# Cryptography/Mathematical Background: Wikis

Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.

# Wikibooks

Up to date as of January 23, 2010

## Introduction

Modern public-key (asymmetric) cryptography is based upon a branch of mathematics known as number theory, which is concerned solely with the solution of equations that yield only integer results. These type of equations are known as diophantine equations, named after the Greek mathematician Diophantos of Alexandria (ca. 200 CE) from his book Arithmetica that addresses problems requiring such integral solutions.

One of the oldest diophantine problems is known as the Pythagorean problem, which gives the length of one side of a right triangle when supplied with the lengths of the other two side, according to the equation

$a^2 + b^2 = c^2 \$

where $c \$ is the length of the hypotenuse. While two sides may be known to be integral values, the resultant third side may well be irrational. The solution to the Pythagorean problem is not beyond the scope, but is beyond the purpose of this chapter. Therefore, example integral solutions (known as Pythagorean triplets) will simply be presented here. It is left as an exercise for the reader to find additional solutions, either by brute-force or derivation.

Pythagorean Triplets
$a \$ $b \$ $c \$
3 4 5
5 12 13
7 24 25
8 15 17

## Prime Numbers

### Description

Asymmetric encryption methods rely heavily on the use of prime numbers, usually exceedingly long primes, for their operation. By definition, prime numbers are divisible only by themselves and 1. In other words, letting the symbol | denote divisibility (i.e. - a | b means "b divides into a"), a prime number strictly adheres to the following mathematical definition

$p \$ | $b \$ Where $b = 1 \$ or $p \$ only

The Fundamental Theorem of Arithmetic states that all numbers can be decomposed into a unique prime factorization. Any number greater than 1 is considered either prime or composite. A composite number is composed of more than one prime factor

$c \$ | $b \$ where ultimately $b = p^{e_0}_{0} p^{e_1}_{1} \cdot \cdot \cdot p^{e_n}_{n} \$

in which $p_n \$ is a unique prime number and $e_n \$ is the exponent.

#### Numerical Examples

543,312 = 24 $\cdot$ 32 $\cdot$ 50 $\cdot$ 73 $\cdot$ 111
553,696 = 25 $\cdot$ 30 $\cdot$ 50 $\cdot$ 70 $\cdot$ 113 $\cdot$ 131


As can be seen, according to this systematic decomposition, each factorization is unique.

In order to deterministically verify whether an integer $a \$ is prime or composite, only the primes $p \le \sqrt c \$ need be examined. This type of systematic, thorough examination is known as a brute-force approach. Primes and composites are noteworthy in the study of cryptography since, in general, a public key is a composite number which is the product of two or more primes. One (or more) of these primes may constitute the private key.

There are several types and categories of of prime numbers, three of which are of importance to cryptography and will be discussed here briefly.

### Fermat Primes

Fermat primes take the following form

$F_n = 2^{2^n} + 1 \$

Note that not all Fermat "primes" are, in fact, prime. The [1] Wolfram Alpha engine reports Fermat Primes, an example input request being "4th Fermat Prime".

#### Numerical Examples

$F_0 = 2^{2^0} + 1= 3 \$
$F_1 = 2^{2^1} + 1= 5 \$
$F_2 = 2^{2^2} + 1= 17 \$
$F_3 = 2^{2^3} + 1= 257 \$
$F_4 = 2^{2^4} + 1= 65,537 \$
$F_5 = 2^{2^5} + 1= 4,294,967,297 \$


The Fermat primes $F_0-F_4 \$ are indeed prime numbers. However, the absolute "primeness" of Fermat primes was disproven by Euler when he showed $F_5 = 641 \cdot 6,700,297$ demonstrating that this Fermat prime was in fact, composite.

### Mersenne Primes

Mersenne primes - another type of formulaic prime generation - follow the form

$M_p = 2^p - 1 \$

where $p \$ is a prime number. The [2] Wolfram Alpha engine reports Mersenne Primes, an example input request being "4th Mersenne Prime".

#### Numerical Examples

The first five Mersenne primes are as follows

$M_2 = 2^2 - 1 = 3 \$
$M_3 = 2^3 - 1 = 7 \$
$M_5 = 2^5 - 1 = 31 \$
$M_7 = 2^7 - 1 = 127 \$
$M_11 = 2^11 - 1 = 2,047 \$


It is easily shown that $M_11 = 23 \cdot 89$, meaning that all Mersenne primes are not in fact prime, as was the case with Fermat primes.

