TITLE: $-1$ as the only negative prime.
QUESTION [46 upvotes]: I was recently thinking about prime numbers, and at the time I didn't know that they had to be greater than $1$. This got me thinking about negative prime numbers though, and I soon realized that, for example, $-3$ could not be prime because $3 \cdot (-1) = -3$. In some sense $-1$ could be though because its only factors that are integers are $-1$ and $1$, and this is allowed for primes. Is there some way, by this logic, that $-1$ can be considered a prime then?

REPLY [5 votes]: I think the answers so far don't give the OP's intuitions enough credit. There is, indeed, something quite special about the set $$X = \{-1,2,3,5,7,\ldots\},$$ and the OP deserves an answer that makes this explicit. I'll do my best to give one.
Firstly, observe the following.

Every non-zero integer can be written as a product of the elements of $X.$
If $Y$ is a proper subset of $X$, then it's not the case that every non-zero integer can be written as a product of the elements of $Y$.

In other words, $X$ is a minimal generating subset for the monoid $(\mathbb{Z}_{\neq 0}, \times,1).$
Of course, so too is the subset $$\{-1,-2,3,5,7,\ldots\}.$$ It follows that $X$, as defined above, isn't altogether that special. Nonetheless, the presence of $-1$ is essential:

Proposition. Every minimal generating subset of $\mathbb{Z}_{\neq 0}$ contains $-1$.

So $-1$ is actually pretty important.
At this point, it's worth observing something cool about $\mathbb{N}_{\neq 0} = \{1,2,3,4,\ldots\}.$ In particular, notice that $\mathbb{N}_{\neq 0}$ has a unique minimal generating subset, namely the (usual) prime numbers.
Finally, there's something else cool about $-1$ that makes it a bit primelike. If $p \in \mathbb{N}$ is a prime number in the usual sense and $n$ is an integer, we can ask: "what is the minimum number of copies of $p$ needed to build $n$?" This is usually denoted $\nu_p(n).$ Following this, it makes sense to make the following definition. Suppose $Z$ is a subset of $\mathbb{Z}_{\neq 0}$ that generates it multiplicatively. Suppose also that $q \in Z$. Then it makes sense to write $\nu_q^Z(n)$ for the least number of occurrences of $q$ in any factorization of $n$ as a product of elements of $Z$.
Now recall that we defined $$X = \{-1,2,3,5,7,\cdots\}.$$
It follows that given a non-zero integer $n$, the expression $\nu_{-1}^X(n)$ equals $0$ if $n$ is positive, and it equals $1$ if $n$ is negative. We can also define $$Y = \{-1,-2,-3,-5,-7,\cdots\}.$$ It follows that given an integer $n>0$, the expression $\nu_{-1}^Y(n)$ equals $0$ if the number of primes in the factorization of $n$ is even, and it equals $1$ if the number of primes in the factorization of $n$ is odd.