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From my understanding it's not prime because it's not greater than $0$. So my followup question is why did mathematicians exclude $-1$?

The definition of prime is having only two factors.

$-1 \cdot 1 = -1$

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    $\begingroup$ An alternate definition of a prime number is that it is a positive integer larger than $1$ whose only factors are $1$ and itself. $\endgroup$ – N. F. Taussig Mar 4 '15 at 17:19
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    $\begingroup$ either $-1$ is prime or not, then so what ?! :) Prime numbers have more important works to do ! $\endgroup$ – Fardad Pouran Mar 4 '15 at 17:23
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    $\begingroup$ By this interpretation of the definition, $3 = 1·3 = (-1)·(-3)$ had four distinct factors and wasn’t prime. This is a definition only with respect to $ℕ$, the positive integers. All numbers involved are assumed to be positive integers. $\endgroup$ – k.stm Mar 4 '15 at 17:25
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    $\begingroup$ For interest, consider the notion of a prime element: en.wikipedia.org/wiki/Prime_element $\endgroup$ – Jonny Mar 4 '15 at 17:30
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    $\begingroup$ Note that $-1 = (-1)^3 = (-1)^5 = (-1)^7 = \cdots$, so we cannot really say that it has only two factors. $\endgroup$ – N. F. Taussig Mar 4 '15 at 17:35

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The usual definition of "prime" is chosen so that get uniqueness of prime factorizations. At least from this perspective, negative numbers aren't prime, because if they were then we would have, for instance, the "prime factorizations" $4=2 \cdot 2$ and $4=(-2) \cdot (-2)$. It is also why $1$ is not prime, because if it were then we would have the "prime factorizations" $4=2 \cdot 2$ and $4=2 \cdot 2 \cdot 1$ and $4=2 \cdot 2 \cdot 1 \cdot 1$ and ...

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    $\begingroup$ The factorization is unique up to unit multipliers. That's fine. Primes are nonunits by definition, so $1,-1$ are not primes. $\endgroup$ – Pedro Tamaroff Mar 4 '15 at 17:43
  • $\begingroup$ @PedroTamaroff I'm justifying the fact that the definition works the way it does. It would be consistent for it to work the other way, but then the fundamental theorem of arithmetic would have a rather awkward statement. $\endgroup$ – Ian Mar 4 '15 at 17:46
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    $\begingroup$ Not really. It is the usual statement that a ring is UFD. One gets used to it. =D $\endgroup$ – Pedro Tamaroff Mar 4 '15 at 17:48
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    $\begingroup$ @PedroTamaroff Eh...I think that is losing track of some motivation from the UFD that everyone knows about... $\endgroup$ – Ian Mar 4 '15 at 17:53
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    $\begingroup$ The statement that negative numbers aren't prime is a bit misleading conceptually, because by general definitions, if $\,p\,$ is a prime (or irreducible) then so to is $\,-p\ $ (or any other associate = $\,up,\,$ where $\,u\,$ is any unit = invertible). For convenience we simply choose the positive rep. Similar unit normalization can be done in some other domains. $\endgroup$ – Bill Dubuque Mar 4 '15 at 18:29
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By commonly used definitions, units (invertibles) such as $\,\pm 1\,$ are not considered primes, even though they satisfy the divisibility property $\,p\mid ab\,\Rightarrow\, p\mid a\,$ or $\,p\mid b\,$ that is key to proving the uniqueness of factorizations into irreducibles. Excluding units (and zero) as primes simplifies the statement of uniqueness of prime factorizations and many related results (cf. unit normalization)

But - as with any convention - what proves convenient in one context may prove inconvenient in another. For example, when studying quadratic forms and related results, some authors find it convenient to consider certain units to be prime, e.g. $\,-1\,$ is considered prime. This idea was first employed by H. Hasse and has more recently been popularized and extended by John H. Conway. Below are excerpts of some of Conway's remarks explaining why he calls $\,-1$ prime. The context is explained at much greater length in his very beautiful book The Sensual (Quadratic) Form, which was cited in his prestigious 2000 Steele prize for mathematical exposition ("This is a book rich in ideas. They seem to burst forth from almost every page")


From p. vii of $ $ The Sensual (Quadratic) Form:

This reminds me of the fact that some people always smile indulgently when I mention
"the prime $\,-1$", $ $ and continue to use what they presume to be the grown-up name "$\infty$".
But consider:

  • Every nonzero rational number is uniquely a product of powers of prime numbers $\,p$.

  • For distinct odd primes $\left(\frac{p}q\right)$ and $\left(\frac{q}p\right)$ differ just when $\,p \equiv q \equiv -1\pmod 4.$

  • There is an invariant called the $p$-signature whose definition involves summing $p$-parts of numbers.

  • If there are $p$-adically integral root vectors of norms $\,k\,$ and $\,kp,\,$ then $\,p\,$ is in the spinor kernel.

Each of these statements includes the case $\,p = {-}1,\,$ but none of them is even meaningful when we use the silly name "$\infty$". In the future, I shall smile indulgently back!


From a Sep 29, 2000 post to usenet newsgroup geometry.research.independent

I call $\,-1$ a prime for a very real reason, not just for fun. Namely, the theory of quadratic forms largely consists of statements - let me call the typical one "Statement($p$)" - that are usually only asserted for the positive primes $\,p = 2,3,5,\ldots$. Now in my experience, in almost every case when Statement($p$) is true for all positive primes, it's automatically both meaningful and true for $\,p = -1\,$ too, and moreover, that additional assertion is just as useful.

Of course, it's also the case that $-1$ does behave differently to the positive primes, but the way this happens is usually that more is true when $\,p = -1,\,$ not less. If less were true, then it would indeed be silly to count $\,-1$ as a prime, but since it's almost always more, it's silly not too.

