The Euclid's algorithm

Let be two natural numbers \( (a, b) \in \hspace{0.05em}\mathbb{N}^2, \enspace a > b \).

Euclid's algorithm tells us that:

1. If \( b \) divides \( a \)
$$ \exists q \in \mathbb{N}, $$

$$ b / a \ \Longrightarrow \ a = bq \ \Longrightarrow \ PGCD(a, b) = b $$

2. If \( b\) does not divide \( a\)
$$ \exists q \in \mathbb{N}, \enspace \exists R \in [\![0, (b-1) ]\!] , $$

$$ b \nmid a \ \Longrightarrow \ a = bq + R $$

The \( GCD(a, b) \) is the last non-zero \( R_n \) of the Euclidian division of \( a \) by \( b \) such as:

$$ a = \textcolor{#8E5B5B}{b} q_0 + \textcolor{#446e4f}{R_0} $$

As long as the remainder does not divide the previous division divisor, we continue the algorithm.

We then decompose the new element at each step:

$$ b = \textcolor{#8E5B5B}{R_0} q_1 + \textcolor{#446e4f}{R_1} \ \Longrightarrow \ R_0 = \textcolor{#8E5B5B}{R_1} q_2 + \textcolor{#446e4f}{R_2} \ \Longrightarrow \ R_k = \textcolor{#8E5B5B}{R_{k + 1}} q_{k + 2} + \textcolor{#446e4f}{R_{k + 2}} $$
$$ up \ to \ \Longrightarrow R_{n - 3} = q_{n -1 }\textcolor{#8E5B5B}{R_{n - 2}} + \textcolor{#446e4f}{R_{n -1}} $$

As long a the remainder \( R_{n - 1} \) still does not divide \( R_{n - 2} \), the remainder \( R_{n - 2} \) can be written:

$$ R_{n - 2} = q_{n}\textcolor{#8E5B5B}{R_{n - 1}} + \textcolor{#446e4f}{R_{n}} $$

Until the moment where finally \( R_n \) divides \( R_{n - 1} \), so \( R_{n - 1} \) will be written:

$$ R_{n - 1} = q_{n + 1}\textcolor{#8E5B5B}{R_{n}} $$

At this stage, as we trace back the algorithm injecting each previous step's values until \( a \) and \( b \), we realize that we can always factorize by \(R_{n} \). Thus:

$$ GCD(a, b) = R_{n} $$



\( \forall n \in \mathbb{N}\), \(\forall k \in [\![0, n ]\!] \), as long as the remainder \(R_{k} \) does not divide its predecessor \(R_{k -1 } \):

$$ a \nmid b \Longrightarrow GCD(a, b) = GCD(b, R_0) = GCD(R_0, R_1) = \enspace ... \enspace = GCD(R_{n - 1}, R_n) = R_n \neq 0$$

Demonstration

Let be two integers \( (a, b) \in \hspace{0.05em} \mathbb{N}^2 \), with \( a > b \).

1. If \( b \) divides \( a \)

Then, \( a \) can be written:

$$ a = bq_0 \qquad (a) $$

At this stage, as \( b /a \) and \( b/b \), so \( PGCD(a, b) = b\).

$$ \exists q \in \mathbb{N}, $$

$$ b / a \ \Longrightarrow \ a = bq \ \Longrightarrow \ PGCD(a, b) = b $$

2. If \( b \) does not divide \( a \)

In this case, \( a \) can be written as:

$$ \exists q_0 \in \mathbb{N}, \enspace \exists R_0 \in [\![0, (b-1) ]\!] , \enspace a = \textcolor{#8E5B5B}{b}q_0 + \textcolor{#446e4f}{R_0} \qquad (a^*) $$

So in a general way,

$$ \exists q \in \mathbb{N}, \enspace \exists R \in [\![0, (b-1) ]\!] , $$

$$ b \nmid a \ \Longrightarrow \ a = bq + R $$

1. If \( R_0 \) divides \( b \)

Then \( b \) can be written as:

$$ \exists q_1 \in \mathbb{N}, \enspace b = q_1R_0 \qquad (b) $$



We can reassemble the algorithm and inject \( (b) \) into \( (a^*) \):

$$ a = bq_0 + R_0 = q_1R_0q_0 + R_0 = R_0 \underbrace{(q_1 q_0 + 1)} _\text{ \( Q_{a} \in \mathbb{N} \)} $$

