Suppose that there is $S$, a finite set of unit squares. So, $S$ is chosen from a larger grid of unit squares. The unit squares of $S$ are tiled with isoceles right triangles. Each of these triangles has a hypotenuse of $2$. There are $3$ conditions that these triangles meet: none of the triangles overlap each other, none extend past the finite set $S$, and all of $S$ is covered fully by the triangles. Also, the hypotenuse of each triangle lies along any grid line. Also, the vertices of the trianlgles lie at the corners of the squares. Can someone help me prove that the number of triangles there are, have to be of a multiplicity of $4$?
I draw a directed graph. For each unit square we use a vertex. Now start at any vertex and create a Hamiltonian directed cycle, where A->B if one half-square of A and one half-square of B create a hypotenuse-2 triangle.
We obtain a polygon that is the Hamiltonian directed cycle. If a side of the polygon is between two turns in opposite directions, it has even length, otherwise it has odd length.
Now for each 270º corner, substitute it by three 90º corners, this will add 4 units to the total perimeter so the total perimeter modulo 4 hasn't changed. Now we have a (maybe self-intersecting and non-convex) polygon where all the corners are 90º. All the sides measure an odd integer. Now it is trivial that the perimeter is a multiple of 4, because the polygon is equivalent to a rectangle with all sides odd.
Here are the major steps. You can look at them one at a time.
Step 1
Follow the chain of shorter sides of the triangles, since each triangle's shorter side must be shared with another triangle. The chains must be finite and don't overlap. [You already did something similar.]
Step 2
The chain of shorter sides must alternate in direction and form a closed polygon with every angle being 90 or 270 degrees.
Step 3
The polygon has the same number of sides of each direction. And the number of sides of each direction is even.