linear-diophantine: move identity

jxu · web-flow · commit 5617b117b888 · 2024-10-13T21:29:12.000-04:00
diff --git a/src/algebra/linear-diophantine-equation.md b/src/algebra/linear-diophantine-equation.md
@@ -23,14 +23,6 @@ In this article, we consider several classical problems on these equations:
 
 A degenerate case that need to be taken care of is when $a = b = 0$. It is easy to see that we either have no solutions or infinitely many solutions, depending on whether $c = 0$ or not. In the rest of this article, we will ignore this case.
 
-## Bézout's lemma
-
-Bézout's lemma (also called Bézout's identity) is a useful result that can be used to understand and prove the following analytic solution. 
-
-> Let $g = \gcd(a,b)$. Then there exist integers $x,y$ such that $ax + by = g$.
-> 
-> Moreover, $g$ is the least such positive integer that can be written as $ax + by$; all integers of the form $ax + by$ are multiples of $g$. 
-
 ## Analytic solution
 
 When $a \neq 0$ and $b \neq 0$, the equation $ax+by=c$ can be equivalently treated as either of the following:
@@ -59,6 +51,12 @@ y = \frac{c-ax}{b}.
 
 ## Algorithmic solution
 
+**Bézout's lemma** (also called Bézout's identity) is a useful result that can be used to understand and prove the following solution. 
+
+> Let $g = \gcd(a,b)$. Then there exist integers $x,y$ such that $ax + by = g$.
+> 
+> Moreover, $g$ is the least such positive integer that can be written as $ax + by$; all integers of the form $ax + by$ are multiples of $g$. 
+
 To find one solution of the Diophantine equation with 2 unknowns, you can use the [Extended Euclidean algorithm](extended-euclid-algorithm.md). First, assume that $a$ and $b$ are non-negative. When we apply Extended Euclidean algorithm for $a$ and $b$, we can find their greatest common divisor $g$ and 2 numbers $x_g$ and $y_g$ such that:
 
 $$a x_g + b y_g = g$$
