I have a solution to the following brainteaser, which I think is the correct answer, but I haven't been able to come up with a way to prove that it's the right answer. I know very little about graph theory, and even less about Ramsey theory, but this feels applicable.
You are camping with your friends. You have a flashlight for any emergency and have brought 8 batteries along with you. Your brother calls to tell you that four of those batteries are already dead. Your flashlight requires two working batteries to run. What is the least number of pairs you will need to test to guarantee that you can get the flashlight on?
Note that the final time you load the working batteries into flashlight counts as a test, even if you already know that the pair works.
My solution is as follows (spoiler):
Partition the batteries into four disjoint pairs. Test each pair. If any of them work, great, you're done. If none of them work, you know that each pair has exactly one good battery, and one bad battery.
Now take two of the four pairs. Test each battery from one pair with each battery of the other pair. I.e., if the pairs are $(A,B)$ and $(C,D)$, then test $(A,C)$, $(A,D)$, $(B,C)$, and $(B,D)$. Since one of $A$ and $B$ works, and one of $C$ and $D$ work, one of these four trials will work.
So the total number of tests is eight, and you're guaranteed to find at least one pair of good batteries.
Note: Sisi has offered a better solution in the comments which can do it in seven tests.
I know my solution works to find a working pair, but how would I prove that there isn't a way to do it in fewer tries (or, if there is a better solution, how would you prove that it is the best solution)?