$-1$ as the only negative prime.

If we define prime so that $-1$ is prime, unique factorization into primes fails, since $6=3\cdot 2=3\cdot 2\cdot (-1)^2$. So it is not useful to define $-1$ as a prime.

When we get to higher math, we find that when we talk about "primes" in other systems, we are required to treat any pair of numbers that divide each other as "equivalent." That is, if $a$ divided $b$ and $b$ divides $a$, then we treat $a$ and $b$ as "equivalent" for the purposes of primeness and factorization.

In particular, any number that divides $1$ is equivalent to $1$, since $1$ divides everything. The numbers that are equivalent to $1$ are called "units."

In the integers, the only units are $+1$ and $-1$, so we can just avoid this complication by only talking about the positive integers. In other rings, we are not so lucky.

Your logic argument for $-1$ being the only negative prime:

• $-3$, for example, could not be prime because it is equal to $3\cdot(-1)$
• $-1$, however, is prime because it is divisible only by itself and by $1$

According to this logic, $+1$ is the only positive prime, since:

• $3$, for example, could not be prime because it is equal to $(-3)\cdot(-1)$

Instead of primes, consider the set $S = \{p^{2^n} \space\vert\space n,p \in \mathbb{N}, p \space \text{prime}\}$. These are the primes as well as the squares of primes, fourth powers, eighth powers, etc. Every positive integer can be represented uniquely as a product of distinct elements of $S$, in other words multiplication is a bijection between $\mathbb{N} \setminus \{0\}$ and the set of finite subsets of $S$, this can be seen by writing a prime factorization with each exponent in base $2$. Clearly, multiplication is also a bijection between $\mathbb{Z} \setminus \{0\}$ and the set of finite subsets of $S \cup \{-1\}$. So while $-1$ isn't quite like a prime it seems fair to say that it is like a prime to the power of a power of $2$.