Question

12．求方程2x²-2xy+2y²-4x-4y+6=0的整數解．

Answer 1

分析 將方程變形為（x-y）²+（x-2）²+（y-2）²=2，根據整數解的定義分三個代數式x-y，x-2，y-2中兩個絕對值為1，一個為0，可得方程組$\left\{\begin{array}{l}{x-y=-1}\\{x-2=-1}\\{y-2=0}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=-1}\\{x-2=0}\\{y-2=-1}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=-1}\\{x-2=1}\\{y-2=0}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=-1}\\{x-2=0}\\{y-2=1}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=1}\\{x-2=-1}\\{y-2=0}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=1}\\{x-2=0}\\{y-2=-1}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=1}\\{x-2=1}\\{y-2=0}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=1}\\{x-2=0}\\{y-2=1}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=0}\\{x-2=-1}\\{y-2=-1}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=0}\\{x-2=-1}\\{y-2=1}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=0}\\{x-2=1}\\{y-2=-1}\end{array}\right.$，$\left\{\begin{array}{l}{x-y=0}\\{x-2=1}\\{y-2=1}\end{array}\right.$，解方程組求得x，y的值．

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

點評 本題考查了非一次不定方程（組）中方程整數解的求法：把方程進行變形，使方程左邊分解為含未知數的3個式子，右邊為常數，然后利用整數的整除性求解．