Geometric proof that a Pythagorean triple has a number divisible by $3$?

A slightly detailed(irrelevant) approach: If any of, m, n are divisible by 3, then we are done as $\ (a,b,c)=({ m }^{ 2 }-{ n }^{ 2 },\quad 2mn,\quad { m }^{ 2 }+{ n }^{ 2 }) $ and $\ 3|b $. Hence, assume both m and n aren't multiples of 3. Therefore $$\ m\equiv 1,2\quad (mod\quad3)\\ n\equiv 1,2\quad (mod\quad3)$$ This implies that $$\ { m }^{ 2 }\equiv { n }^{ 2 }\equiv 1(mod\quad 3) $$ Hence $$\ { m }^{ 2 }-{ n }^{ 2 }=a\equiv 0(mod\quad 3)$$

Simple graphical solution to the problem.

Red region: Has unit squares divisible by 3.

Green region: All unit squares together are a multiple of 3.

Yellow squares: Left out square(s).

If you have two squares of type 1, you will have 2 yellow squares, which don't fit into any type of square.