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---
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share: true
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toc: true
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math: true
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categories:
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- Lecture Notes
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- Internet Security
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tags:
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- lecture-note
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- security
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- cryptography
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title: 04. Modular Arithmetic (1)
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date: 2023-09-25
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github_title: 2023-09-25-modular-arithmetic-1
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---
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**Number theory** is a branch of mathematics devoted primarily to the study of the integers. **Modular arithmetic** is heavily used in cryptography.
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## Divisibility
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> **Definition.** Let $a, b, c \in \mathbb{Z}$ such that $a = bc$. Then,
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>
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> 1. $b$ and $c$ are said to **divide** $a$, and are called **factors** of $a$.
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> 2. $a$ is said to be a **multiple** of $b$ and $c$.
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> **Notation.** For $a, b \in \mathbb{Z}$, we write $a \mid b$ if $a$ divides $b$. If not, we write $a \nmid b$.
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These are simple lemmas for checking divisibility.
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> **Lemma.** Let $a, b, c \in \mathbb{Z}$.
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>
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> 1. If $a \mid b$ and $a \mid c$, then $a \mid (b + c)$.
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> 2. If $a \mid b$, then $a \mid bc$.
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> 3. If $a \mid b$ and $b \mid c$, then $a \mid c$.
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## Prime Numbers
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> **Definition.** Integer $n \geq 2$ is **prime** if it is only divisible by $1$ and itself. If it is not prime, then it is **composite**.
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Note that $1$ is neither prime nor composite.
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### Primality Tests
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It is hard to verify if some given number is prime. Many encryption schemes heavily rely on this fact.
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The following is a simple algorithm to check if a given integer is prime.
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```c
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bool naive_prime_test(int n) {
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if (n < 2) {
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return false;
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}
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for (int i = 2; i < sqrt(n); ++i) {
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if (n % i == 0) {
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return false;
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}
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}
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return true;
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}
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```
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However, this algorithm has complexity $\mathcal{O}(\sqrt{n})$, which is slow. We have better algorithms like Fermat's test, Miller-Rabin test, Pollard's rho algorithm... (Not covered in this lecture)
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## Division Algorithm
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> **Theorem.** (Euclidean Division) For $a, b \in \mathbb{Z}$ with $b \neq 0$, there exist unique integers $q, r$ with $0 \leq r < \left\lvert b \right\rvert$ such that $a = bq + r$.
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*Proof.* [By induction](https://en.wikipedia.org/wiki/Euclidean_division#Proof).
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Other proofs use the well-ordering principle.
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## Modulo Operation
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There are two ways to think about 'mod': as a function, and as a congruence.
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### Modulo as a Function
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As a function, $a \bmod b$ return the remainder of $a$ divided by $b$. This operation is commonly denoted `%` in many programming languages.[^1]
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### Modulo as a Congruence
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As a congruence, it means that $a, b$ are in the same *equivalence class*.[^2]
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> **Definition.** For $a, b, n \in \mathbb{Z}$ and $n \neq 0$, $a \equiv b \pmod n$ if and only if $n \mid (a - b)$.
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Properties of modulo operation.
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> **Lemma.** Suppose that $a \equiv b \pmod n$ and $c \equiv d \pmod n$. Then, the following hold.
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>
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> 1. $a + c \equiv (b + d) \pmod n$.
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> 2. $ac \equiv bd \pmod n$.
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> 3. $a^k \equiv b^k \pmod n$.
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> 4. $a \equiv (a \bmod n) \pmod n$.
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*Proof.* Trivial. :)
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The last one is very useful in computing. For example, if $a, b$ are very large integers, using the identity
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$$
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(a + b)^k \equiv ((a + b) \bmod n)^k \pmod n
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$$
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allows us to reduce the size of the numbers before exponentiation.
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## Modular Arithmetic
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For modulus $n$, **modular arithmetic** is operation on $\mathbb{Z}_n$.
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### Residue Classes
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For each positive integer $n$, we can partition $\mathbb{Z}$ into $n$ cells according to whether the remainder is $0, 1, 2, \dots, n - 1$ when the integer is divided by $n$. These cells are the **residue classes modulo $n$ in $\mathbb{Z}$**.
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We write each residue class as follows.
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$$
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\overline{k} = [k] = \left\lbrace m \in \mathbb{Z} : m \bmod n = k\right\rbrace
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$$
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Consider the relation
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$$
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R = \left\lbrace (a, b) : a \equiv b \pmod m \right\rbrace \subset \mathbb{Z} \times \mathbb{Z}
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$$
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then $R$ has the following properties.
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- **Reflexive**: $\forall a \in \mathbb{Z}$, $(a, a) \in R$.
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- **Symmetric**: $\forall a, b \in \mathbb{Z}$, if $(a, b) \in R$, then $(b, a) \in R$.
