Find the minimum value of $f=x^2+y^2+x(1-y)+y(1-x)$, holds on $x,y$ are integers. Let $x$ and $y$ are integers such that $-2\le x\le3$ and $-8\le y\le4$
Find the minimum value of $f=x^2+y^2+x(1-y)+y(1-x)\tag{1}$
From $(1)$, I get $f=(x-y)^2+x+y$ and I don't know what I should do next.
I had tried to use differential of f then I get no critical point (the minimum is on the bound $x=-2,x=3,y=-8$ or $y=4$). I found the minimum on $x=-2$ but $y=-2.5$ that isn't integer, the equation f that holds on $x=-2$ is $(y+2.5)^2-4.25,$ So no conclusion for $x,y$ are integers or Can it conclude that $y=2,3$ minimize f ? (I think this solution is suitable for real numbers).
I had tried to substitute all the possible cases then I got the minimum is $-4$, when $(x,y)$=$(-2,-2)$, $(-2,-3)$.  
All help would be appreciated.

Brute-forcing in Haskell:
Prelude> let f x y = x**2 + y**2 + x*(1-y) + y*(1-x)
Prelude> let graph = [ ((x,y),f x y) | x <- [-2..3], y <- [-8..4] ]
Prelude> minimum (map snd graph)
-4.0
Prelude> filter (\((x,y),z)->(z==(-4.0))) graph
[((-2.0,-3.0),-4.0),((-2.0,-2.0),-4.0)]

Thus, the minimum is $-4$ and the minimizers are indeed $(-2,-3)$ and $(-2,-2)$.
