Question

12．求方程2x²-2xy+2y²-4x-4y+6=0的整数解．

Answer 1

分析 将方程变形为（x-y）²+（x-2）²+（y-2）²=2，根据整数解的定义分三个代数式x-y，x-2，y-2中两个绝对值为1，一个为0，可得方程组$\left\{\begin{array}{l}{x-y=-1}\\{x-2=-1}\\{y-2=0}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=-1}\\{x-2=0}\\{y-2=-1}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=-1}\\{x-2=1}\\{y-2=0}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=-1}\\{x-2=0}\\{y-2=1}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=1}\\{x-2=-1}\\{y-2=0}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=1}\\{x-2=0}\\{y-2=-1}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=1}\\{x-2=1}\\{y-2=0}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=1}\\{x-2=0}\\{y-2=1}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=0}\\{x-2=-1}\\{y-2=-1}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=0}\\{x-2=-1}\\{y-2=1}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=0}\\{x-2=1}\\{y-2=-1}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=0}\\{x-2=1}\\{y-2=1}\end{array}\right.$，解方程组求得x，y的值．

解答 解：2x²-2xy+2y²-4x-4y+6=0，
（x-y）²+（x-2）²+（y-2）²=2，
∵求方程的整数解，
∴$\left\{\begin{array}{l}{x-y=-1}\\{x-2=-1}\\{y-2=0}\end{array}\right.$，解得$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$；
$\left\{\begin{array}{l}{x-y=-1}\\{x-2=0}\\{y-2=-1}\end{array}\right.$，无解；
$\left\{\begin{array}{l}{x-y=-1}\\{x-2=1}\\{y-2=0}\end{array}\right.$，无解；
$\left\{\begin{array}{l}{x-y=-1}\\{x-2=0}\\{y-2=1}\end{array}\right.$，解得$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$；
$\left\{\begin{array}{l}{x-y=1}\\{x-2=-1}\\{y-2=0}\end{array}\right.$，无解；
$\left\{\begin{array}{l}{x-y=1}\\{x-2=0}\\{y-2=-1}\end{array}\right.$，无解；
$\left\{\begin{array}{l}{x-y=1}\\{x-2=1}\\{y-2=0}\end{array}\right.$，解得$\left\{\begin{array}{l}{x=3}\\{y=2}\end{array}\right.$；
$\left\{\begin{array}{l}{x-y=1}\\{x-2=0}\\{y-2=1}\end{array}\right.$，无解；
$\left\{\begin{array}{l}{x-y=0}\\{x-2=-1}\\{y-2=-1}\end{array}\right.$，解得$\left\{\begin{array}{l}{x=1}\\{y=1}\end{array}\right.$；
$\left\{\begin{array}{l}{x-y=0}\\{x-2=-1}\\{y-2=1}\end{array}\right.$，无解；
$\left\{\begin{array}{l}{x-y=0}\\{x-2=1}\\{y-2=-1}\end{array}\right.$，无解；
$\left\{\begin{array}{l}{x-y=0}\\{x-2=1}\\{y-2=1}\end{array}\right.$，解得$\left\{\begin{array}{l}{x=3}\\{y=3}\end{array}\right.$．
故方程2x²-2xy+2y²-4x-4y+6=0的整数解为$\left\{\begin{array}{l}{x=1}\\{y=2}\end{array}\right.$；$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$；$\left\{\begin{array}{l}{x=3}\\{y=2}\end{array}\right.$；$\left\{\begin{array}{l}{x=1}\\{y=1}\end{array}\right.$；$\left\{\begin{array}{l}{x=3}\\{y=3}\end{array}\right.$．

点评 本题考查了非一次不定方程（组）中方程整数解的求法：把方程进行变形，使方程左边分解为含未知数的3个式子，右边为常数，然后利用整数的整除性求解．