Bézout's Identity/Proof 3

Proof
Consider the Euclidean algorithm in action:

First it will be established that there exist $x_i, y_i \in \Z$ such that:
 * $(1): \quad a x_i + b y_i = r_i$

for $i \in \set{1, 2, \ldots, n - 1}$.

The proof proceeds by induction.

Basis for the Induction
When $i = 1$, let $x_1 = 1, y_1 = -q_1$.

When $i = 2$, let $x_2 = -q_2, y_2 = 1 + q_1 q_2$.

This is the basis for the induction.

Induction Hypothesis
Our induction hypothesis is that the integer solutions to $(1)$ have been found for all $i$ such that $i \le k$ where $k < n - 1$.

Induction Step
We know that:
 * $r_{k - 1} = r_k q_{k + 1} + r_{k + 1}$

Thus, by the induction hypothesis:


 * $\paren {a x_{k - 1} + b y_{k - 1} } - \paren {a x_k + b y_k} q_{k + 1} = r_{k + 1}$

which can be rewritten as:


 * $\paren {x_{k - 1} - x_k q_{k + 1} } a + \paren {y_{k - 1} - y_k q_{k + 1} } b = r_{k + 1}$

Hence we have the following solutions to $(1)$ when $i = k + 1$:


 * $x_{k + 1} = x_{k - 1} - x_k q_{k + 1}$


 * $y_{k + 1} = y_{k - 1} - y_k q_{k + 1}$

The result follows by the Principle of Mathematical Induction.