Size of closed loop on a (bipartite) hexagonal lattice with equal number of enclosed A and B sublattice sites. If I draw closed loops on a hexagonal lattice such that it always encloses equal number of A and B sublattice sites, I seem to get loops of sizes 4n+2.
Is there a way this can be proved in general?

Shown in figure is a sample loop (in red) enclosing one A and one B sublattice points. The loop size is 14. (4n+2 | n=3).
 A: Yes, this is true. In fact this is also true for any closed loops which enclose an even number of interior lattice points.
Let $p$ be the number of edges of a nontrivial simple loop on a hexagonal lattice. Let $k$ be the number of interior hexagons, and $x$ be the number of interior lattice points.
Lemma: $2x+p-2 = 4k$.
We can simply prove this by induction over $k$. Clearly the statement is true for a loop containing exactly one hexagon. Now consider a loop containing more than one hexagon. Choose an interior hexagon that touches the loop in exactly one connected subpath, i.e. an interior hexagon that does not separate the interior of the loop. Such a hexagon always exists.
Now consider changing the path to exclude that hexagon, leaving all nonadjacent path edges intact. In doing so, $p$ changes by $-4, -2, 0, 2$ or $4,$ and $x$ changes accordingly, namely by $0, -1, -2, -3$ and $-4$ respectively. This means that $2x+p-2$ always changes by $-4$, just like $4k$ changes by $-4$. Apply the lemma to the new path (by induction), and we get $2x+p-2-4 = 4k-4$, hence $2x+p-2 = 4k$. $\square$
Corollary: $p$ is always even, and $x$ is even iff $p/2$ is odd. This answers your question.
EDIT: There is a much nicer way to prove the lemma which does not require induction or involve postulating the existence of a hexagonal face with certain properties:
Let $z$ denote the number of obtuse loop vertices, and let $y$ denote the number of reflex loop vertices. We know that $z+y = p$ and that $z-y = 6$.
Consider cutting each interior hexagon into 12 right triangles by cutting along all its axes of symmetry. We can count the total number $t$ of such right triangles in two ways: $t = 12k$, but also $t = 6x + 4y + 2z$.
So we have $12k = t = 6x+4y+2z = 6x+3(y+z)+(y-z) = 6x+3p-6$. Dividing by three yields $4k = 2x + p - 2$. $\square$