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- **Transitive**: $\forall a, b, c \in \mathbb{Z}$, if $(a, b), (b, c) \in R$ then $(a, c) \in R$.
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Thus, $R$ is an **equivalence relation** and each residue class $[k]$ is an **equivalence class**.
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We write the set of residue classes modulo $n$ as
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$$
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\mathbb{Z}_n = \left\lbrace \overline{0}, \overline{1}, \overline{2}, \dots, \overline{n-1} \right\rbrace.
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$$
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Note that $\mathbb{Z}_n$ is closed under addition and multiplication.
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### Identity
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> **Definition.** For a binary operation $\ast$ defined on a set $S$, $e$ is the **identity** if
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>
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> $$
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> \forall a \in S,\, a * e = e * a = a.
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> $$
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In $\mathbb{Z}_n$, the additive identity is $0$, the multiplicative identity is $1$.
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### Inverse
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> **Definition.** For a binary operation $\ast$ defined on a set $S$, let $e$ be the identity. $x$ is the **inverse of $a$** if
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>
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> $$
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> x * a = a * x = e.
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> $$
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>
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> We write $x = a^{-1}$.
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In the language of modular arithmetic, $x$ is the inverse of $a$ if
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$$
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ax \equiv 1 \pmod n.
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$$
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The inverse exists if and only if $\gcd(a, n) = 1$.
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> **Lemma**. For $n \geq 2$ and $a \in \mathbb{Z}$, its inverse $a^{-1} \in \mathbb{Z}_n$ exists if and only if $\gcd(a, n) = 1$.
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*Proof*. We use the Extended Euclidean Algorithm. There exists $u, v \in \mathbb{Z}$ such that
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$$
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au + nv = \gcd(a, n).
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$$
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($\impliedby$) If $\gcd(a, n) = 1$, then $au + nv = 1$, so $au = 1 - nv \equiv 1 \pmod n$. Thus $a^{-1} = u$.
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($\implies$) Suppose that $x = a^{-1}$ exists. Then $ax \equiv 1 \pmod n$, so $ax = 1 + kn$ for some $n \in \mathbb{Z}$. Then $ax - nk = 1$. $\gcd(a, n)$ must divide the LHS, so $\gcd(a, n) = 1$.
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## Euclidean Algorithm
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### Greatest Common Divisor
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> **Definition.** Let $a, b \in \mathbb{Z} \setminus \left\lbrace 0 \right\rbrace$ . The **greatest common divisor** of $a$ and $b$ is the largest integer $d$ such that $d \mid a$ and $d \mid b$. We write $d = \gcd(a, b)$.
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> **Definition.** If $\gcd(a, b) = 1$, we say that $a$ and $b$ are **relatively prime**.
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### Euclidean Algorithm
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Euclidean Algorithm is an efficient way to find $\gcd(a, b)$. It relies on the following lemma.
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> **Lemma.** For $a, b \in \mathbb{Z}$ and $b \neq 0$, $\gcd(a, b) = \gcd(b, a \bmod b)$.
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*Proof*. By the division algorithm, there exists $q, r \in \mathbb{Z}$ such that $a = bq + r$. Here, $r = a \bmod b$.
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Let $d = \gcd(a, b)$. Then $d \mid a$ and $d \mid b$, so $d \mid (a - bq)$. Thus $d \leq \gcd(b, r)$. Conversely, let $d' = \gcd(b, r)$. Then $d' \mid b$ and $d' \mid (a - bq)$, so $d' \mid a$. Thus $d' \leq \gcd(a, b)$. Thus $d = d'$.
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The following code computes the greatest common divisor.
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```c
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int gcd(int a, int b) {
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if (b == 0) {
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return a;
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} else {
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return gcd(b, a % b);
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}
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}
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```
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### Extended Euclidean Algorithm
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We can extend the Euclidean algorithm to compute $u, v \in \mathbb{Z}$ such that
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$$
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ua + vb = \gcd(a, b).
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$$
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Basically, we use the Euclidean algorithm and solve for the remainder (which is the $\gcd$).
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#### Calculating Modular Multiplicative Inverse
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We can use the extended Euclidean algorithm to find modular inverses. Suppose we want to calculate $a^{-1}$ in $\mathbb{Z}_n$. We assume that the inverse exist, so $\gcd(a, n) = 1$.
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Therefore, we use the extended Euclidean algorithm and find $x, y \in \mathbb{Z}$ such that
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$$
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ax + ny = 1.
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$$
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Then $ax \equiv 1 - ny \equiv 1 \pmod n$, thus $x$ is the inverse of $a$ in $\mathbb{Z}_n$.
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[^1]: Note that in C standards, `(a / b) * b + (a % b) == a`.
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[^2]: $a$ and $b$ are in the same coset of $\mathbb{Z}/n\mathbb{Z}$.
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