\Phi _{d}(x)\equiv x^{n}-1{\pmod {q}}.$

Thus $N$  must be a root of the derivative so

$\left.{\frac {d(x^{n}-1)}{dx}}\right|_{N}\equiv nN^{n-1}\equiv 0{\pmod {q}}.$

But $q\nmid N$  and therefore $q\nmid n.$  This is a contradiction so $m=n$ . The order of $N{\pmod {q}},$  which is $n$ , must divide $q-1$ . Thus $q\equiv 1{\pmod {n}}.$