### Coprimes (Relatively Prime Numbers)

Two numbers are said to be coprime if the largest integer that divides evenly into both of them is 1. Mathematically, this is written

$\gcd(a,b) = 1 \$

where $\gcd \$ is the greatest common divisor. Two rules can be derived from the above definition

If $ab \$ | $c \$ and $\gcd(b,c) = 1 \$, then $a \$ | $c \$
If $ab = c^2 \$ with $\gcd(a,b) = 1 \$, then both $a \$ and $b \$ are squares, i.e. - $a = a^2_{0} \$, $b = b^2_{0} \$

### The Prime Number Theorem

The Prime Number Theorem estimates the probability that any integer, chosen randomly will be prime. The estimate is given below, with $\pi (x) \$ defined as the number of primes $\le x \$

$\pi (x) \approx \frac {x}{\ln x} \$

$\pi (x) \$ is asymptotic to $\frac {x}{\ln x} \$, that is to say $\quad\lim_{x\to \infty} \frac {\pi (x)}{\ln x} = 1 \$. What this means is that generally, a randomly chosen number is prime with the approximate probabilty $\tfrac {1}{x} \$.

## The Euclidean Algorithm

### Introduction

The Euclidean Algorithm is used to discover the greatest common divisor of two integers. In cryptography, it is most often used to determine if two integers are coprime, i.e. - $\gcd (a,b) = 1 \$.

In order to find $\gcd (a,b) \$ where $a > b \$ efficiently when working with very large numbers, as with cryptosystems, a method exists to do so. The Euclidean algorithm operates as follows - First, divide $a \$ by $b \$, writing the quotient $q_1 \$, and the remainder $r_1 \$. Note this can be written in equation form as $a = q_1b + r_1 \$. Next perform the same operation using $b \$ in $a \$'s place: $b = q_2r_1 + r_2 \$. Continue with this pattern until the final remainder is zero. Numerical examples and a formal algorithm follow which should make this inherent pattern clear.

### Mathematical Description

$a = q_1b + r_1 \$
$b = q_2r_1 + r_2 \$
$r_1 = q_3r_2 + r_3 \$
$r_2 = q_4r_3 + r_4 \$
$\cdot \$
$\cdot \$
$\cdot \$
$r_{n-2} = q_nr_{n-1} + r_n \$


When $r_n = 0 \$, stop with $\gcd (a,b) = r_{n-1} \$.

### Numerical Examples

Example 1 - To find gcd(17,043,12,660)

17,043 = 1 $\cdot$ 12,660 + 4383
12,660 = 2 $\cdot$ 4,383 + 3894
4,383 = 1 $\cdot$ 3,894 + 489
3,894 = 7 $\cdot$ 489 + 471
489 = 1 $\cdot$ 471 + 18
471 = 26 $\cdot$ 18 + 3
18 = 6 $\cdot$ 3 + 0


gcd (17,043,12,660) = 3 \ [/itex]

Example 2 - To find gcd(2,008,1,963)

2,008 = 1 $\cdot$ 1,963 + 45
1,963 = 43 $\cdot$ 45 + 28
45 = 1 $\cdot$ 28 + 17
28 = 1 $\cdot$ 17 + 11
17 = 1 $\cdot$ 11 + 6
11 = 1 $\cdot$ 6 + 5
6 = 1 $\cdot$ 5 + 1
5 = 5 $\cdot$ 1 + 0


gcd (2,008,1963) = 1 Note: the two number are coprime.

### Algorithmic Representation

Euclidean Algorithm(a,b)
Input:     Two integers a and b such that a > b
Output:    An integer r = gcd(a,b)
1.   Set a0 = a, r1 = r
2.   r = a0 mod r1
3.   While(r1 mod r $\ne$ 0) do:
4.      a0 = r1
5.      r1 = r
6.      r = a0 mod r1
7.   Output r and halt


## The Extended Euclidean Algorithm

In order to solve the type of equations represented by Bézout's identity, as shown below

$au + bv = \gcd (a,b) \$

where $a \$, $b \$, $u \$, and $v \$ are integers, it is often useful to use the extended Euclidean algorithm. Equations of the form above occur in public key encryption algorithms such as RSA (Rivest-Shamir-Adleman) in the form $ed + w(p-1)(q-1) = 1 \$ where $\gcd (e,(p-1)(q-1)) = 1 \$. There are two methods in which to implement the extended Euclidean algorithm; the iterative method and the recursive method.

As an example, we shall solve an RSA key generation problem with e = 216 + 1, p = 3,217, q = 1,279. Thus, 62,537d + 51,456w = 1.

### Methods

#### The Iterative Method

This method computes expressions of the form ri = axi + byi for the remainder in each step i of the Euclidean algorithm. Each modulus can be written in terms of the previous two remainders and their whole quotient as follows:

$r_i = r_{i-2} - \left \lfloor \frac{r_{i-2}}{r_{i-1}} \right \rfloor \cdot r_{i-1}$

By substitution, this gives:

$r_i = (ax_{i-2} + by_{i-2}) - \left \lfloor \frac{r_{i-2}}{r_{i-1}} \right \rfloor \cdot (ax_{i-1} + by_{i-1})$
$r_i = a(x_{i-2} - \left \lfloor \frac{r_{i-2}}{r_{i-1}} \right \rfloor \cdot x_{i-1}) + b(y_{i-2} - \left \lfloor \frac{r_{i-2}}{r_{i-1}} \right \rfloor \cdot y_{i-1})$

The first two values are the initial arguments to the algorithm:

$r_1 = a = a(1) + b(0) \$
$r_2 = b = a(0) + b(1) \$

The expression for the last non-zero remainder gives the desired results since this method computes every remainder in terms of a and b, as desired.

