# Root of unity modulo n

In mathematics, namely ring theory, a k-th root of unity modulo n for positive integers k, n ≥ 2, is a solution x to the equation (or congruence) ${\displaystyle x^{k}\equiv 1{\pmod {n}}}$. If k is the smallest such exponent for x, then x is called a primitive k-th root of unity modulo n.[1] See modular arithmetic for notation and terminology.

Do not confuse this with a primitive element modulo n, where the primitive element must generate all units of the residue class ring ${\displaystyle \mathbb {Z} /n\mathbb {Z} }$ by exponentiation. That is, there are always roots and primitive roots of unity modulo n for n ≥ 2, but for some n there is no primitive element modulo n. Being a root or a primitive root modulo n always depends on the exponent k and the modulus n, whereas being a primitive element modulo n only depends on the modulus n — the exponent is automatically ${\displaystyle \varphi (n)}$.

## Roots of unity

### Properties

${\displaystyle (x-1)\cdot \sum _{j=0}^{k-1}x^{j}\equiv x^{k}-1\equiv 0{\pmod {n}}.}$

### Number of k-th roots

For the lack of a widely accepted symbol, we denote the number of k-th roots of unity modulo n by ${\displaystyle f(n,k)}$. It satisfies a number of properties:

## Primitive roots of unity

### Number of primitive k-th roots

For the lack of a widely accepted symbol, we denote the number of primitive k-th roots of unity modulo n by ${\displaystyle g(n,k)}$. It satisfies the following properties:

${\displaystyle f={\mathbf {1}}*g}$, i.e. ${\displaystyle f(n,k)=\sum _{d\mid k}g(n,d)}$
You can compute values of ${\displaystyle g}$ recursively from ${\displaystyle f}$ using this formula, which is equivalent to the Möbius inversion formula.

### Testing whether x is a primitive k-th root of unity modulo n

By fast exponentiation you can check that ${\displaystyle x^{k}\equiv 1{\pmod {n}}}$. If this is true, x is a k-th root of unity modulo n but not necessarily primitive. If it is not a primitive root, then there would be some divisor ℓ of k, with ${\displaystyle x^{\ell }\equiv 1{\pmod {n}}}$. In order to exclude this possibility, one has only to check for a few ℓ's equal k divided by a prime. That is, what needs to be checked is:

${\displaystyle \forall p{\text{ prime dividing}}\ k,\quad x^{k/p}\not \equiv 1{\pmod {n}}.}$

### Finding a primitive k-th root of unity modulo n

Among the primitive k-th roots of unity, the primitive ${\displaystyle \lambda (n)}$-th roots are most frequent. It is thus recommended to try some integers for being a primitive ${\displaystyle \lambda (n)}$-th root, what will succeed quickly. For a primitive ${\displaystyle \lambda (n)}$-th root x, the number ${\displaystyle x^{\lambda (n)/k}}$ is a primitive ${\displaystyle k}$-th root of unity. If k does not divide ${\displaystyle \lambda (n)}$, then there will be no k-th roots of unity, at all.

### Finding multiple primitive k-th roots modulo n

Once you have a primitive k-th root of unity x, every power ${\displaystyle x^{l}}$ is a ${\displaystyle k}$-th root of unity, but not necessarily a primitive one. The power ${\displaystyle x^{l}}$ is a primitive ${\displaystyle k}$-th root of unity if and only if ${\displaystyle k}$ and ${\displaystyle l}$ are coprime. The proof is as follows: If ${\displaystyle x^{l}}$ is not primitive, then there exists a divisor ${\displaystyle m}$ of ${\displaystyle k}$ with ${\displaystyle (x^{l})^{m}\equiv 1{\pmod {n}}}$, and since ${\displaystyle k}$ and ${\displaystyle l}$ are coprime, there exists an inverse ${\displaystyle l^{-1}}$ of ${\displaystyle l}$ modulo ${\displaystyle k}$. This yields ${\displaystyle 1\equiv ((x^{l})^{m})^{l^{-l}}\equiv x^{m}{\pmod {n}}}$, which means that ${\displaystyle x}$ is not a primitive ${\displaystyle k}$-th root of unity because there is the smaller exponent ${\displaystyle m}$.

That is, by exponentiating x one can obtain ${\displaystyle \varphi (k)}$ different primitive k-th roots of unity, but these may not be all such roots. However, finding all of them is not so easy.

### Finding an n with a primitive k-th root of unity modulo n

You may want to know, in what integer residue class rings you have a primitive k-th root of unity. You need it for instance if you want to compute a Discrete Fourier Transform (more precisely a Number theoretic transform) of a ${\displaystyle k}$-dimensional integer vector. In order to perform the inverse transform, you also need to divide by ${\displaystyle k}$, that is, k shall also be a unit modulo ${\displaystyle n}$.

A simple way to find such an n is to check for primitive k-th roots with respect to the moduli in the arithmetic progression ${\displaystyle k+1,2k+1,3k+1,\dots }$. All of these moduli are coprime to k and thus k is a unit. According to Dirichlet's theorem on arithmetic progressions there are infinitely many primes in the progression, and for a prime ${\displaystyle p}$ it holds ${\displaystyle \lambda (p)=p-1}$. Thus if ${\displaystyle mk+1}$ is prime then ${\displaystyle \lambda (mk+1)=mk}$ and thus you have primitive k-th roots of unity. But the test for primes is too strong, you may find other appropriate moduli.

### Finding an n with multiple primitive roots of unity modulo n

If you want to have a modulus ${\displaystyle n}$ such that there are primitive ${\displaystyle k_{1}}$-th, ${\displaystyle k_{2}}$-th, ..., ${\displaystyle k_{m}}$-th roots of unity modulo ${\displaystyle n}$, the following theorem reduces the problem to a simpler one:

For given ${\displaystyle n}$ there are primitive ${\displaystyle k_{1}}$-th, ..., ${\displaystyle k_{m}}$-th roots of unity modulo ${\displaystyle n}$ if and only if there is a primitive ${\displaystyle \mathrm {lcm} (k_{1},...,k_{m})}$-th root of unity modulo n.
Proof

Backward direction: If there is a primitive ${\displaystyle \mathrm {lcm} (k_{1},...,k_{m})}$-th root of unity modulo ${\displaystyle n}$ called ${\displaystyle x}$, then ${\displaystyle x^{\mathrm {lcm} (k_{1},\dots ,k_{m})/k_{l}}}$ is a ${\displaystyle k_{l}}$-th root of unity modulo ${\displaystyle n}$.

Forward direction: If there are primitive ${\displaystyle k_{1}}$-th, ..., ${\displaystyle k_{m}}$-th roots of unity modulo ${\displaystyle n}$, then all exponents ${\displaystyle k_{1},\dots ,k_{m}}$ are divisors of ${\displaystyle \lambda (n)}$. This implies ${\displaystyle \mathrm {lcm} (k_{1},\dots ,k_{m})\mid \lambda (n)}$ and this in turn means there is a primitive ${\displaystyle \mathrm {lcm} (k_{1},...,k_{m})}$-th root of unity modulo ${\displaystyle n}$.