Question

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

Answer 1

$x^2+y^2=2(x+y)+xy$
$\Rightarrow x^2-x(2+y)+y^2-2y=0$
It is a quadratic equation in the variable $x$.
Discriminant of the above equation is given by
$D=(2+y{)}^2-4(1\left)\right(y^2-2y)$
$=4+12y-3y^2=16-3(y-2{)}^2$
Since $x$ is an integer, $D$ must be a perfect square and thus it is non-negative.
Hence, $16-3(y-2{)}^2\ge 0$
$\Rightarrow (y-2{)}^2\le \frac{16}{3}$
$\Rightarrow |y-2|\le \frac{4}{\sqrt{3}}<3$
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 $x^2-2x=0$
$\Rightarrow x=0,2$
When $y=2,$
we have $x^2-4x=0$
$\Rightarrow x=0,4$
When $y=4,$
we have $x^2-6x+8=0$
$\Rightarrow x=2,4$
Thus, there are $6$ solutions,
$(0,0),(2,0),(0,2),(4,2),(2,4),(4,4)$