So,

$$ a = Q_{a}R_0 \qquad (a') $$

We then have,

$$ \Biggl \{ \begin{align*} a = Q_{a}R_0 \qquad (a') \\ b = q_1R_0 \qquad (b) \end{align*} $$

Ant at that moment, as \( R_0 /a \), \( R_0/b \) and \( R_0/R_0 \), so:

$$ GCD(a, b) =GCD(b, R_0) = R_0 $$
2. If \( R_0 \) does not divide \( b \)

In this case, \( b \) can be written as:

$$ \exists q_1 \in \mathbb{N}, \enspace \exists R_1 \in [\![0, (R_0-1) ]\!], \enspace b = \textcolor{#8E5B5B}{R_0} q_1 + \textcolor{#446e4f}{R_1} \qquad (b^*) $$


1. - If \( R_1 \) divides \( R_0 \)

Then \( R_0 \) can be written:

$$ \exists q_2 \in \mathbb{N}, \enspace R_0 = q_2R_1 \qquad (R_0) $$



We can reassemble the algorithm and inject \( (R_0) \) into \( (b^*) \):

$$ b = R_0 q_1 + R_1 = q_2R_1 q_1 + R_1 = R_1 \underbrace{(q_2 q_1 + 1)} _\text{ \( Q_{b} \in \mathbb{N} \)} $$

So,

$$ b = Q_{b}R_1 \qquad (b') $$

We trace back the algorithm another time injecting the new value of \( (b') \) and \( (R_0) \) into \( (a^*) \):

$$a = bq_0 + R_0 = Q_{b}R_1q_0 + q_2R_1 = R_1 \underbrace{(Q_{b} q_0 + q_2)} _\text{ \( Q_{a} \in \mathbb{N} \)} $$
$$a = Q_{a}R_1 \qquad (a') $$

We then have,

$$ \Biggl \{ \begin{align*} a = Q_{a}R_1 \qquad (a') \\ b = Q_{b}R_1 \qquad (b') \end{align*} $$

Now, as \( R_1 /a \), \( R_1/b \), \( R_1/R_0 \) and \( R_1/R_1 \), so:

$$ GCD(a, b) =GCD(b, R_0) =GCD(R_0, R_1) = R_1 $$
2. - If \( R_1 \) does not divide \( R_0 \)

In this cas, \( R_0 \) can be written as:

$$ \exists q_2 \in \mathbb{N}, \enspace \exists R_2\in [\![0, (R_1-1) ]\!], \enspace R_0 = \textcolor{#8E5B5B}{R_1} q_2 + \textcolor{#446e4f}{R_2} \qquad (R_0^*) $$



We continue the algorithm this way until \( R_n \) divides \( R_{n- 1} \) and then:

$$ R_{n - 1} = q_{n + 1}\textcolor{#8E5B5B}{R_n} $$

Thus, by tracing back the algorithm we get:

$$ R_{n - 2} = q_{n} \textcolor{#8E5B5B}{R_{n - 1}} + \textcolor{#446e4f}{R_n} $$
$$ R_{n - 2} = q_{n}\textcolor{#8E5B5B}{q_{n + 1}R_n} + \textcolor{#446e4f}{R_n} = R_n\underbrace{(q_{n}q_{n + 1} + 1)} _ \text{ \( Q_{n - 2} \in \mathbb{N} \) } $$
$$ R_{n - 2} = R_nQ_{n - 2} $$

As a result, we have a chain reaction until the beginning of the algorithm:

$$ \Biggl[ R_{n - 3} = R_nQ_{n - 3} \Biggr] \Longrightarrow ... \Longrightarrow \Biggl[ R_{n - k} = R_nQ_{n - k} \Biggr] \Longrightarrow ... \Longrightarrow \Biggl[ R_{2} = R_nQ_{2} \Biggr] $$
$$ \Biggl[ R_{1} = R_nQ_{1}\Biggr] \Longrightarrow \Biggl[ R_{0} = R_nQ_{0} \Biggr] \Longrightarrow \Biggl[ b = R_nQ_{b}\Biggr] \Longrightarrow \Biggl[ a = R_nQ_{a}\Biggr] $$

