Problem 7 Solution: By Our Powers Combined

Choose an arbitrary hero H. There are 23 other heroes that H can be paired with, and each pairing must have one of the three fusion types. Therefore, there must exist at least one fusion type such that at least 6 of H's pairings have that type. (In fact you only need 16 other heroes, or 17 including H, for this to be true.) Without loss of generality assume this is true of the targeted fusions.

Now consider any 6 heroes that have a targeted fusion when paired with H. Given these six heroes, form a complete graph with the six heroes as vertices. Color each pairing with a guided fusion red, each pairing with a clearing fusion blue, and the targeted fusions can be colored either red or blue. Then by the special case of Ramsey's Theorem that was previously mentioned, there must exist a group of three whose pairings all have the same color. Without loss of generality assume that this color is red. Now those three have only targeted and guided fusions between them. If you form a team consisting of those three and H, they still only have targeted and guided fusions (because adding H just adds targeted fusions) so they have no clearing fusions available. QED.

Note that the Ramsey's Theorem article also includes a "multicolour example" with three colors that, coincidentally, also requires 17 vertices. However this is not the same as our problem. In that problem they are looking for groups of three that have only one edge color, in this problem we are looking for groups of four that have two edge covers. In fact since R(4,4)=18 (as explained in the article) with 18 or more heroes it would be impossible to make sure every team of 4 had one of each fusion type even if there were only two fusion types.