A couple of days ago I was at a nice seminar given by Christian Reiher, during which he told us about a short proof of the following special case of a theorem of Olson.

Theorem. Let $(a_1,b_1),\dots,(a_n,b_n)$ be a sequence of points in ${\mathbb{Z}}_p^2$ with $n\geq 2p-1$. Then there is a non-empty subset $A\subset\{1,2,\dots,n\}$ such that $\sum_{i\in A}(a_i,b_i)=(0,0)$.

The short proof wasn't short in absolute terms, but was short if you were prepared to accept the following result of Noga Alon, known as the combinatorial nullstellensatz.

Theorem. Let F be a field and let P be a polynomial in n variables $x_1,\dots,x_n$ over F. Let $x_1^{t_1}\dots x_n^{t_n}$ be a monomial of maximal total degree $t_1+\dots+t_n$ that occurs in P with a non-zero coefficient, and let $S_1,\dots,S_n$ be subsets of F such that $|S_i|>t_i$ for every i. Then there exist $s_i\in S_i$ such that $P(s_1,\dots,s_n)\ne 0.$

Once you have the combinatorial nullstellensatz, the special case of Olson's theorem (and I think the whole theorem) is reduced to a nice exercise: basically, once you sit down and think about it you quickly see that it makes sense to choose each $S_i$ to be {0,1}, and then a few simple tricks using Fermat's little theorem (the polynomial $1-x^{p-1}$ is zero if $x\ne 0$ and 1 otherwise) you can finish off quite easily.

This method is known as the polynomial method. My question is not how to apply the combinatorial nullstellensatz. It's how to recognise, when you see a problem, that the polynomial method might work. In this case, once you have that clue, it's easy to finish off. But how do you manage if there's nobody there to give you the clue?

I'm interested in this question in general: I've always found spotting that mathematical results can be used quite a difficult process -- somehow I have to do it for myself in one problem before I truly understand how to do it in other problems. And here's a good example where I have never spotted how to use the result. And had I been faced with the task of proving Olson's theorem, I don't think it would have occurred to me to use it.