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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.

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2 is the oddest prime of them all... :) – J. M. Sep 8 '11 at 19:09
I recommend checking out the Mathoverflow questions… and – Jonas Meyer Sep 8 '11 at 19:10
Did you mean to say $2-1$ divides $2$? For no $n$ (except $1$) does $n$ divide $n-1$ – Ross Millikan Sep 8 '11 at 19:11
Well, there's Optimus Prime. – Curiosity Sep 8 '11 at 19:22
I think there might be something to the idea that $2$ is an exceptional prime. But what I cannot understand is some people attempt to justify it by saying $2$ is the only even prime. Well, true; but isn't $17$ the only prime divisible by $17$? (I find Iasafro's answer more acceptable than this :-).) – Srivatsan Sep 8 '11 at 19:23
up vote 5 down vote accepted

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.

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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.

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I will say that the reason why $2$ is special is that $2$ is the smallest prime in the natural numbers.

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I downvoted this, because I found this a very strange reason before reading Buzzard's and Carnahan's answers on MO. Now that I want to take it back, the system is not allowing me to do so saying that my vote is locked in until the answer is edited. :( – Soarer Sep 8 '11 at 19:44
Don't worry, we all are humans ;) – Josué Tonelli-Cueto Sep 8 '11 at 19:52
@Soarer: I made a trivial edit to allow undownvoting. – Jonas Meyer Sep 8 '11 at 20:22
@Jonas: Done. $\hspace{1mm}$ – Soarer Sep 8 '11 at 20:59

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).

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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.

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