According to Pick's Theorem, the size of an area $A$ can be calculated by the sum of
the interior lattice points located in the polygon $i$ and the number of lattice points on the boundary placed on the polygons perimeter $b$ divided by two, minus 1.
My question is - can I use this sentence to prove that a polygon with an Area size $5$ has at least $6$ lattice points in its perimeter? (that the shape is actually lying on $6$ lattice points)? I'm asking this because when I set A = $5$ and b = $6$, I get the result that i= $3$ - but I couldn't draw a polygon with $3$ interior points.
A = i + b/2 -1
than for area size 5:
5 = i + 6/2 -1
i = 3
is it possible to draw a polygon with i=3 anyway ?