$-1$ as the only negative prime.

Solution 1:

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.

Solution 2:

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 $(-3)\cdot(-1)$

Solution 3:

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 $2$.

Solution 4:

If you think negative numbers are weird, try imaginary numbers.

There's not really a good way to think of positive numbers. You could say an imaginary number is positive if its real part is positive, but then you still get results like $(1+2i)(1+2i) = -3+4i$ so you immediately lose the fact that the product of two positive numbers is positive. That means once you start thinking about unique factorization you can escape the "positive" numbers very easily, and you're back to the problem you started with.

Unique factorization seems to fail, because $17 = (4+i)(4-i) = (-1+4i)(-1-4i)$. But does that really count? Play with more complicated examples, you'll see the pattern that the factorizations you write down seem suspiciously like each other.

I would say the simplest reason $6 = 2\cdot3 = -2\cdot-3$ doesn't violate prime factorization is we are interested in finding structure in numbers, and if that means our unique factorization theorem must protect itself from "noise" that the negative numbers introduce. It should be clear that the negative signs are obfuscating structure, not true counterexamples.

If you play with the complex integers you will find that the "noise" that interferes with unique factorization are $1$ $-1$, $i$ and $-i$. These are the units: the numbers that divide $1$, and therefore divide anything however well they please.

The natural numbers in the integer get to cheat a little bit because they are closed under multiplication, so they have enough mathematical structure to be factored neatly and avoid the unit-noise without any extra work. But once we look at other number systems we have to think about units and watch out for how they interfere with factorizations.

Also, we do not say $-1$ is prime because we do not want primes to include numbers that divide every number. We need prime numbers to describe the multiplicative structure of the integers or of another ring, and units are the trivial part of this structure and need to be treated separately.