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