Well if [math]x+y=xy[/math] then [math]y=xy-x=x(y-1)[/math] and so as long as [math]y[/math] isn't equal to 1, this gives [math]x=\dfrac{y}{y-1}[/math]. We then manipulate this fraction slightly as so: [math]\dfrac{y}{y-1}=\dfrac{y-1+1}{y-1}=1+\dfrac{1}{y-1}=x[/math]. Since [math]x[/math] is an integer, [math]1+\dfrac{1}{y-1}[/math] is an integer, and so [math]\dfrac{1}{y-1}[/math] must be an...
Không có nhận xét nào:
Đăng nhận xét