##### Example
Step Quotient Remainder Substitute Combine terms
1 4,110,048 = a 4,110,048 = 1a + 0b
2 65,537 = b 65,537 = 0a + 1b
3 62 46,754 = 4,110,048 - 65,537 $\cdot$ 62 46,754 = (1a + 0b) - (0a + 1b) $\cdot$ 62 46,754 = 1a - 62b
4 1 18,783 = 65,537 - 46,754 $\cdot$ 1 18,783 = (0a + 1b) - (1a - 62b) $\cdot$ 1 18,783 = -1a + 63b
5 2 9,188 = 46,754 - 18,783 $\cdot$ 2 9,188 = (1a - 62b) - (-1a + 62b) $\cdot$ 2 9,188 = 3a - 188b
6 2 407 = 18,783 - 9,188 $\cdot$ 2 407 = (-1a + 63b) - (3a - 188b) $\cdot$ 2 407 = -7a + 439b
7 22 234 = 9,188 - 407 $\cdot$ 22 234 = (3a - 188b) - (-7a + 439b) $\cdot$ 22 234 = 157a - 9,846b
8 1 173 = 407 - 234 $\cdot$ 1 173 = (-7a + 439b) - (157a - 9,846b) $\cdot$ 1 173 = -164a + 10,285b
9 1 61 = 234 - 173 $\cdot$ 1 61 = (157a - 9,846b) - (-164a + 10,285b) $\cdot$ 1 61 = 321a + 20,131b
10 2 51 = 173 - 61 $\cdot$ 2 51 = (-164a + 10,285b) - (321a +20,131b) $\cdot$ 2 51 = -806a + 50,547b
11 1 10 = 61 - 51 $\cdot$ 1 61 = (321a +20,131b) - (-806a + 50,547b) $\cdot$ 1 10 = 1,127a - 70,678b
12 5 1 = 51 -10 $\cdot$ 5 1 = (-806a + 50,547b) - (1,127a - 70,678b) $\cdot$ 5 1 = -6,441a + 403,937b
13 10 0 End of algorithm

Putting the equation in its original form $ed + w(p - 1)(q - 1) = 1 \$ yields $(65,537)(403,937) + (-6,441)(3,217 - 1)(1,279 - 1) = 1 \$, it is shown that $d = 403,937 \$ and $w = -6,441 \$. During the process of key generation for RSA encryption, the value for w is discarded, and d is retained as the value of the private key In this case

d = 0x629e1 = 01100010100111100001


#### The Recursive Method

This is a direct method for solving Diophantine equations of the form $au + bv = \gcd (a,b) \$. Using this method, the dividend and the divisor are reduced over a series of steps. At the last step, a trivial value is substituted into the equation, and is then worked backward until the solution is obtained.

##### Example

Using the previous RSA vales of $(p - 1)(p - 1) = 4,110,048 \$ and $e = 2^{16} + 1 = 65,537 \$

