I’m late, but I’m quite proud of this proof for #4:
Call the large triangle a graph and the n2 triangles simply triangles. First, note that for any size, there is a graph where the top node is colored red, the remaining nodes on the right diagonal are colored green, and all nodes not on the right diagonal are colored blue. This graph meets the conditions, and has exactly one trichromatic triangle, namely the one at the top (no other triangle contains a red node). It is trivial to see that this graph can be changed into an arbitrary graph by re-coloring finitely many nodes. This will affect up to six triangles; we say that a triangle has changed iff it was trichromatic before the recoloring but not after, or vice versa, and we shall show that re-coloring any node leads to an even number of triangles being changed. This proves that the number of trichromatic triangles never stops being odd.
Label the three colors x,y, and z. Let v be the node being recolored, wlog from x to y. Suppose first that v has six neighbors. It is easy to see that a triangle between v and two neighbors has changed if and only if one of the neighbors has color z and the other has color x or y. Thus, we must show that the number of such triangles is even. If all neighbors have color z, or if none of them do, then no triangles have changed. If exactly one node has color z, then the two adjacent triangles have changed. Otherwise, let j and k denote two different neighbors of color z. There are two paths using only neighbors of v between j and k. Both start and end at a node of color z. By the 1-D Sperner Lemma (assuming the more general result), it follows that both paths have an even number of edges between nodes of color z and {x,y}; these correspond to the triangles that have changed.
If v is a node on one of the graph’s boundaries changing color from x to y, it has exactly 4 neighbors and three adjacent triangles. The two neighbors that are also on the boundary cannot have color z, so either none, one, or both of the ones that aren’t do. If it’s none, no triangle has changed; if it’s one, the two neighboring triangles have changed; and if it’s both, then the two triangles with two nodes on the graph’s boundary have changed.
I’m late, but I’m quite proud of this proof for #4:
Call the large triangle a graph and the n2 triangles simply triangles. First, note that for any size, there is a graph where the top node is colored red, the remaining nodes on the right diagonal are colored green, and all nodes not on the right diagonal are colored blue. This graph meets the conditions, and has exactly one trichromatic triangle, namely the one at the top (no other triangle contains a red node). It is trivial to see that this graph can be changed into an arbitrary graph by re-coloring finitely many nodes. This will affect up to six triangles; we say that a triangle has changed iff it was trichromatic before the recoloring but not after, or vice versa, and we shall show that re-coloring any node leads to an even number of triangles being changed. This proves that the number of trichromatic triangles never stops being odd.
Label the three colors x,y, and z. Let v be the node being recolored, wlog from x to y. Suppose first that v has six neighbors. It is easy to see that a triangle between v and two neighbors has changed if and only if one of the neighbors has color z and the other has color x or y. Thus, we must show that the number of such triangles is even. If all neighbors have color z, or if none of them do, then no triangles have changed. If exactly one node has color z, then the two adjacent triangles have changed. Otherwise, let j and k denote two different neighbors of color z. There are two paths using only neighbors of v between j and k. Both start and end at a node of color z. By the 1-D Sperner Lemma (assuming the more general result), it follows that both paths have an even number of edges between nodes of color z and {x,y}; these correspond to the triangles that have changed.
If v is a node on one of the graph’s boundaries changing color from x to y, it has exactly 4 neighbors and three adjacent triangles. The two neighbors that are also on the boundary cannot have color z, so either none, one, or both of the ones that aren’t do. If it’s none, no triangle has changed; if it’s one, the two neighboring triangles have changed; and if it’s both, then the two triangles with two nodes on the graph’s boundary have changed.