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MathNerd
#13
Feb19-04, 07:41 AM
P: n/a
Integer solutions of [tex] xy = x + y [/tex] where [tex] x,y \epsilon Z [/tex]

[tex] xy - x - y = 0 [/tex]

[tex] (x-1)(y-1) = 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]