Euclidean Expansion Collect Terms Substitute Retrograde Substitution Solve For dx
4,110,048 w0 + 65,537d0 = 1
(62 $\cdot$ 65,537 + 46,754) w0 + 65,537d0 = 1
65,537 (62w0 + d0) + 46,754w0 = 1 w1 = 62w0 + d0 4,595 = (62)(-6441) + d0 d0 = 403,937
65,537 w1 + 46,754d1 = 1 d1 = w0 w1 = -6,441
(1 $\cdot$ 46,754 + 18,783) w1 + 46,754d1 = 1
46,754 (w1 + d1) + 18,783w1 = 1 w2 = w1 + d1 -1,846 = 4,595 + d1 d1 = -6,441
46,754 w2 + 18,783d2 = 1 d2 = w1
(2 $\cdot$ 18,783 + 9,188) w2 + 18,783d2 = 1
18,783 (2w2 + d2) + 9,188w2 = 1 w3 = 2w2 + d2 903 = (2)(-1,846) + d2 d2 = 4,595
18,783 w3 + 9,188d3 = 1 d3 = w2
(2 $\cdot$ 9,188 + 407) w3 + 9,188d3 = 1
9,188 (2w3 + d3) + 407w3 = 1 w4 = 2w3 + d3 -40 = (2)(903) + d3 d3 = -1846
9,188 w4 + 407d4 = 1 d4 = w3
(22 $\cdot$ 407 + 234) w4 + 407d4 = 1
407 (22w4 + d4) + 234w4 = 1 w5 = 22w4 +d4 23 = (22)(-40) + d4 d4 = 903
407 w5 + 234d5 = 1 d5 = w4
(1 $\cdot$ 234 + 173) w5 + 234d5 = 1
234 (w5 + d5) + 173w5 = 1 w6 = w5 +d5 -17 = 23 + d5 d5 = -40
234 w6 + 173d6 = 1 d6 = w5
(1 $\cdot$ 173 + 61) w6 + 173d6 = 1
173 (w6 + d6) + 61w6 = 1 w7 = w6 +d6 6 = -17 + d6 d6 = 23
173 w7 + 61d7 = 1 d7 = w6
(2 $\cdot$ 61 + 51) w7 + 61d7 = 1
61 (2w7 + d7) + 51w7 = 1 w8 = 2w7 +d7 -5 = (2)(6) + d7 d7 = -17
61 w8 + 51d8 = 1 d8 = w7
(1 $\cdot$ 51 + 10) w8 + 51d8 = 1
51 (w8 + d8) + 10w8 = 1 w9 = w8 +d8 1 = -5 + d8 d8 = 6
51 w9 + 10d9 = 1 d9 = w8
(5 $\cdot$ 10 + 1) w9 + 10d9 = 1
10 (5w9 + d9) + 1w9 = 1 w10 = 5w9 +d9 0 = (5)(1) + d9 d9 = -5
10 w10 + 1d10 = 1 d10 = w9
(1 $\cdot$ 10 + 0) w10 + 1d10 = 1
1 (10w10 + d10) + 0w10 = 1 w11 = 10w10 +d10 1 = (10)(0) + d10 d10 = 1
1 w11 + 0d11 = 1 d11 = w10 w11 = 1, d11 = 0

## Euler's Totient Function

Significant in cryptography, the totient function (sometimes known as the phi function) is defined as the number of nonnegative integers $a \$ less than $n \$ that are coprime to $n \$. Mathematically, this is represented as

$\phi (n) = \left | \bigg\{ 0 \le a \le n | \gcd (a, n) = 1 \bigg\} \right |$

Which immediately suggests that for any prime $p \$

$\phi (p) = p - 1 \$

The totient function for any exponentiated prime is calculated as follows

$\phi (p^\alpha) \$
$= p^\alpha - p^{\alpha - 1} \$
$= p^\alpha \left ( 1 - \tfrac{1}{p} \right ) \$

The Euler totient function is also multiplicative

$\phi (ab) = \phi (a) \phi (b) \$

where $\gcd (a,b) = 1 \$

## Finite Fields and Generators

A field is simply a set $\mathbb{F}$ which contains numerical elements that are subject to the familiar addition and multiplication operations. Several different types of fields exist; for example, $\mathbb{R}$, the field of real numbers, and $\mathbb{Q}$, the field of rational numbers, or $\mathbb{C}$, the field of complex numbers. A generic field is usually denoted $\mathbb{F}$.

### Finite Fields

Cryptography utilizes primarily finite fields, nearly exclusively composed of integers. The most notable exception to this are the Gaussian numbers of the form $a + bi \$ which are complex numbers with integer real and imaginary parts. Finite fields are defined as follows

$\left( \mathbb{Z} / n \mathbb{Z} \right) = \mathbb{Z}_n \$ The set of integers modulo $n \$
$\left( \mathbb{Z} / p \mathbb{Z} \right) = \mathbb{Z}_p \$ The set of integers modulo a prime $p \$

Since cryptography is concerned with the solution of diophantine equations, the finite fields utilized are primarily integer based, and are denoted by the symbol for the field of integers, $\mathbb{Z}$.

A finite field $\mathbb{F}_n \$ contains exactly $n \$ elements, of which there are $n - 1 \$ nonzero elements. An extension of $\mathbb{Z}_n \$ is the multiplicative group of $\mathbb{Z}_n \$, written $\left( \mathbb{Z} / n \mathbb{Z} \right)^* = \mathbb{Z}^*_n \$, and consisting of the following elements

$a \in \mathbb{Z}^*_n \$ such that $\gcd (a,n) = 1 \$

in other words, $\mathbb{Z}^*_n \$ contains the elements coprime to $n \$

Finite fields form an abelian group with respect to multiplication, defined by the following properties

$\centerdot$ The product of two nonzero elements is nonzero $\left( ab = c | c \ne 0 \right) \$
$\centerdot$ The associative law holds $\left( \left( ab \right) c = a \left( bc \right) \right) \$
$\centerdot$ The commutative law holds $\left( ab = ba \right) \$
$\centerdot$ There is an identity element $\left( I \cdot a = a \right) \$
$\centerdot$ Any nonzero element has an inverse $\left( a \cdot a^{-1} = 1 \right) \$


A subscript following the symbol for the field represents the set of integers modulo $n \$, and these integers run from $0 \$ to $n - 1 \$ as represented by the example below

