TITLE: Neat solutions to tedious questions
QUESTION [11 upvotes]: In mathematics there are often two ways of answering a question,

Lots of thinking and little working,
Little thinking and lots of working.

I was wondering what the most extreme cases of this would be? Questions where the "usual" answer is just to plough through the working and hope everything turns out all right, but a much more subtle solution exists (and ideally this subtle solution would point out what makes the question "tick", as it were).
For example,
Question: Prove that the symmetric difference operation is associative.
The symmetric difference of two sets, $S \triangle T$, is defined to be $(S\cup T)\setminus (S \cap T)$. Now, one could simply plough through the working, but the working is tedious and there is a "trick" half way through it. However, a much more subtle solution exists. It requires a bit of setting-up, but is overall much neater.
Proof: The idea is that $S\triangle T$ "looks like" addition mod $2$. To see this, notice that,

If $x\in S$, $x\in T$ then $x\not\in S\triangle T$, which corresponds to $1+1=0 \text{ mod }2$
If $x\in S$, $x\not\in T$ then $x\in S\triangle T$, which corresponds to $1+0=1\text{ mod }2$
If $x\not\in S$, $x\in T$ then $x\in S\triangle T$, which corresponds to $0+1=1\text{ mod }2$
If $x\not\in S$, $x\not\in T$ then $x\not\in S\triangle T$, which corresponds to $0+0=0\text{ mod }2$

This means that showing $(S\triangle T)\triangle U=S\triangle (T\triangle U)$ now comes down to showing that addition mod $2$ is associative. And it is, which is easily checked. So we're done.

REPLY [5 votes]: Here's another example that's even about associativity. On any (projective) algebraic curve with (affine) equation
$$y^2 = x^3 + ax^2 + bx + c$$
it is possible to define a group multiplication (see elliptic curve) on the points of the curve uniquely defined as follows: given two points $P, Q$, their sum $R$ is defined to be $(x, -y)$ where $(x, y)$ is obtained by drawing a line through $P, Q$ and taking the (unique, if multiplicities are taken into account) third point on the curve intersecting the line. 
It is not at all obvious from the above definition that this operation is associative. There is a messy geometric proof and a possibly even messier algebraic proof, but a (to my mind) much cleaner proof is to prove that the group defined above is isomorphic to the divisor class group of the curve (the group of degree-zero divisors modulo the group of principal divisors), which is defined in such a way that it is obviously a group. The isomorphism sends a point $P$ to the divisor $P - O$ where $O$ is the point at infinity.

REPLY [5 votes]: Elementary probability has lots of such questions.  One of my favourites: 
Shuffle a standard deck of cards (52 cards, of which four are aces).  What is the probability that the card directly below the second ace from the top is another ace?
[ For those who might want to puzzle it out, I'll wait a day or two before posting the elegant solution ]