Question

16．求下列不定方程的整数解：
①111x+321y=75；
②37x+41y=1；
③25x+13y+7z=4；
④x²-4xy+5y²=169；
⑤x²-y²=88．

Answer 1

分析 ①首先将方程做适当变形，根据解为整数确定其中一个未知数的取值，再进一步求得方程的另一个解；
②首先将方程做适当变形，根据解为整数确定其中一个未知数的取值，再进一步求得方程的另一个解；
③设25x+13y=t，于是t+7z=4．于是原方程可化为$\left\{\begin{array}{l}{25x+8y=t①}\\{t+7z=4②}\end{array}\right.$，用前面的方法可以求得①的解为：$\left\{\begin{array}{l}{x=3t-8u}\\{y=-t+3u}\end{array}\right.$，u是整数，继而可由②得出$\left\{\begin{array}{l}{t=2000+5v}\\{z=1000+3v}\end{array}\right.$，v是整数，消去t即可得；
④原方程整理得：x²-4xy+5y²-169=0，由△=（-4y）²-4（5y²-169）=4（169-y²）可知169-y²是完全平方数即可，满足条件的y值有0，5，-5，12，-12，据此可得方程组的整数解；
⑤正整数x、y满足方程时必有x＞y＞0知x+y＞x-y＞0．又x+y与x-y有相同的奇偶性，根据原方程（x-y）（x+y）=88，右边为偶数知x+y与x-y均为偶数，根据x+y，x-y是88的因数可得$\left\{\begin{array}{l}{x-y=2}\\{x+y=44}\end{array}\right.$或$\left\{\begin{array}{l}{x-y=4}\\{x+y=22}\end{array}\right.$或$\left\{\begin{array}{l}{x-y=-2}\\{x+y=-44}\end{array}\right.$或$\left\{\begin{array}{l}{x-y=-4}\\{x+y=-22}\end{array}\right.$或$\left\{\begin{array}{l}{x-y=44}\\{x+y=2}\end{array}\right.$或$\left\{\begin{array}{l}{x-y=22}\\{x+y=4}\end{array}\right.$或$\left\{\begin{array}{l}{x-y=-44}\\{x+y=-2}\end{array}\right.$或$\left\{\begin{array}{l}{x-y=-22}\\{x+y=-4}\end{array}\right.$，解之即可．

解答 解：①111x+321y=75，
x=$\frac{75-321y}{111}$=$\frac{75-99y}{111}$-2y①，
∵原方程的解为整数，
∴当y=3时，x=-8，是原方程的一组解，故y=111t+3，代入①式得x=-8-321t（t为整数），
故原方程的解为$\left\{\begin{array}{l}{x=-8-321t}\\{y=111t+3}\end{array}\right.$（t为整数）．

②37x+41y=1，
x=$\frac{1-41y}{37}$=$\frac{1-4y}{37}$-y①，
∵原方程的解为整数，
∴当y=-9时，x=10，是原方程的一组解，故y=37t-9，代入①式得x=10-41t（t为整数），
故原方程的解为$\left\{\begin{array}{l}{x=10-41t}\\{y=37t-9}\end{array}\right.$（t为整数）．

③25x+13y+7z=4，
设25x+13y=t，于是t+7z=4．
于是原方程可化为$\left\{\begin{array}{l}{25x+8y=t①}\\{t+7z=4②}\end{array}\right.$，
用前面的方法可以求得①的解为：$\left\{\begin{array}{l}{x=3t-8u}\\{y=-t+3u}\end{array}\right.$，u是整数；
②的解为$\left\{\begin{array}{l}{t=2000+5v}\\{z=1000+3v}\end{array}\right.$，v是整数．
消去t，得$\left\{\begin{array}{l}{x=6000-8u+15v}\\{y=-2000+3u-5v}\\{z=1000+3v}\end{array}\right.$，u，v是整数．
即当u、v取不同整数的时候，会得到相应的x、y、z的整数值．

④原方程整理得：x²-4xy+5y²-169=0，
∵△=（-4y）²-4（5y²-169）=4（169-y²），
∴169-y²是完全平方数即可，满足条件的y值有0，5，-5，12，-12，
由此适合原方程的全部整数解为：$\left\{\begin{array}{l}{x=22}\\{y=5}\end{array}\right.$或$\left\{\begin{array}{l}{x=-2}\\{y=5}\end{array}\right.$或$\left\{\begin{array}{l}{x=2}\\{y=-5}\end{array}\right.$或$\left\{\begin{array}{l}{x=-22}\\{y=-5}\end{array}\right.$或$\left\{\begin{array}{l}{x=29}\\{y=12}\end{array}\right.$或$\left\{\begin{array}{l}{x=-29}\\{y=12}\end{array}\right.$或$\left\{\begin{array}{l}{x=-19}\\{y=-12}\end{array}\right.$或$\left\{\begin{array}{l}{x=19}\\{y=-12}\end{array}\right.$或$\left\{\begin{array}{l}{x=13}\\{y=0}\end{array}\right.$或$\left\{\begin{array}{l}{x=-13}\\{y=0}\end{array}\right.$；

⑤∵正整数x、y满足方程时，必有x＞y＞0．
∴x+y＞x-y＞0．
又∵x+y与x-y有相同的奇偶性，
∵原方程（x-y）（x+y）=88，右边为偶数，
∴从而x+y与x-y均为偶数，
又∵x+y，x-y是88的因数，
∴有$\left\{\begin{array}{l}{x-y=2}\\{x+y=44}\end{array}\right.$或$\left\{\begin{array}{l}{x-y=4}\\{x+y=22}\end{array}\right.$或$\left\{\begin{array}{l}{x-y=-2}\\{x+y=-44}\end{array}\right.$或$\left\{\begin{array}{l}{x-y=-4}\\{x+y=-22}\end{array}\right.$或$\left\{\begin{array}{l}{x-y=44}\\{x+y=2}\end{array}\right.$或$\left\{\begin{array}{l}{x-y=22}\\{x+y=4}\end{array}\right.$或$\left\{\begin{array}{l}{x-y=-44}\\{x+y=-2}\end{array}\right.$或$\left\{\begin{array}{l}{x-y=-22}\\{x+y=-4}\end{array}\right.$，
由此可解得$\left\{\begin{array}{l}{x=43}\\{y=41}\end{array}\right.$或$\left\{\begin{array}{l}{x=13}\\{y=9}\end{array}\right.$或$\left\{\begin{array}{l}{x=-43}\\{y=-41}\end{array}\right.$或$\left\{\begin{array}{l}{x=-13}\\{y=-9}\end{array}\right.$或$\left\{\begin{array}{l}{x=23}\\{y=-21}\end{array}\right.$或$\left\{\begin{array}{l}{x=13}\\{y=-9}\end{array}\right.$或$\left\{\begin{array}{l}{x=21}\\{y=-23}\end{array}\right.$或$\left\{\begin{array}{l}{x=-13}\\{y=-17}\end{array}\right.$．

点评 本题主要考查了一次不定方程（组），本题是求不定方程的整数解，先将方程做适当变形，然后列举出其中一个未知数的适合条件的所有整数值，再求出另一个未知数的值．