$\mathbb{Z}_{12} = \{ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 \} \$

The multiplicative order of $\mathbb{Z}_n$ is represented $\mathbb{Z}^*_n$ and consists of all elements $a \in \mathbb{Z}_n$ such that $\gcd (a,n) = 1 \$. An example for $\mathbb{Z}_{12}$ is given below

$\mathbb{Z}^*_{12} = \{ 1, 5, 7, 11 \} \$

If $p \$ is prime, the set $\mathbb{Z}^*_p$ consists of all integers $a \$ such that $1 \le a \le p \$. For example

Composite n Prime p
$\mathbb{Z}_9 = \{ 0, 1, 2, 3, 4, 5, 6, 7, 8 \}$ $\mathbb{Z}_{11} = \{ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \}$
$\mathbb{Z}^*_9 = \{ 1, 2, 4, 5, 7, 8 \}$ $\mathbb{Z}^*_{11} = \{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \}$

### Generators

Every finite field has a generator. A generator $g \$ is capable of generating all of the elements in the set $\mathbb{Z}_n$ by exponentiating the generator $g\,\bmod\,n \$. Assuming $g \$ is a generator of $\mathbb{Z}^*_n$, then $\mathbb{Z}^*_n$ contains the elements $g^i\,\bmod\,n \$ for the range $0 \le i \le \phi (n) - 1$. If $\mathbb{Z}^*_n$ has a generator, then $\mathbb{Z}_n$ is said to be cyclic.

The total number of generators is given by

$\phi \left( \phi \left( n \right) \right)$

#### Examples

For $n = p = 13 \$ (Prime)

$\mathbb{Z}_{13} = \{ 0,1,2,3,4,5,6,7,8,9,10,11,12 \}$
$\mathbb{Z}^*_{13} = \{ 1,2,3,4,5,6,7,8,9,10,11,12 \}$

Total number of generators $\phi \left( \phi \left( 13 \right) \right) = \phi \left( 12 \right) = 4$ generators

Let $g = 2 \$, then $g = \{ 2,4,8,3,6,12,11,9,5,10,7,1 \} \$, $g = 2 \$ is a generator

Since $2 \$ is a generator, check if $\gcd (i, p - 1) = 1 \$
$2^2 = 4 \$, and $i = 2 \$, $\gcd \left( 2, 12 \right) = 2 \ne 1 \$, therefore, $4 \$ is not a generator
$2^3 = 8 \$, and $i = 3 \$, $\gcd \left( 3, 12 \right) = 3 \ne 1 \$, therefore, $4 \$ is not a generator

Let $g = 6 \$, then $g = \{ 6,10,8,9,2,12,7,3,5,4,11,1 \} \$, $g = 6 \$ is a generator
Let $g = 7 \$, then $g = \{ 7,10,5,9,11,12,6,3,8,4,2,1 \} \$, $g = 7 \$ is a generator
Let $g = 11 \$, then $g = \{ 11,4,5,3,7,12,2,9,8,10,6,1 \} \$, $g = 11 \$ is a generator

There are a total of $4 \$ generators, $\left( 2,6,7,11 \right)$ as predicted by the formula $\phi \left( \phi \left( n \right) \right).$

For $n = 10 \$ (Composite)

$\mathbb{Z}_9 = \{ 0,1,2,3,4,5,6,7,8,9 \} \$
$\mathbb{Z}^*_9 = \{ 1,3,7,9 \} \$

Total number of generators $\phi \left( \phi \left( 10 \right) \right) = \phi \left( 4 \right) = 2 \$ generators

Let $g = 3 \$, then $g = \{ 3,9,7,1,3,9,7,1,3 \} \$, $g = 3 \$ is a generator
Let $g = 7 \$, then $g = \{ 7,9,3,1,7,9,3,1,7 \} \$, $g = 7 \$ is a generator

There are a total of $2 \$ generators $\left( 3,7, \right) \$ as predicted by the formula $\phi \left( \phi \left( n \right) \right).$


## Congruences

### Description

Number theory contains an algebraic system of its own called the theory of congruences. The mathematical notion of congruences was introduced by Karl Friedrich Gauss in Disquisitiones (1801).

### Definition

If $a \$ and $b \$ are two integers, and their difference is evenly divisible by $m \$, this can be written with the notation

$\left( a - b \right) | m \$

This is expressed by the notation for a congruence

$a \equiv b\,\bmod\,m$

where the divisor $m \$ is called the modulus of congruence. $a \equiv b\,\bmod\,m$ can equivalently be written as

$a - b = mk \$

where $k \$ is an integer.

Note in the examples that for all cases in which $a \equiv 0\,\bmod\,m$, it is shown that $a | m \$. with this in mind, note that

$a \equiv 0\,\bmod\,2$ Represents that $a \$ is an even number.

$a \equiv 1\,\bmod\,2$ Represents that $a \$ is an odd number.