\( R_n \) will be the greatest common divisor of all these numbers.



$$ GCD(a, b) = GCD(b, R_0) = GCD(R_0, R_1) = \enspace ... \enspace = GCD(R_{n - 1}, R_n) = R_n \neq 0$$

And finally,

\( \forall n \in \mathbb{N}\), \(\forall k \in [\![0, n ]\!] \), tant que le reste \(R_{k} \) ne divise pas le reste \(R_{k -1 } \):

$$ a \nmid b \Longrightarrow GCD(a, b) = GCD(b, R_0) = GCD(R_0, R_1) = \enspace ... \enspace = GCD(R_{n - 1}, R_n) = R_n \neq 0$$

This remainder \( R_n \) can itself be broken into different factors, which will make up the set of divisors of \( a \) and \( b \):

Examples

1. Calculation of \( GCD(1365, 25) \)

We have seen that it is necessary to carry out successive divisions, in order to recover the last non-zero remainder of this division series.

1. Calculation of \( 1365 / 25 \)



\( 25 \) does not divide \( 1365 \), because there is a remainder. We first recover its floor.

$$ \left \lfloor{\frac{1365}{25}} \right \rfloor = 54 \hspace{5em} (1^{st} \ division) $$

So \( 1365 \) can be written:

$$ 1365 = \textcolor{#8E5B5B}{25} \times 54 + \textcolor{#446e4f}{R_0} $$

Let's calculate the remainder \( \textcolor{#446e4f}{R_0} \).

$$ \textcolor{#446e4f}{R_0} = 1365 -\textcolor{#8E5B5B}{25} \times 54 = \textcolor{#446e4f}{15} \hspace{5em} (R_0) $$

So,

$$ 1365 = \textcolor{#8E5B5B}{25} \times 54 + \textcolor{#446e4f}{15} \qquad (a) $$
$$ (a = \textcolor{#8E5B5B}{b} q_0 + \textcolor{#446e4f}{R_0}) $$

2. Calculation of \( 25 / 15 \)



\( 15 \) does not divide \( 25 \).

$$ \left \lfloor{\frac{25}{15}} \right \rfloor = 1 \hspace{5em} (2^{nd} \ division) $$

Well \( 25 \) can be written:

$$ 25 = \textcolor{#8E5B5B}{15} \times 1 + \textcolor{#446e4f}{10} \qquad (b) $$
$$ (b = \textcolor{#8E5B5B}{R_0} q_1 + \textcolor{#446e4f}{R_1}) $$

3. Calculation of \( 15 / 10 \)



\( 10 \) does not divide \( 15 \).

$$ \left \lfloor{\frac{15}{10}} \right \rfloor = 1 \hspace{5em} (3^{rd} \ division) $$

Then \( 15 \) can be written:

$$ 15 = \textcolor{#8E5B5B}{10} \times 1 + \textcolor{#446e4f}{5} \qquad (R_0) $$
$$ (R_0 = \textcolor{#8E5B5B}{R_1} q_2 + \textcolor{#446e4f}{R_2}) $$

4. Calculation of \( 10 / 5 \)



Now, \( 5 \) divides \( 10 \). The remainder is zero.

$$ \frac{10}{5} = 2 \hspace{5em} (4^{th} \ division) $$

Then \( 10 \) can be written:

$$ 10 = \textcolor{#8E5B5B}{5} \times 2 \qquad (R_1) $$
$$ (R_1 = \textcolor{#8E5B5B}{R_2} q_3) $$

We therefore have a final result:

$$ GCD(1365, 25) = 5 $$
$$ (GCD(a, b) = R_2) $$


2. Algorithm feedback

We can eventually go back through the algorithm and find two nuumbers that fit the Bézout's identity.

