Is there a single or best reason that 2 is an exceptional prime? I've recently been studying some elementary number theory, and I've frequently come across the fact that there are a fair number of results (the main one being the law of quadratic reciprocity) for which $2$ has to be treated as a special case, separately from the odd primes.
My question is why this is.  I can see that there are reasons in each proof that the theorem doesn't hold for $2$ (a pretty common one is the fact that $2-1$ is not divisible by $2$), but I'm curious if there's an overarching theoretical reason.  My best guess is that it might have to do with the fact that $\left ( \mathbb{Z}/2\mathbb{Z} \right )^\times$ is trivial, although that might just be a fancier way of saying that $2-1 = 1$.
I am, of course, open to the possibility that there are a variety of reasons that $2$ is an exceptional prime, but I'm curious what those are and if there's a best example, or a most common example.
 A: Many of the more meaningful exceptions seem to come about because $1\equiv -1$ modulo 2. So the reasoning "$x=-x$, therefore $x=0$" is not available in characteristic 2.
A: I will say that the reason why $2$ is special is that $2$ is the smallest prime in the natural numbers.
A: To pick up on your example, quadratic reciprocity has everything to do with roots of quadratic equations - and the presence of a 2 in the denominator of the quadratic formula $$ x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} $$ is what screws things up. 
Similarly, the correspondence between quadratic forms and bilinear forms involves dividing by 2, which screws things up at 2 in algebraic number theory, for example.
A: I believe that another important source of exceptionality for the prime $2$ is that if $n$ is odd, then in fact $n^2\equiv1\bmod4$ while for every prime $p>2$ there are numbers $n$ such that $n^2\equiv1\bmod p$ but $n^2\not\equiv1\bmod p^2$ (i.e., there are no squares $\equiv3\bmod4$).
Whether this, philosophically, is a "manifestation of smallness" I'm not entirely sure.
A: At a lesser elementary level, when the characteristic of a field is $2$, and only in this case, the relation between the tensor product $V\otimes V$ and the symmetric product ${\rm Sym}_2(V)$ is less straightforward. This has important implications for the theory of quadratic forms (which is so basic in Number Theory).