#### Examples

$a \equiv b\,\bmod\,m$ $a - b = mk \$
$14 \equiv 5\,\bmod\,3$ $14 - 5 = 3 \cdot 3 \$
$-13 \equiv 7\,\bmod\,4$ $(-13) - 7 = 4 \cdot (-5) \$
$90 \equiv 0\,\bmod\,18$ $90 - 0 = 18 \cdot 5 \$

### Properties of Congruences

All congruences (with fixed $m \$) have the following properties in common

$a \equiv a\,\bmod\,m$
$a \equiv b\,\bmod\,m$ if and only if $b \equiv a\,\bmod\,m$
If $a \equiv a\,\bmod\,m$ and $b \equiv c\,\bmod\,m$ then $a \equiv c\,\bmod\,m$
$a \equiv a\,\bmod\,1$ implies that $a = b \$
Given $a \equiv a\,\bmod\,m$ there exists a unique $b \$ such that $0 \le b \le m - 1 \$

These properties represent an equivalence class, meaning that any integer is congruent modulo $m \$ to one specific integer in the finite field $\mathbb{Z}_m$.

### Congruences as Remainders

If the modulus of an integer $m > 2 \$, then for every integer $a \$

$a = mk + r \, \left( r \in \mathbb{Z}_m \right)$

which can be understood to mean $r \$ is the remainder of $a \$ divided by $m \$, or as a congruence

$a \equiv r\,\bmod\,m$

Two numbers that are incongruent modulo $m \$ must have different remainders. Therefore, it can be seen that any congruence $a \equiv b\,\bmod\,m$ holds if and only if $a \$ and $b \$ are integers which have the same remainder when divided by $m \$.

#### Example

$10 \equiv 3\,\bmod\,7$ is equivalent to
$10 = \left( 7 \cdot 1 \right) + 3 \$ implies
$3 \$ is the remainder of $10 \$ divided by $7 \$


### The Algebra of Congruences

Suppose for this section we have two congruences, $a \equiv a\,\bmod\,m$ and $c \equiv d\,\bmod\,m$. These congruences can be added or subtracted in the following manner

$a + c \equiv b + d\,\bmod\,m$
$a - c \equiv b - d\,\bmod\,m$

If these two congruences are multiplied together, the following congruence is obtained

$ac \equiv bd\,\bmod\,m$

or the special case where $c = d \$

$ac \equiv bc\,\bmod\,m$

Note: The above does not mean that there exists a division operation for congruences. The only possibility for simplifying the above is if and only if $c \$ and $m \$ are coprime. Mathematically, this is represented as

$ac \equiv bc\,\bmod\,m$ implies that $a \equiv b\,\bmod\,m$ if and only if $\gcd \left( c,m \right) = 1$

The set of equivalence classes defined above form a commutative ring, meaning the residue classes can be added, subtracted and multiplied, and that the operations are associative, commutative and have additive inverses.

### Reducing Modulo m

Often, it is necessary to perform an operation on a congruence $a \equiv b\,\bmod\,m$ where $b > m \$, when what is desired is a new integer $d \$ such that $0 \le d \le m - 1 \$ with the resultant $d \$ being the least nonnegative residue modulo m of the congruence. Reducing a congruence modulo $m \$ is based on the properties of congruences and is often required during exponentiation of a congruence.

#### Algorithm

Input: Integers $b \$ and $m \$ from $a \equiv b\,\bmod\,m$ with $b > m \$
Output: Integer $d \$ such that $0 \le d \le m - 1 \$

1. Let $q = \left \lfloor \tfrac{b}{m} \right \rfloor$
2.     $c = qm \$
3.     $d = b - c \$
4. Output $d \$


#### Example

Given $289 \equiv 49\,\bmod\,5$

$9 = \left \lfloor \tfrac{49}{5} \right \rfloor$
$45 = 9 \cdot 5 \$
$4 = 49 - 45 \$

∴ $289 \equiv 49 \equiv 4\,\bmod\,5$


Note that $4 \$ is the least nonnegative residue modulo $5 \$

### Exponentiation

Assume you begin with $a \equiv b\,\bmod\,m$. Upon multiplying this congruence by itself the result is $a^2 \equiv b^2\,\bmod\,m$. Generalizing this result and assuming $n \$ is a positive integer

$a^n \equiv b^n\,\bmod\,m$

#### Example

$9 \equiv 4\,\bmod\,13$
$81 \equiv 16\,\bmod\,13$
$729 \equiv 64\,\bmod\,13$

This simplifies to

$81 \equiv 16\,\bmod\,13$ implies $16 \equiv 3\,\bmod\,13$
$729 \equiv 64\,\bmod\,13$ implies $256 \equiv 9\,\bmod\,13$


### Repeated Squaring Method

Sometimes it is useful to know the least nonnegative residue modulo $m \$ of a number which has been exponentiated as $a^2 \equiv\,\bmod\,m$. In order to find this number, we may use the repeated squaring method which works as follows:

1. Begin with $a \equiv\,\bmod\,m$
2. Square $a \$ and $b \$ so that $a^2 \equiv b^2\,\bmod\,m$
3. Reduce $b \$ modulo $m \$ to obtain $a^ \equiv b_1\,\bmod\,m$
4. Continue with steps 2 and 3 until $a^{2^n} \equiv b_n\,\bmod\,m$ is obtained.
Note that $n \$ is the integer where $2^{n+1} \$ would be just larger than the exponent desired
5. Add the successive exponents until you arrive at the desired exponent
6. Multiply all $b_i \$'s associated with the $a \$'s of the selected powers
7. Reduce the resulting $b\,\bmod\,m$ for the desired result


#### Example

To find $6^{149}\bmod\,11$:

$6 \equiv 6\,\bmod\,11$
$6^2 = 36 \equiv 3\,\bmod\,11$
$6^4 \equiv 9\,\bmod\,11$
$6^8 \equiv 81 \equiv 4\,\bmod\,11$
$6^{16} \equiv 16 \equiv 5\,\bmod\,11$
$6^{32} \equiv 25 \equiv 3\,\bmod\,11$
$6^{64} \equiv 9\,\bmod\,11$
$6^{128} \equiv 81 \equiv 4\,\bmod\,11$

$128 + 16 + 4 + 1 \$

Multiplying least nonnegative residues associated with these exponents:

$4 \cdot 5 \cdot 9 \cdot 6 = 1080 \$
$1080\,\bmod\,11 = 2$

Therefore:

$6^{149} \equiv 2\,\bmod\,11$


### Inverse of a Congruence

#### Description

While finding the correct symmetric or asymmetric keys is required to encrypt a plaintext message, calculating the inverse of these keys is essential to successfully decrypt the resultant ciphertext. This can be seen in cryptosystems Ranging from a simple affine transformation

$C \equiv aP + b\,\bmod\,N$

Where

$P \equiv a^{-1}C + b^{-1}\,\bmod\,N$

To RSA public key encryption, where one of the deciphering (private) keys is

$d_A = e^{-1}_A\,\bmod\,\phi \left( n_A \right)$

#### Definition

For the elements $a \in \mathbb Z_m$ where $\gcd \left( a, m \right) = 1$, there exists $b \in \mathbb Z_m$ such that $ab \equiv 1\,\bmod\,m$. Thus, $b \$ is said to be the inverse of $a \$, denoted $a^{-n}\,\bmod\,m$ where $n \$ is the $n^{th} \$ power of the integer $b \$ for which $ab \equiv 1\,\bmod\,m$.

##### Example
Find $633^{-1}\,\bmod\,2801$

This is equivalent to saying $633b \equiv 1\,\bmod\,2801$

First use the Euclidean algorithm to verify $\gcd \left( 633, 2801 \right) = 1 \$.
Next use the Extended Euclidean algorithm to discover the value of $b \$.
In this case, the value is $177 \$.

Therefore, $633^{-1}\,\bmod\,2801 = 177$

It is easily verified that $\left( 633 \right) \left( 177 \right) \equiv 1\,\bmod\,2801$


### Fermat's Little Theorem

#### Definition

Where $p \$ is defined as prime, any integer will satisfy the following relation:

$a^p \equiv a\,\bmod\,p$

#### Example

When $a = 2 \$ and $p = 19 \$

$2^2 \equiv 23\,\bmod\,19$
$2^4 \equiv 529 \equiv 16\,\bmod\,19$
$2^8 \equiv 256 \equiv 9\,\bmod\,19$
$2^{16} \equiv 81 \equiv 5\,\bmod\,19$
$16 + 2 + 1 = 19 \$ implies that $5 \cdot 23 \cdot 2 = 230 \equiv 2\,\bmod\,19$

#### Conditions and Corollaries

An additional condition states that if $a \$ is not divisible by $p \$, the following equation holds

$a^{p-1} \equiv 1\,\bmod\,p$

Fermat's Little Theorem also has a corollary, which states that if $a \$ is not divisible by $p \$ and $n \equiv m\,\bmod\,\left( p - 1 \right)$ then

$a^n \equiv a^m\,\bmod\,p$

#### Euler's Generalization

If $\gcd \left( a, m \right) = 1 \$, then $a^{\phi \left( m \right)} \equiv 1\,\bmod\,m$

### Chinese Remainder Theorem

If one wants to solve a system of congruences with different moduli, it is possible to do so as follows:

$x \equiv a_1\,\bmod\,m_1$
$x \equiv a_2\,\bmod\,m_2$
$\cdots$
$x \equiv a_k\,\bmod\,m_k$

A simultaneous solution $x \$ exists if and only if $\gcd \left( m_i, m_j \right) = 1$ with $\left( i \ne j \right) \$, and any two solutions are congruent to one another modulo $M = m_1m_2 \ldots m_k \$.

