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# Extended Euclidean Algorithm | Brilliant Math & Science Wiki

Without loss of generality we can assume that $a$ and $b$ are non-negative integers, because we can always do this: $\gcd(a,b)=\gcd\big(\lvert a \rvert, \lvert b \rvert\big)$.

Let's define the sequences $\{q_i\},\{r_i\},\{s_i\},\{t_i\}$ with $r_0=a,r_1=b$. Set $i \gets 2$, and increase it at the end of every iteration. We're going to find in every iteration $q_i, r_i, s_i, t_i$ such that $r_{i-2}=r_{i-1}q_i+r_i$, $0 \leq r_i < r_{i-1}$ using the division algorithm. We also want to write $r_i$ as a linear combination of $a$ and $b$, i.e., $r_i=s_i a+t_i b$. But $r_i=r_{i-2}-r_{i-1}q_i$, so

$r_i=s_{i-2}a+t_{i-2}b-(s_{i-1}a+t_{i-1}b)q_i=(s_{i-2}-s_{i-1}q_i)a+(t_{i-2}-t_{i-1}q_i)b.$

Hence, we obtain $s_i=s_{i-2}-s_{i-1}q_i$ and $t_i=t_{i-2}-t_{i-1}q_i$.

Now, we have to find the initial values of the sequences $\{s_i\}$ and $\{t_i\}$. By the definition of $r_i,$ we have

\begin{aligned} a=r_0=s_0 a+t_0 b &\implies s_0=1, t_0=0\\ b=r_1=s_1 a+t_1 b &\implies s_1=0, t_1=1. \end{aligned}

Finally, we stop at the iteration in which we have $r_{i-1}=0$. Let's call this the $n^\text{th}$ iteration, so $r_{n-1}=0$. That means that $\gcd(a,b)=\gcd(r_0,r_1)=\gcd(r_1,r_2)=\cdots=\gcd(r_{n-2},r_{n-1})=\gcd(r_{n-2},0)=r_{n-2}$, so we found our desired linear combination:

$\gcd(a,b)=r_{n-2}=s_{n-2} a + t_{n-2} b.$

This algorithm is always finite, because the sequence $\{r_i\}$ is decreasing, since $0 \leq r_i < r_{i-1}$ for $2 \leq i < n-1$, so $r_2 > r_3 > \cdots > r_{n-2} > r_{n-1} = 0$. In some moment we reach the value of zero, because all of the $r_i$ are integers.

To implement the algorithm, note that we only need to save the last two values of the sequences $\{r_i\}$, $\{s_i\}$ and $\{t_i\}$.