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Question

How many ordered pairs of integers (x,y) satisfy the equation x2+y2=2(x+y)+xy?
(correct answer + 5, wrong answer 0)

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Solution

x2+y2=2(x+y)+xy
x2x(2+y)+y22y=0
It is a quadratic equation in the variable x.
Discriminant of the above equation is given by
D=(2+y)24(1)(y22y)
=4+12y3y2=163(y2)2
Since x is an integer, D must be a perfect square and thus it is non-negative.
Hence, 163(y2)20
(y2)2163
|y2|43<3
Since y is an integer, it follows that its only possible values are given by y=0,1,2,3,4.

When y=1,3,
D=13, which is not a perfect square.
Thus, y=0,2,4
When y=0,
we have x22x=0
x=0,2
When y=2,
we have x24x=0
x=0,4
When y=4,
we have x26x+8=0
x=2,4
Thus, there are 6 solutions,
(0,0),(2,0),(0,2),(4,2),(2,4),(4,4)

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