求方程2x^2-xy-3x+y+2008=0的正整数解

2(x^2)-xy-3x+y+2008=0可化为：（x-1)*y=2(x^2)-3x+2008;故：y=(2(x^2)-3x+2008)/(x-1) =2x-1+{2007/(x-1)}因此,当x-1是2007的约数时且x取值是正整数时即可满足题意.这其中x不等于1.当x=1时,2*1-y-3+y+2008=2007≠0,...