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本帖最后由 hbghlyj 于 2023-2-14 16:53 编辑 《简明数论》63页 习题八
7. 设 $a_{1}, a_{2}, c$ 是正整数, $\left(a_{1}, a_{2}\right)=1$. 对于方程 $a_{1} x_{1}+a_{2} x_{2}=c$ 有以下结论：
(i) $c<a_{1}+a_{2}$ 时一定没有正解;
(ii) 全体非负解和全体正解相同的充要条件是 $a_{1} \nmid c$ 且 $a_{2} \nmid c$;
(iii) 若 $a_{1} \mid c, a_{1} a_{2} \nmid c$, 则正解的个数等于 $\left[c /\left(a_{1} a_{2}\right)\right]$;
(iv) 若 $a_{1} a_{2} \mid c$, 则正解个数等于 $-1+c /\left(a_{1} a_{2}\right)$.
8. 设 $a_{1}, a_{2}, a_{3}$ 是两两既约的正整数. 证明：不定方程 $a_{2} a_{3} x_{1}+a_{3} a_{1} x_{2}+a_{1} a_{2} x_{3}=c$,
(1)当 $c>2 a_{1} a_{2} a_{3}-a_{1} a_{2}-a_{2} a_{3}-a_{3} a_{1}$ 时一定有非负解;
(2)当 $c=2 a_{1} a_{2} a_{3}-a_{1} a_{2}-a_{2} a_{3}-a_{3} a_{1}$ 时无非负解.
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7. (i) 若$a_1,a_2⩾1$,则$c=a_{1} x_{1}+a_{2} x_{2}⩾x_1+x_2$.
(ii) $a_1=0⇒a_2x_2=c⇒a_2\mid c$.
(iii)由$a_1\mid c$知$\left(\frac c{a_1},0\right)$是一组解. 由定理3知,通解为$\cases{x_1=\frac c{a_1}-a_2t\\x_2=a_1t}$
$x_1,x_2>0⇔t<\frac c{a_1a_2}∧t>0$. 又$a_1a_2\nmid c$, 因此$t=1,2,⋯,\left[c\over a_1a_2\right]$, 正解个数等于 $\left[c\over a_1a_2\right]$.
(iv)通解和(iii)相同,同样有$x_1,x_2>0⇔t<\frac c{a_1a_2}∧t>0$.
又$a_1a_2\mid c$,因此$t=1,2,⋯,{c\over a_1a_2}-1$, 正解个数等于 ${c\over a_1a_2}-1$.
8.(1)$a_1,a_2,a_3$两两既约, 所以存在一组整数解 $x_{1}^{0}, x_{2}^{0}, x_{3}^{0}$, 任意一组整数解 $x_{1}, x_{2}, x_{3}$ 必满足$$a_{2} a_{3}\left(x_{1}-x_{1}^{0}\right)+a_{3} a_{1}\left(x_{2}-x_{2}^{0}\right)+a_{1} a_{2}\left(x_{3}-x_{3}^{0}\right)=0$$因此,$a_{1} \mid x_{1}-x_{1}^{0}$, $a_{2}\left|x_{2}-x_{2}^{0}, a_{3}\right| x_{3}-x_{3}^{0}$, 即有$$\array{x_{1}=x_{1}^{0}+a_{1} t_{1}\\ x_{2}=x_{2}^{0}+a_{2} t_{2}\\ x_{3}=x_{3}^{0}+a_{3} t_{3}\\t_{1}+t_{2}+t_{3}=0}$$若有非负解, 那么, 当 $x_{2}, x_{3}$ 取最小非负值时, $x_{1}$ 的值必为非负.
[注意,只要确定$t_1,t_2,t_3$中的两个,就确定了第三个,也就确定了一组整数解$(x_1,x_2,x_3)$.]
所以取 $t_{2}, t_{3}$ 使\begin{gathered}0 \leqslant x_{2}=x_{2}^{0}+a_{2} t_{2} \leqslant a_{2}-1\\0 \leqslant x_{3}=x_{3}^{0}+a_{3} t_{3} \leqslant a_{3}-1\end{gathered}得到\begin{aligned}a_{2} a_{3}x_1&=c-a_3a_1x_2-a_1a_2x_3\\
&⩾c-a_3 a_1 \left(a_2-1\right)-a_1 a_2 \left(a_3-1\right)\\
&=c-2 a_{1} a_{2} a_{3}+a_{3} a_{1}+a_{1} a_{2}\end{aligned}由此就推出当 $c>2 a_{1} a_{2} a_{3}-a_{1} a_{2}-a_{2} a_{3}-a_{3} a_{1}$ 时, $x_{1}>-1$, 即必有非负解.
(2)当 $c=2 a_{1} a_{2} a_{3}-a_{1} a_{2}-a_{2} a_{3}-a_{3} a_{1}$ 时, 若有非负解 $x_{1}$, $x_{2}, x_{3}$, 则$$a_{2} a_{3}\left(x_{1}+1\right)+a_{3} a_{1}\left(x_{2}+1\right)+a_{1} a_{2}\left(x_{3}+1\right)=2 a_{1} a_{2} a_{3}$$由此推出 $$a_{1}\left|x_{1}+1 \geqslant a_{1}, a_{2}\right| x_{2}+1 \geqslant a_{2}, a_{3} \mid x_{3}+1 \geqslant a_{3}$$代入上式,$$2a_1a_2a_3⩾3a_1a_2a_3$$矛盾.
(1)的另证:
由定理3, 这个不定方程等价于4个变数,2个方程的不定方程组\begin{cases}a_3y+a_{1} a_{2} x_{3}=c&①\\
a_{2}x_{1}+a_{1} x_{2}=y&②\end{cases}若$c>2 a_{1} a_{2} a_{3}-a_{1} a_{2}-a_{2} a_{3}-a_{3} a_{1}$,
因为$2 a_{1} a_{2} a_{3}-a_{1} a_{2}-a_{2} a_{3}-a_{3} a_{1}=(a_1a_2a_3-a_1a_2-a_3)+\underbrace{(a_1-1) (a_2-1) a_3}_{⩾0}$
所以$c>a_1a_2a_3-a_1a_2-a_3$. 由定理4, 不定方程①一定有非负解$(x_{3,0},y_0)$.
作带余除法$x_{3,0}=a_3t+x_{3,1},0⩽x_{3,1}<a_3,t⩾0$,
令$y_1=y_0+a_1a_2t$,则 $(x_{3,1},y_1)$是不定方程①的非负解.
我们有$$x_{3,1}⩽a_3-1\implies2 a_1 a_2 a_3 - a_1 a_2 - a_2 a_3 - a_3 a_1 ⩾ a_3 (a_1 a_2 - a_1 - a_2) + a_1 a_2 x_{3,1}$$
而$$c>2 a_1 a_2 a_3 - a_1 a_2 - a_2 a_3 - a_3 a_1$$
所以$$c>a_3 (a_1 a_2 - a_1 - a_2) + a_1 a_2 x_{3,1}$$
因此$$y_1=\frac{c-a_1a_2x_{3,1}}{a_3}>a_1a_2-a_1-a_2$$由定理4, 不定方程②一定有非负解$(x_{1,1},x_{2,1})$, 因此$(x_{1,1},x_{2,1},x_{3,1})$是原不定方程的一组非负解.
PS:
发现Chrome的内置pdf阅读器,缩放到25%就不能再小了...
发现$a|b$ (a|b)和$a\mid b$ (a\mid b)效果不一样. |
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