Question

1．求不定方程x²-5xy+6y²-3x+5y-11=0的整数解．

Answer 1

分析 原方程变形为：（x-2y+1）（x-3y-4）=7，根据整数的整除性得到$\left\{\begin{array}{l}{x-2y+1=1}\\{x-3y-4=7}\end{array}\right.$；$\left\{\begin{array}{l}{x-2y+1=-1}\\{x-3y-4=-7}\end{array}\right.$；$\left\{\begin{array}{l}{x-2y+1=7}\\{x-3y-4=1}\end{array}\right.$；$\left\{\begin{array}{l}{x-2y+1=-7}\\{x-3y-4=-1}\end{array}\right.$；从而求得x，y的值．

解答 解：x²-5xy+6y²-3x+5y-11
=（x-2y）（x-3y）-3x+5y-11
=[（x-2y）+1][（x-3y）-4]-7
=（x-2y+1）（x-3y-4）-7
=0，
（x-2y+1）（x-3y-4）=7，
则$\left\{\begin{array}{l}{x-2y+1=1}\\{x-3y-4=7}\end{array}\right.$，解得$\left\{\begin{array}{l}{x=-22}\\{y=-11}\end{array}\right.$；
$\left\{\begin{array}{l}{x-2y+1=-1}\\{x-3y-4=-7}\end{array}\right.$，解得$\left\{\begin{array}{l}{x=0}\\{y=1}\end{array}\right.$；
$\left\{\begin{array}{l}{x-2y+1=7}\\{x-3y-4=1}\end{array}\right.$，解得$\left\{\begin{array}{l}{x=8}\\{y=1}\end{array}\right.$；
$\left\{\begin{array}{l}{x-2y+1=-7}\\{x-3y-4=-1}\end{array}\right.$，解得$\left\{\begin{array}{l}{x=-30}\\{y=-11}\end{array}\right.$．
故不定方程x²-5xy+6y²-3x+5y-11=0的整数解为$\left\{\begin{array}{l}{x=-22}\\{y=-11}\end{array}\right.$；$\left\{\begin{array}{l}{x=0}\\{y=1}\end{array}\right.$；$\left\{\begin{array}{l}{x=8}\\{y=1}\end{array}\right.$；$\left\{\begin{array}{l}{x=-30}\\{y=-11}\end{array}\right.$．

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两个式子，右边为常数，然后利用整数的整除性求解．