Let $a,b \in \mathbb{Z}$, we say that $a$ divides $b$, written $a \given b$, if there’s an integer $n$ so that $$b = na$$

If $a$ divides $b$ then $b$ is divisible by $a$ and $a$ is a divisor or factor of $b$, also $b$ is called a multiple of $a$.

Additional properties of the relation $|$:

1. if $a \given b$ and $b \given c$ then $a \given c$
2. if $a \given b$ and $c \given d$ then $ac \given bd$
3. if $d \given a$ and $d \given b$ then $d \given a + b$
4. if $d \given a$ and $d \given b$ then $d \given ax + by$ for any integers $x,y$

Proof.

1. if $b=ma$ and $c=nb$ then $c=(nm)a$
2. if $b=ma$ and $d=nc$ then $bd=(nm)ac$
3. if $a=md$ and $b=nd$ then $a + b=(m + n)d$
4. if $a=md$, $b=nd$ then $ax=(mx)d$, $by=(ny)d$ therefore $ax + by = (mx + ny)d$

## Division algorithm

Let $a, b \in \mathbb{Z}$ with $b > 0$, then there exists $q, r \in \mathbb{Z}$ such that $$a = bq + r, \quad \text{where 0 \leq r \lt b}$$

Proof. if $bq$ is the largest multiple of $b$ that does not exceed $a$ then $r = a - bq$ is positive and since $b(q + 1) > a$ then $r \lt b$.

Also, if $r = 0$ then $a = bq$ which implies that $q \given a$.

## Greatest common divisor

Let $a, b \in \mathbb{N}$, the greatest common divisor of $a$ and $b$, written as $gcd(a,b)$ or $(a,b)$, is the element $d$ in $\mathbb{N}$ such that $d \given a$ and $d \given b$ and every common divisor of $a$ and $b$ also divides $d$.

Let $a$ and $b$ be two numbers in $\mathbb{N}$, the value of $(a,b)$ is a linear combination of $a$ and $b$ i.e. there exists $s,t$ in $\mathbb{Z}$ such that $$sa + tb = (a, b)$$

Proof.

Let $d$ be the least positive integer that is a linear combination of $a$ and $b$

$$d = sa + tb$$

First lets show that $d \given a$, by the division algorithm we know that

$$a = dq + r, \quad \text{where 0 \le r \lt d}$$

It follows that

\begin{align*} r &= a - dq \\ &= a - (sa + tb)q \\ &= a - saq - tbq \\ &= (1 - sq)a + (-tq)b \\ \end{align*}

We can see that $r$ is a linear combination of $a$ and $b$. Since $0 \le r \lt d$ and considering that we defined $d$ as the least positive linear combination of $a$ and $b$ it follows that $r = 0$ (if $0 \lt r \lt d$ then $r$ would be the least possible linear combination which is a contradiction), therefore $d \given a$.

In a similar fashion $d \given b$, therefore by the divisibility property #4

$$d \given sa + tb$$

The next thing to prove is that $d$ is the greatest common divisor of $a$ and $b$. To prove this lets show that if $d'$ is any other common divisor of $a$ and $b$ then $d' \le d$.

If $d' \given a$ and $d' \given b$ then by the divisibility property #4 it divides any other linear combination of $a$ and $b$, since $d = sa + bt$ is one linear combination of $a$ and $b$ it follows that $d' \given d$ so either $d' \lt d$ or $d' = d$, finally we can conclude that

$$d = (a,b)$$

### Euclidean Algorithm

A very efficient method to compute the greatest common denominator

Suppose $a, b$ be integers with $a \ge b \gt 0$

1. Apply the division algorithm $a = bq + r, 0 \le r \lt b$
2. Rename $b$ as $a$ and $r$ as $b$ and repeat 1 until $r = 0$ The last nonzero remainder is the greatest common divisor of $a$ and $b$

The euclidean algorithm depends on the following lemma

Let $a, b$ be integers with $a \ge b \gt 0$. Let $r$ be the remainder of dividing $a$ by $b$ then $$(a,b) = (b, r)$$

Proof. Let $q$ be the quotient of dividing $a$ by $b$ so that $a = bq + r$. If $d = (a,b)$ then it must divide any other linear combination of $a$ and $b$ like $r = a - bq$, therefore $d \given r$. Finally we can conclude that $d = (b,r)$.

Proof of the theorem If we keep on repeating the division algorithm we have:

\begin{align*} a &= bq_1 + r_1, \quad (a,b) = (b, r_1) \\ b &= r_1q_2 + r_2, \quad (b, r_1) = (r_1, r_2) \\ r_1 &= r_2q_3 + r_3, \quad (r_1, r_2) = (r_2, r_3) \\ r_2 &= r_3q_4 + r_4, \quad (r_2, r_3) = (r_3, r_4) \\ & \; \vdots \\ r_{n-3} &= r_{n-2}q_{n-1} + r_{n-1}, \quad (r_{n-3}, r_{n-2}) = (r_{n-2}, r_{n-1}) \\ r_{n-2} &= r_{n-1}q_n + r_n, \quad (r_{n-2}, r_{n-1}) = (r_{n-1}, r_n) \\ r_{n-1} &= r_n q_{n+1}, \quad \quad (r_{n-1}, r_n) = r_n \end{align*}

Therefore

$$(a,b) = (b,r_1) = (r_1,r_2) = (r_2, r_3) = (r_3, r_4) = \ldots = (r_{n-3}, r_{n-2}) = (r_{n-2}, r_{n-1}) = (r_{n-1}, r_n) = r_n$$

### Extended Euclidean Algorithm

One of the applications of the euclidean algorithm is the calculation of the integers $x,y$ satisfying $d = (a,b) = ax + by$

First note that if $b=0$ then $(a,b) = (a,0) = a$, now assume that there are integers $x'$ and $y'$ so that

$$(a,b) = (b,r) = bx' + ry'$$

Since

\begin{align*} r &= a - bq \\ &= a - b \left \lfloor \frac{a}{b} \right \rfloor \end{align*}

Then

\begin{align*} (a,b) &= bx' + \Big( a - \left \lfloor \frac{a}{b} \right \rfloor b \Big) y' \\ &= bx' + ay' - \left \lfloor \frac{a}{b} \right \rfloor by' \\ &= a(y') + b \Big(x' - \left \lfloor \frac{a}{b} \right \rfloor y'\Big) \end{align*}

Comparing it to $(a,b) = ax + by$ we obtain the required coefficients $x$ and $y$ based on the following recursive equations

\begin{align*} x &= \begin{cases} 1, & \text{when r = 0} \\ y', & \text{otherwise} \end{cases} \\ y &= \begin{cases} 0, & \text{when r = 0} \\ x' - \left \lfloor \frac{a}{b} \right \rfloor y', & \text{otherwise} \end{cases} \end{align*}