Vtyjkyuk

Do we have negative prime numbers?

$..., -7, -5, -3, -2, ...$

I don't know why this question has a down vote, because it identifies a subtle point about arithmetic which becomes particularly significant when the notion of "prime" is extended to other contexts, and is relevant so far as the integers are concerned when looking at issues like unique factorisation.

When we first encounter prime numbers, we do so in the context of the positive integers. The significant point about this context is that $+1$ is the only unit (the only positive integer with an multiplicative inverse). So the question here does not really arise. Often, when the main focus of work is the positive integers, the word prime will be used to imply a positive integer.

As soon as we start to extend this to the integers, and in particular, to consider the integers as having the structure of a ring, we add in a second unit $-1$ with $(-1)^2=1$. Even in this context it is possible to define the prime numbers as positive integers without too much inconvenience.

But if we extend further and add $i$ with $i^2=-1$ as another unit - note that $icdot -i=1$, we are in a different world. For example, $2=(1+i)(1-i)$ and $(1+i)=i(1-i)$ so that $2=i(1-i)^2$. Are these factorisations of $2$ to be taken as the same or different?

So very soon, in the context of ring theory and the theory of algebraic integers, we start talking about prime ideals (initially thought of as all the multiples of some prime $p$ - but extended beyond that idea too - an ideal which consists of all the multiples of a single element is called principal). And it is somewhat natural, if the ideal is principal, to call the generator a prime element of the ring. However, the primes are then only identified up to multiplication by units - $1+i$ generates the same ideal as $1-i$. One of the reasons for using ideals is that the uniqueness of factorisation can be maintained in this larger context. In $mathbb Z$ both $2$ and $-2$ generate the same ideal.

A prime element of a ring is a nonunit $p$ with the property that if $p$ divides a product $ab$ then it divides $a$ or $b$. In the ring $mathbb Z$ of integers, this property is shared by the (positive) primes $2,3,5,7,11,ldots$ and also the negative primes $-2,-3,-5,-7,ldots$, and even by $0$. However, the term prime number is conventionally used only for the positve prime elements of $mathbb Z$, and there are good reasons for this convention, for example $15=3cdot 5=(-3)cdot(-5)$ shows that the prime factorization would be less unique than we are used to.

Some more context: In general ring theory one looks at prime elements or irreducible elements as the analogues of prime numbers.

A (nonzero) element $a$ of a ring $R$ is called prime if it generates a prime ideal, or equivalently, if it is not a unit and $a|cb$ implies $a|b$ or $a|c$, i.e. if $a$ divides a product, it divides one of the factors.

An element $ain R$ is called irreducible if it is not a unit and $a=bc$ implies $b$ or $c$ is a unit, i.e. it cannot be factored into two other nonunits.

In nice rings (such as $mathbbZ$), these coincide and one has a unique factorisation up to units ($-1,1$ in $mathbbZ$) of every element into prime/irreducible elements.

I just wanted to supply a specific quote from a text that defines prime numbers for negatives in the way that several of the other answers already affirm. This is from Hungerford's Abstract Algebra: An Introduction (sec. 1.3):

DEFINITION. An integer $p$ is said to be prime if $p ne 0, pm1$ and the only divisors of $p$ are $pm1$ and $pm p$.

EXAMPLE. $3, -5, 7, -11, 13,$ and $-17$ are prime, but $15$ is not (because $15$ has divisors other than $pm1$ and $pm15$, such as $3$
and $5$). The integer $4567$ is prime; to prove this from the definition
requires a tedious check of all its possible divisors.

It is not difficult to show that there are infinitely many distinct
primes (Exercise 25). Because an integer $p$ has the same divisors as
$-p$, we see that

$p$ is prime if and only if $-p$ is prime.

In this case the Fundamental Theorem of Arithmetic is stated in terms of both positive and negative prime factors (Theorem 1.1), with the standard natural-number statement given thereafter as a corollary (Corollary 1.2).

Note that this is before rings (et. al.) appear in this text (although of course they do later), so the definition is not restricted to that domain.

This is a question that is going to get lots of duplicates. That's how I came across it.

Let me answer your question with another question:

Is $-14$ prime?

Of course it isn't! Since $$frac-142 = -7,$$ for starters. It is also divisible by $-7, -2, 7$, as well as by its obvious divisors, $-1, 1, 14$ and itself. Clearly $-14$ is a composite number.

But is $-7$ prime? That's the tougher question. It is trivially divisible by itself as well as by $-1, 1, 7$, but it's not divisible by any other numbers (we're operating in the domain of the integers, $textbfZ$).

My terminology "obvious divisor" is of course not standard. What I mean by it is that if $n$ is any integer, we automatically know that it is divisible by itself, by $-n$ and by $1$ and $-1$.

Let's agree that if $n$ is divisible only by these numbers that we automatically know to be divisors but not by any others, then it is a prime number, but if it's divisible by these "automatic divisors" and by other integers (with all the divisors forming a finite set), then it is a composite number.

Invariably this causes people to whine about breaking the fundamental theorem of arithmetic. But if we recognize that $1$ and $-1$ are special numbers in that they are units, then the big and melodramatic problem of the fundamental theorem becomes irrelevant nonsense. In $textbfZ$, each nonunit nonzero number has unique factorization regardless of units.

Don't worry about the prime counting function either. If $x$ is a positive real number, then $pi(x)$ counts how many numbers are between $0$ and $x$. But if $x$ is negative, then $pi(x)$ counts how many numbers are between $0$ and $|x|$ or between $0$ and $x$; either way the answer will be the same.

Picture the prime numbers as dots on the real number line. They are symmetrical, with the positive primes reflected in the negative primes on the other side of $0$.

I had come across this page before, but today I'm here because of a duplicate, which I will now answer here.

Generally, the definition of prime numbers is all those natural numbers greater than 1, having only two divisiors [sic], the number itself and 1. But, can the negative integers also be thought of in the same way?

Almost, but not quite. We could say that a negative integer is prime if it has only two divisors ($-1$ and itself) among the negative integers.

This is very similar to the positive integers, for which we say that a positive integer is prime if it has only two divisors among the positive integers.

For example, 163 is divisible by 1 and itself. It's also divisible by $-163$ and $-1$, but we usually don't bother with that when we are trying to see if 163 is prime.

Similarly to check whether $-163$ is prime, we could try dividing it by $-2, -3, -5, -7, -11$. We don't need to try dividing by $-13$ because $$-13 < (-1) fracsqrt-163i.$$

More commonly though, what people usually do is multiply the number by $-1$ and check whether that is prime, and that way they sidestep the issue of imaginary numbers. For example, try Divisors[-163] in Wolfram Alpha.