线性方程整数解，扩展欧几里得解分析

Posted 2020-12-21 lxzbky

符号说明

(y_{0}=left ( a_{1},a_{2},...,a_{n} ight ) 即y_{0}是a_{1},a_{2},...,a_{n})的最大公因数
(d|a_{i}（d整除a_{i}，a_{i}是d的倍数）)

定理

(设a_{i}in Z ,A=left { y|y=sum_{i=1}^{n}a_{i}x_{i},x_{i}in Z,1leq i leq n ight })
如果(y_{0})是集合A中最小的正数，则(y_{0}=left ( a_{1},a_{2},...,a_{n} ight ))

证明

设d是(a_{1},a_{2},...,a_{n})的一个公因数，(d|a_{i})，则(a_{1},a_{2},...,a_{n})的线性组合都是d的倍数
(d|y_{0}=sum_{i=1}^{n}a_{i}x‘_{i},故而dleq y_{0})
(y_{0}=sum_{i=1}^{n}a_{i}x‘_{i},对任意的y=sum_{i=1}^{n}a_{i}x_{i},存在q,rin Z,这里两个式子的a_{i}相同,使得y=qy_{0}+r)
(r=y-qy_{0}=a_{1}(x_{1}-qx‘_{1})+...+a_{n}(x_{n}-qx‘_{n})in A)
因为r小于(y_{0},若r eq 0)则r是A中比(y_{0})还小的正整数，这与(y_{0})是A中最小的正整数矛盾，故而r=0，(y_{0})|y
显然(a_{i}in A,在y=sum_{i=1}^{n}a_{i}x_{i}中取a_{i}系数x_{i}=1,其他所有x=0,故y_{0}|a_{i})
由此可得(y_{0}也是a_{1},a_{2},...,a_{n}的公因数而dleq y_{0}，所以y_{0}=left ( a_{1},a_{2},...,a_{n} ight ))