I solved two math problems today. The solutions to both were uniquely disappointing.
The first problem was the first problem from IMO 1986:
Let be a number that is not or . Prove that out of the set , we can select two different numbers such that is not a square.
A quick check would show that and are all squares. This shows that of the two numbers that we select from , one of them has to be . We essentially need to prove that for , and are not all squares (a maximum of two of them might be. But all three cannot).
I tried figuring out some of the implications of all of them being squares, but couldn’t come up with anything concrete, as I didn’t write anything at all, and just tried to devise a strategy mentally. Ultimately, I came up with possibly the most mundane proof of all time.
Proof: Consider squares (I couldn’t come up with any contradiction up until ). The possible residues for squares are . Now my strategy was to plug in all possible residues in place of , and checking that in every such instance, at least one of and gave a residue that was not . This I could successfully verify. Hence proved.
The solution that I could find in the IMO compendium was much smarter, but involved some algebra that would obviously involve putting pen to paper.
The second question that I solved today was proving the Koszul formula, which states that
I’d seen proofs before in textbooks, that would involve writing down expansions, and then combining all of them in some clever way. The way that I came up with was pretty algorithmic, and perhaps even more elementary. It is simply the following: Take . Use the torsion free condition to write it as . Now use the formula for connection acting on a tensor to write as . Performing the two operations, we have . Performing this operation of first using the torsion free condition and then the formula for a connection on a tensor three times, we get the Koszul formula.