we have
$x^2-xy+y^2 = 13$
we can complete square by addition of some thing from $y^2$ to $x^2-xy$
to get $(x^2-xy + \frac{y^2}{4}) + \frac{3y^2}{4} = 13$
or $(x-\frac{y}{2})^2 + \frac{3y^2}{4} = 13$
or multiplying by 4 we get $(2x-y)^2 + 3 y^2 = 52$
as it is sum of positive numbers we get $3y^2 <=52$ or $ y <=4$'
$y =1$ gives $(2x-y)^2 = 49$ or $2x-y=7$ as -7 shall give -ve x so x = 4, y= 1
$y=2$ gives $(2x-y)^2= 40$ not a square
$y=3$ gives $(2x-y)^2 = 25$ or $2x-y = 5$ giving x = 4 , y = 3 and $2x-y = -5$ gives -ve x
$y=4$ gives (2x-y)^2 = 4 $, $2x-y=2$ gives x = 3, y = 4
$2x-y=-2$ gives x = 1, y = 4
So we have solutions (4,1),(4,3),(1,4),(1,3)
