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## 初等数论--同余方程--二元一次不定方程的通解形式
* 不定方程：变量个数>方程个数
若二元一次不定方程$ax + by = n$有解，$x_0, y_0$为它的一组整数解，则通解为：
$$\large
\left\{\begin{matrix}
x=x_0 + \frac{b}{(a,b)} \cdot t\\
y=y_0-\frac{a}{(a,b)}\cdot t
\end{matrix}\right.
\ \ \ \ t \in Z
$$
证明：
* * *该形式确实是二元一次方程的解**
将$x,y$代入原方程，得：
$\large \displaystyle a(x_0+\frac{b}{(a,b)}\cdot t) + b(y_0-\frac{a}{(a,b)}\cdot t)$
$\large \displaystyle =ax_0+a\frac{b}{(a,b)}\cdot t + by_0-b\frac{a}{(a,b)}\cdot t$
$\large \displaystyle =ax_0+by_0$
$\large \displaystyle =n$
* **二元一次不定方程的解都可以表达成这种形式**
已知
$$\large
\left\{\begin{matrix}
ax+by=n
\\
ax_0+by_0=n
\end{matrix}\right.
$$
联立方程，相减得：
$$\large a(x-x_0)+b(y-y_0)=0$$
$$\large a(x-x_0)=-b(y-y_0)$$
$$\large \frac{a}{(a,b)}(x-x_0)=-\frac{b}{(a,b)}(y-y_0)$$
$$\large \because \frac{a}{(a,b)}\nmid \frac{b}{(a,b)}$$
且
$$\large \frac{b}{(a,b)} | \frac{a}{(a,b)}(x-x_0)$$
$$\large \therefore \frac{b}{(a,b)} | x-x_0$$
即
$$\large x-x_0=\frac{b}{(a,b)}\cdot t$$
同理,$\large \displaystyle \frac{a}{(a,b)}|y-y_0$,即
$$\large y-y_0=-\frac{a}{(a,b)}\cdot t$$
$$\huge Q.E.D$$
