What is the motivation for finding this solution? I was looking at Putnam problem A1 from 1995:

Let S be a set of real numbers which is closed under
  multiplication (that is, if a and b are in S, then so is ab).
  Let T and U be disjoint subsets of S whose union is
  S. Given that the product of any three (not necessarily
  distinct) elements of T is in T and that the product of
  any three elements of U is in U, show that at least one of the two subsets T,U is closed under multiplication.

I had tried a few different things, but I couldn't solve it and I looked up the solution:

Suppose on the contrary that there exist $t_1$, $t_2$ ∈ T
  with $t_1t_2$ ∈ U and $u_1,u_2$ ∈ U with $u_1u_2$ ∈ T. Then
  $(t_1t_2)u_1u_2$ ∈ U while $t_1t_2(u_1u_2)$ ∈ T, contradiction.

My question is, what train of thought should I have been following to find this (or another) solution? I of course realized myself that if if $t_1 t_2 \not \in T,$ then  $t_1 t_2 \in U$. But I never thought to take elements from both sets simultaneously and multiply all four elements at the same time. Even if I ha thought of randomly multiplying all four terms, I'm not sure I would've realized the contradiction until I stared at it for a while (I didn't even think it was a good idea to try to prove this one by contradiction).
 A: First of all, contradiction is a very natural approach to take here because you are trying to prove a disjunction.  One of the most common ways to prove "$P$ or $Q$" (without proving $P$ or proving $Q$ individually) is to assume that both fail and get a contradiction.  The reason this is a good strategy is that you can't really hope to prove $P$ or $Q$ directly (how could you, since you can't prove either one of the statements individually?).  On the other hand, the assumption that both fail is something that is often easy to use.
Now when doing a proof by contradiction, you start by taking elements from both sets simultaneously, since what you are trying to prove is that either $T$ or $U$ is closed under multiplication.  The negation of this is that neither $T$ nor $U$ is closed under multiplication, so to get a contradiction you would assume you have a counterexample to both.  This gives you the elements $t_1,t_2,u_1$, and $u_2$ as in the solution.  There's not really anything you can do with these elements other than multiply them in some way, and so it's natural enough to consider $t_1t_2u_1u_2$.  Also, note that you haven't yet used the assumption that $T$ and $U$ are disjoint, so it is natural to try to find some element that is forced to be in both $T$ and $U$.  And in order to do so, the obvious thing to try to use is that $T$ and $U$ are closed under triple products.  And given $t_1,t_2,u_1$, and $u_2$, the only three elements you know of in $T$ are $t_1,t_2$, and $u_1u_2$, and similarly for $U$.
Ultimately, I doubt there's any deep secret that makes this solution obvious: this is a Putnam problem, and it's meant to force you to be clever.  But I hope I've explained why the steps of the solution are reasonably natural things to try.
