# Modular multiplicative inverse

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In modular arithmetic, the modular multiplicative inverse of an integer a modulo m is an integer x such that

$a\,x \equiv 1 \pmod{m}.$

That is, it is the multiplicative inverse in the ring of integers modulo m, denoted $\mathbb{Z}_m$.

Once defined, x may be noted $a^{-1}$, where the fact that the inversion is m-modular is implicit.

The multiplicative inverse of a modulo m exists if and only if a and m are coprime (i.e., if gcd(a, m) = 1). If the modular multiplicative inverse of a modulo m exists, the operation of division by a modulo m can be defined as multiplying by the inverse, which is in essence the same concept as division in the field of reals.

## Example

Suppose we wish to find modular multiplicative inverse x of 3 modulo 11.

$x \equiv 3^{-1} \pmod{11}$

This is the same as finding x such that

$3x \equiv 1 \pmod{11}$

Working in $\mathbb{Z}_{11}$ we find one value of x that satisfies this congruence is 4 because

$3 (4) = 12 \equiv 1 \pmod{11}$

and there are no other values of x in $\mathbb{Z}_{11}$ that satisfy this congruence. Therefore, the modular multiplicative inverse of 3 modulo 11 is 4.

Once we have found the inverse of 3 in $\mathbb{Z}_{11}$, we can find other values of x in $\mathbb{Z}$ that also satisfy the congruence. They may be found by adding multiples of m = 11 to the found inverse. Generalizing, all possible x for this example can be formed from

$4 + (11 \cdot z ), z \in \mathbb{Z}$

yielding {..., −18, −7, 4, 15, 26, ...}.

## Computation

### Extended Euclidean algorithm

The modular multiplicative inverse of a modulo m can be found with the extended Euclidean algorithm. The algorithm finds solutions to Bézout's identity

$ax + by = \gcd(a, b)\,$

where a and b are given and xy and gcd(ab) are the integers that the algorithm discovers. So, since the modular multiplicative inverse is the solution to

$ax \equiv 1 \pmod{m},$

by the definition of congruence, m | ax − 1, which means that m is a divisor of ax − 1. This, in turn, means that

$ax - 1 = qm.\,$

Rearranging produces

$ax - qm = 1,\,$

with a and m given, x the inverse, and q an integer multiple that will be discarded. This is the exact form of equation that the extended Euclidean algorithm solves—the only difference being that gcd(a, m) = 1 is predetermined instead of discovered. Thus, a needs to be coprime to the modulus, or the inverse won't exist.

This algorithm runs in time O(log(m)2), assuming |a| < m, and is generally more efficient than exponentiation.

### Using Euler's theorem

As an alternative to the extended Euclidean algorithm, Euler's theorem may be used to compute modular inverse:[1]

According to Euler's theorem, if a is coprime to m, that is, gcd(a, m) = 1, then

$a^{\varphi(m)} \equiv 1 \pmod{m}$

where φ(m) is Euler's totient function. This follows from the fact that a belongs to the multiplicative group (Z/mZ)× iff a is coprime to m. Therefore the modular multiplicative inverse can be found directly:

$a^{\varphi(m)-1} \equiv a^{-1} \pmod{m}$

In the special case when m is a prime, the modular inverse is given by the below equation as:

$a^{-1} \equiv a^{m-2} \pmod{m}$

This method is generally slower than the extended Euclidean algorithm, but is sometimes used when an implementation for modular exponentiation is already available. Some disadvantages of this method include:

• The value φ(m) must be known, whose most efficient computation requires m's factorization. Factorization is widely believed to be a computationally hard problem. However, calculating φ(m) is straightforward when the prime factorisation of m is known.
• The relative cost of exponentiation. Though it can be implemented more efficiently using modular exponentiation, when large values of m are involved this is most efficiently computed with the Montgomery reduction method. This algorithm itself requires a modular inverse mod m, which is what was to be calculated in the first place. Without the Montgomery method, we're left with standard binary exponentiation which requires division mod m at every step, a slow operation when m is large. Furthermore, any kind of modular exponentiation is a taxing operation with computational complexity O(log φ(m)) = O(log m).

## Applications

The modular multiplicative inverse has many applications in algorithms, particularly those related to number theory, since many such algorithms rely heavily on the theory of modular arithmetic. As a simple example, consider the exact division problem where you have a list of odd word-sized numbers each divisible by k and you wish to divide them all by k. One solution is as follows:

1. Use the extended Euclidean algorithm to compute k−1, the modular multiplicative inverse of k mod 2w, where w is the number of bits in a word. This inverse will exist since the numbers are odd and the modulus has no odd factors.
2. For each number in the list, multiply it by k−1 and take the least significant word of the result.

On many machines, particularly those without hardware support for division, division is a slower operation than multiplication, so this approach can yield a considerable speedup. The first step is relatively slow but only needs to be done once.