If #4 is true, changing the color of a single node (except to forbidden colors on edges) cannot change the parity of the trichromatic triangle count, and this would be checkable through a finite case analysis of graphs of size <=7. Given that lemma, we can recolor one corner to red, the remainder of one large edge blue and the remaining nodes green, producing the odd count 1.
#8:
We are looking for a surface [0,1]²->[0,1] whose intersection with the plane x=y does not contain a function of t. It suffices to show that the intersection looks like the letter s, with exactly the endpoints reaching t=0 or t=1. It suffices to show that the intersection can be any continous function of x including the points t=x=y=0 and t=x=y=1. Within each plane of constant x, define the surface as 0 for small t, 1 for large t, and rapidly rising through the plane x=y wherever we want the intersection.
If #4 is true, it is provable:
If #4 is true, changing the color of a single node (except to forbidden colors on edges) cannot change the parity of the trichromatic triangle count, and this would be checkable through a finite case analysis of graphs of size <=7. Given that lemma, we can recolor one corner to red, the remainder of one large edge blue and the remaining nodes green, producing the odd count 1.
#8:
We are looking for a surface [0,1]²->[0,1] whose intersection with the plane x=y does not contain a function of t. It suffices to show that the intersection looks like the letter s, with exactly the endpoints reaching t=0 or t=1. It suffices to show that the intersection can be any continous function of x including the points t=x=y=0 and t=x=y=1. Within each plane of constant x, define the surface as 0 for small t, 1 for large t, and rapidly rising through the plane x=y wherever we want the intersection.