Integer solutions of [tex] xy = x + y [/tex] where [tex] x,y \epsilon Z [/tex]
[tex] xy  x  y = 0 [/tex]
[tex] (x1)(y1) = 1 [/tex]
let [tex] x' = x 1[/tex] and [tex] y' = y  1 [/tex] so [tex] x',y' \epsilon Z [/tex]
So [tex] x' = 1 / y' [/tex]
The only integers whose reciprocals are also integers are 1 and 1
So [tex] y' = 1 [/tex] and [tex] x' = 1 [/tex]
So [tex] y = 2 [/tex] and [tex] x = 2 [/tex]
AND
So [tex] y' = 1 [/tex] and [tex] x' = 1 [/tex]
So [tex] y = 0 [/tex] and [tex] x = 0 [/tex]
Therefore there are only two solutions to this diophantine equation [tex] x = y = 2 [/tex] and [tex] x = y = 0 [/tex]