This theorem tell us thtat:

$$ \forall (a, b) \in \hspace{0.05em}\mathbb{N}^2, \enspace a > b, $$

$$ \delta = a \wedge b \hspace{0.2em} \Longrightarrow \hspace{0.2em} \exists (u, v) \in \hspace{0.05em}\mathbb{Z}^2, \enspace au + bv = \delta \qquad (Bachet-Bézout)$$

To do this, we will each time reinject the values into the result of the previous calculation.

$$ \Biggl \{ \begin{gather*} \underline{10} \hspace{0.1em} = \textcolor{#8E5B5B}{5} \times 2 \qquad (R_1) \\ 15 = \hspace{0.1em} \underline{\textcolor{#8E5B5B}{10}} \hspace{0.1em} \times 1 + \textcolor{#446e4f}{5} \qquad (R_0) \end{gather*} \enspace \Longrightarrow \enspace 15 = \textcolor{#8E5B5B}{(5 \times 2)} \times 1 + \textcolor{#446e4f}{5} \qquad (R_0') $$

We isolate the \( GCD \) value, because this is what we want in the end as a result.

$$ 15 = \textcolor{#8E5B5B}{10} \times 1 + \textcolor{#446e4f}{5} \qquad (R_0) \enspace \Longleftrightarrow \enspace \textcolor{#446e4f}{5} = 15 - \textcolor{#8E5B5B}{10} \times 1 \qquad (R_2) $$

Then, we do have this:

$$ 25 = \textcolor{#8E5B5B}{15} \times 1 + \textcolor{#446e4f}{10} \qquad (b) \enspace \Longleftrightarrow \enspace \textcolor{#446e4f}{10} = 25 - \textcolor{#8E5B5B}{15} \times 1 \qquad (R_1') $$

Injecting \( (R_1') \) into \( (R_2) \), we get:

$$ \Biggl \{ \begin{gather*} \textcolor{#446e4f}{5} = 15 - \hspace{0.1em} \underline{ \textcolor{#8E5B5B}{10} } \hspace{0.1em} \times 1 \qquad (R_2) \\ \hspace{0.1em} \underline{ \textcolor{#446e4f}{10} } \hspace{0.1em} = 25 - \textcolor{#8E5B5B}{15} \times 1 \qquad (R_1') \end{gather*} \enspace \Longrightarrow \enspace \textcolor{#446e4f}{5} = 15 - ( 25 - \textcolor{#8E5B5B}{15} \times 1 ) \times 1 \qquad (R_2') $$

Once again,

$$ 1365 = \textcolor{#8E5B5B}{25} \times 54 + \textcolor{#446e4f}{15} \qquad (a) \enspace \Longleftrightarrow \enspace \textcolor{#446e4f}{15} = 1365 - \textcolor{#8E5B5B}{25} \times 54 \qquad (R_0'') $$

Now injecting twice \( (R_0'') \) into \( (R_2') \), we get:

$$ \Biggl \{ \begin{gather*} \hspace{0.1em} \underline{ \textcolor{#446e4f}{15} } \hspace{0.1em} = 1365 - \textcolor{#8E5B5B}{25} \times 54 \qquad (R_0'') \\ \textcolor{#446e4f}{5} = \hspace{0.1em} \underline{ 15 } \hspace{0.1em} - ( 25 - \hspace{0.1em} \underline{ \textcolor{#8E5B5B}{15} } \hspace{0.1em} \times 1 ) \times 1 \qquad (R_2') \end{gather*} \enspace \Longrightarrow \enspace \textcolor{#446e4f}{5} = 1365 - \textcolor{#8E5B5B}{25} \times 54 - ( 25 - ( 1365 - \textcolor{#8E5B5B}{25} \times 54 ) \times 1 ) \times 1 \qquad (R_2'') $$

So,

$$ \textcolor{#446e4f}{5} = 1365 - \textcolor{#8E5B5B}{25} \times 54 - 25 + 1365 - \textcolor{#8E5B5B}{25} \times 54 \qquad (R_2''') $$
$$ \textcolor{#446e4f}{5} = 1365 \times 2 + \textcolor{#8E5B5B}{25} \times (-54 - 54 -1) $$
$$ \textcolor{#446e4f}{5} = 1365 \times 2 + \textcolor{#8E5B5B}{25} \times (-109) $$

We definitely have:

$$ 5 = 1365 \wedge 25 \hspace{0.2em} \Longrightarrow \hspace{0.2em} \exists (u, v) \in \hspace{0.05em}\mathbb{Z}^2, \enspace 1365u + 25v = 5 $$

$$ with \ \Biggl \{ \begin{gather*} u = 2 \\ v = -109 \end{gather*} $$