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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.

Divisibility

Definition. Let a, b, c \in \mathbb{Z} such that a = bc. Then,

1. b and c are said to divide a, and are called factors of a.
2. a is said to be a multiple of b and c.

Notation. For a, b \in \mathbb{Z}, we write a \mid b if a divides b. If not, we write a \nmid b.

These are simple lemmas for checking divisibility.

Lemma. Let a, b, c \in \mathbb{Z}.

1. If a \mid b and a \mid c, then a \mid (b + c).
2. If a \mid b, then a \mid bc.
3. If a \mid b and b \mid c, then a \mid c.

Prime Numbers

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.

Note that 1 is neither prime nor composite.

Primality Tests

It is hard to verify if some given number is prime. Many encryption schemes heavily rely on this fact.

The following is a simple algorithm to check if a given integer is prime.

bool naive_prime_test(int n) {
    if (n < 2) {
        return false;
    }

    for (int i = 2; i < sqrt(n); ++i) {
        if (n % i == 0) {
            return false;
        }
    }
    return true;
}

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)

Division Algorithm

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.

Proof. By induction.

Other proofs use the well-ordering principle.

Modulo Operation

There are two ways to think about 'mod': as a function, and as a congruence.

Modulo as a Function

As a function, a \bmod b return the remainder of a divided by b. This operation is commonly denoted % in many programming languages.¹

Modulo as a Congruence

As a congruence, it means that a, b are in the same equivalence class.²

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).

Properties of modulo operation.

Lemma. Suppose that a \equiv b \pmod n and c \equiv d \pmod n. Then, the following hold.

1. a + c \equiv (b + d) \pmod n.
2. ac \equiv bd \pmod n.
3. a^k \equiv b^k \pmod n.
4. a \equiv (a \bmod n) \pmod n.

Proof. Trivial. :)

The last one is very useful in computing. For example, if a, b are very large integers, using the identity

(a + b)^k \equiv ((a + b) \bmod n)^k \pmod n

allows us to reduce the size of the numbers before exponentiation.

Modular Arithmetic

For modulus n, modular arithmetic is operation on \mathbb{Z}_n.

Residue Classes

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 $\mathbb{Z}$.

We write each residue class as follows.

\overline{k} = [k] = \left\lbrace m \in \mathbb{Z} : m \bmod n = k\right\rbrace

Consider the relation

R = \left\lbrace (a, b) : a \equiv b \pmod m \right\rbrace \subset \mathbb{Z} \times \mathbb{Z}

then R has the following properties.

• Reflexive: \forall a \in \mathbb{Z}, (a, a) \in R.
• Symmetric: \forall a, b \in \mathbb{Z}, if (a, b) \in R, then (b, a) \in R.
• Transitive: \forall a, b, c \in \mathbb{Z}, if (a, b), (b, c) \in R then (a, c) \in R.

Thus, R is an equivalence relation and each residue class [k] is an equivalence class.

We write the set of residue classes modulo n as

\mathbb{Z}_n = \left\lbrace \overline{0}, \overline{1}, \overline{2}, \dots, \overline{n-1} \right\rbrace.

Note that \mathbb{Z}_n is closed under addition and multiplication.

Identity

Definition. For a binary operation \ast defined on a set S, e is the identity if

\forall a \in S,\, a * e = e * a = a.

In \mathbb{Z}_n, the additive identity is 0, the multiplicative identity is 1.

Inverse

Definition. For a binary operation \ast defined on a set S, let e be the identity. x is the inverse of $a$ if

x * a = a * x = e.

We write x = a^{-1}.

In the language of modular arithmetic, x is the inverse of a if

ax \equiv 1 \pmod n.

The inverse exists if and only if \gcd(a, n) = 1.

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.

Proof. We use the Extended Euclidean Algorithm. There exists u, v \in \mathbb{Z} such that

au + nv = \gcd(a, n).

(\impliedby) If \gcd(a, n) = 1, then au + nv = 1, so au = 1 - nv \equiv 1 \pmod n. Thus a^{-1} = u.

(\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.

Euclidean Algorithm

Greatest Common Divisor

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).

Definition. If \gcd(a, b) = 1, we say that a and b are relatively prime.

Euclidean Algorithm

Euclidean Algorithm is an efficient way to find \gcd(a, b). It relies on the following lemma.

Lemma. For a, b \in \mathbb{Z} and b \neq 0, \gcd(a, b) = \gcd(b, a \bmod b).

Proof. By the division algorithm, there exists q, r \in \mathbb{Z} such that a = bq + r. Here, r = a \bmod b.

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'.

The following code computes the greatest common divisor.

int gcd(int a, int b) {
    if (b == 0) {
        return a;
    } else {
        return gcd(b, a % b);
    }
}

Extended Euclidean Algorithm

We can extend the Euclidean algorithm to compute u, v \in \mathbb{Z} such that

ua + vb = \gcd(a, b).

Basically, we use the Euclidean algorithm and solve for the remainder (which is the \gcd).

Calculating Modular Multiplicative Inverse

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.

Therefore, we use the extended Euclidean algorithm and find x, y \in \mathbb{Z} such that

ax + ny = 1.

Then ax \equiv 1 - ny \equiv 1 \pmod n, thus x is the inverse of a in \mathbb{Z}_n.

---

1. Note that in C standards, (a / b) * b + (a % b) == a. ↩︎

2. a and b are in the same coset of \mathbb{Z}/n\mathbb{Z}. ↩︎