The steps for finding the simultaneous solution using the Chinese Remainder theorem are as follows:

1. Compute $M \$
2. Compute $M_i = M / m_i \$ for each of the different $i \$'s
3. Find the inverse $N \$ of $M_i\,\bmod\,m_i$ for each $i \$ using the Extended Euclidean algorithm
4. Multiply out $a_iM_iN_i \$ for each $i \$
5. Sum all $a_iM_iN_i \$
6. Compute $\sum_{i=1}^k a_iM_iN_i\,\bmod\,M$ to obtain the least nonnegative residue

#### Example

Given:

$x \equiv 1\,\bmod\,11$
$x \equiv 2\,\bmod\,7$
$x \equiv 3\,\bmod\,5$
$x \equiv 4\,\bmod\,9$

$M = 3465 \$

$M_{11} = 315 \$
$M_7 = 495 \$
$M_5 = 693 \$
$M_9 = 385 \$

Using the Extended Euclidean algorithm:

$315N \equiv 1\,\bmod\,11\,\,\,N = -3$
$315N \equiv 1\,\bmod\,7\,\,\,N = 3$
$315N \equiv 1\,\bmod\,5\,\,\,N = 2$
$315N \equiv 1\,\bmod\,9\,\,\,N = 4$

$\sum_{i = 1}^4 = \begin{cases} 1 \cdot 315 \cdot \left( -3 \right) = -945 \ 2 \cdot 495 \cdot 3 = 2970 \ 3 \cdot 639 \cdot 2 = 4158 \ 4 \cdot 385 \cdot 4 = 6160 \end{cases}$

$\sum = 12343$

$x = 12343\,\bmod\,3465 = 1948$


If $p \$ is prime and $> 2 \$, examining the nonzero elements of $\mathbb Z_p = \{ 1, 2, \ldots , p - 1 \}$, it is sometimes important to know which of these are squares. If for some $a \in \mathbb Z_p^*$, there exists a square such that $b^2 = a \$. Then all squares for $\mathbb Z_p^*$ can be calculated by $b^2\,\bmod\,p$ where $b = 1, 2, \ldots , \left( p - 1 \right) / 2 \$. $a \in \mathbb Z_n^*$ is a quadratic residue modulo $n \$ if there exists an $x \in \mathbb Z_n^*$ such that $a \equiv x^2\,\bmod\,n$. If no such $x \$ exists, then $a \$ is a quadratic non-residue modulo $n \$. $a \$is a quadratic residue modulo a prime $p \$ if and only if $a^{\tfrac {p - 1}{2}}\,\mod\,p = 1$.

### Example

For the finite field $\mathbb Z_{19}$, to find the squares $\mathbb Z_{19} = \{ 1, 2, \ldots , 9 \},$, proceed as follows:

$\begin{matrix} 1^2 = 1 & 2^2 = 4 & 3^2 = 9 \ 4^2 = 16 & 5^2 = 6 & 6^2 = 2 \ 7^2 = 11 & 8^2 = 7 & 9^2 = 5 \end{matrix}$



The values above are quadratic residues. The remaining (in this example) 9 values are know as quadratic nonresidues. the complete listing is given below.

$p = 19 \$
Quadratic residues: $1, 2, 4, 5, 6, 7, 9, 11, 16 \$
Quadratic nonresidues: $3, 8, 10, 12, 13, 14, 15, 17, 18 \$


### Legendre Symbol

The Legendre symbol denotes whether or not $a \$ is a quadratic residue modulo the prime $p \$ and is only defined for primes $p \$ and integers $a \$. The Legendre of $a \$ with respect to $p \$ is represented by the symbol $L \left( \tfrac{a}{p} \right)$. Note that this does not mean $a \$ divided by $p \$. $L \left( \tfrac{a}{p} \right)$ has one of three values: $0, 1, -1 \$.

$L \left( \tfrac{a}{p} \right) \begin{cases} 0, & \mbox{if }p\mbox{ divides }a\mbox{ evenly} \ 1, & \mbox{if }a\mbox{ is a quadratic residue modulo }p \ -1, & \mbox{if }a\mbox{ is a quadratic nonresidue modulo }p \end{cases}$

### Jacobi Symbol

The Jacobi symbol applies to all odd numbers $n > 3 \$ where $n = p_1^{e_1}p_2^{e_2} \ldots p_m^{e_m} \$, then:

$J \left( \tfrac{a}{n} \right) = L \left( \tfrac{a}{p_1} \right)^{e_1} L \left( \tfrac{a}{p_2} \right)^{e_2} \ldots L \left( \tfrac{a}{p_m} \right)^{e_m}$

If $n \$ is prime, then the Jacobi symbol equals the Legendre symbol (which is the basis for the Solovay-Strassen primality test).