This is a problem from Apostol calculus vol I. It originally asks to prove that the area of any right triangle with vertices at lattice points have area given by $$Z+\frac12B-1$$ where $Z$ is the number of lattice points lying inside the triangle, $B$ the number of lattice points lying on the perimeter.
It is particularly easy to prove that if the triangle has the perpendicular and base parallel to the axes.
My question is:
Is it at all possible for triangles with lattice points as vertices to have a different configuration of the sides (i.e., base and perp. not parallel to the axes)? If not, how do i prove it?