求方程（2x-1)(2y+5)=2009的正整数解

2009＝1×2009
2009＝7×287
2009＝41×49
2x-1=1
2y+5=2009
x=1,y=1002
2x-1=7
2y+5=287
x=4,y=141
2x-1=41
2y+5=49
x=21,y=22
2x-1=49
2y+5=41
x=25,y=18
2x-1=287
2y+5=7
x=144,y=1

2009=7x7x41
因2y+5>=7
因此正整数解有：
2y+5=7, 2x-1=287, 得：y=1, x=144
2y+5=41, 2x-1=49 得：y=18, x=25
2y+5=49, 2x-1=41 得：y=22, x=21
2y+5=287, 2x-1=7得：y=141, x=4
2y+5=2009, 2x-1=1得：y=1002, x=1

（2x-1)(2y+5)=2009=41*7*7=41*49=287*7=2009*1
2x-1=41
2y+5=49,x=21,y=22
2x-1=49
2y+5=41,x=25,y=18
2x-1=287
2y+5=7,x=144,y=1
2x-1=7
2y+5=287,x= 4,y=141
2x-1=1
2y+5=2009,x=1,y=1002

上面说的不错