Let me discuss the simplest such statements in this light.

First, the form of the statement of unique factorization that I prefer is:

The multiplicative group of the non-zero rationals is the direct product of the cyclic subgroups generated by "my" primes, namely $\ {-}1,2,3,5,\,\ldots$.

The additional thing that's true here is that the subgroup generated by $\,-1$ has order $\,2$.

Of course you can still state this with "your" primes, but then have to say "generated by $\,-1$ and the primes", which makes it just a little bit longer.

Second, my preferred statement of quadratic reciprocity is:

If $\,p\,$ and $\,q\,$ are distinct odd primes (in my sense), then $\,\left(\frac{p}q\right)\,$ and $\,\left(\frac{p}q\right)\,$ differ precisely when $\,p\,$ and $\,q\,$ are both congruent to $\,-1$ modulo $4$.

If you want to make the same assertion without counting $\,-1$ as a prime, then you have to add an additional clause:

moreover, $\,\left(\frac{-1}p\right)\,$ is $\ {-}1$ precisely when $\,p\,$ is congruent to $\,-1$ modulo $4$.

Third, perhaps the best of my discoveries in that book is the fact that

for every odd "prime" $\,p\,$ (in my sense), the number

$$ p^a + [A\mid p^a] + p^b + [B\mid p^b] +\, \ldots \pmod 8$$

is an invariant of the quadratic form $\,{\rm diag}(p^a. A,\, p^b.B,\,\ldots )\,$ (with the understanding that $\,p^a\,$ is the $p$-part of $\,p^a.A,\,$ etc.), there being an analogous but slightly different statement for $\,p = 2.$

Once again, if you don't count $\,-1$ as a prime, the statement becomes longer.


From pp. 1-2 of $ $ The genus of a quadratic form:

Our basic problem is to answer the question: when are two quadratic forms equivalent by a rational or integral change of basis? We shall temporarily suppose that all our forms are non-degenerate (i.e., have non-zero determinant), although degenerate forms really give no trouble.

The answer in the rational case is given by the celebrated Hasse-Minkowski theorem, which is usually stated in the form:

Two rational forms are equivalent over the rationals just if they are equivalent over the $\rm\color{#c00}{reals}$, and over the $p$-adic rationals for each positive prime number $p.$

This at first sounds like gobbledigook, because it seems to demand that you must first understand "$p$-adic", which really ain't so. Indeed, Minkowski, who understood the situation very well, did not have the notion of p-adic rational number.

The point really is, that there is a simple invariant for each $p$, and two rational quadratic forms of the same non-zero determinant and dimension are equivalent if and only if they are equivalent over the reals and have the same value of all these invariants.

It is conventional to avoid the exceptional treatment of the reals by regarding them as being the $p$-adic rationals for a rather special "prime number". If one is working over an algebraic extension of the rationals, there may be more than one of these "special primes", which Hasse (who introduced the idea) correctly called "unit primes." Unfortunately, this nomenclature didn't stick, and they are now usually called "infinite primes".

We're going to stay with the rational case, when there is only one such special prime, usually called "infinity." However, we'll go back to Hasse's way of thinking about things, and call it "$\color{#c00}{-1}$" which is really much more natural. The "$\color{#c00}{-1}\!-\!\!$adic" rational or integral numbers will be defined to be the $\color{#c00}{\rm real}$ numbers.

We can now restate the Hasse-Minkowski theorem so as to include $-1.$

Two forms of the same non-zero determinant are equivalent over the rationals just if they are equivalent over the $\,p$-adic rationals for $\,p = \color{#c00}{-1},2,3,5,\ldots$ (in other words, for all the prime numbers, where $\,\color{#c00}{-1}$ is counted as a prime).

This still includes the bogeyman word "$p$-adic," but I said that $p$-adic equivalence was determined by a simple invariant. Traditionally, it's been an invariant taking the two values $1$ and $-1$, and called THE "Hasse-Minkowski invariant." However, the definite article is undeserved, because there is no universal agreement on what "the Hasse-Minkowski invariant" means! There are two systems in use, and topologists further confuse the situation by still using an older version, the "Minkowski unit."

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    $\begingroup$ I really like this idea. Is there accompanying reasoning for why the "$-1$-adic" norm on the integers should be absolute value? $\endgroup$ – jdc Mar 6 '15 at 7:00
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In order to properly consider the question of whether or not $-1$ is a prime number, it is necessary to first ask: how do we extend the definition of prime and composite numbers to the negative numbers?

The definition that I was taught for positive prime numbers is that they have exactly two divisors. But when we broaden our view to include the negative numbers, a positive prime number now has four divisors. For example, $47$ has these divisors: $-47$, $-1$, $1$, $47$. Likewise $-47$ has four divisors and should thus be considered a negative prime number.

But $1$ only has two divisors, and the same is true of $-1$. Now, it would be quite wrong to get stuck on the number of divisors as the defining characteristic of primality if it turns out that $-1$ shares a lot of properties with the negative prime numbers. But it doesn't.

For starters, consider what happens when we multiply an arbitrary integer $n$ by a prime number $p$, be it positive or negative. The absolute value of $np$ is $|np|$. By contrast, $|(-1)n| = |n|$, meaning that multiplication by $-1$ does not change the absolute value of $n$ (this is also true of $1$, by the way).

Here's another: consider the square roots of the negative primes. If $p$ is a positive prime, then $\sqrt{-p} = i\sqrt{p}$, where $i$ is the imaginary unit. So $\sqrt{-p}$ is an irrational imaginary number, just as $\sqrt{p}$ is a real irrational number. Then $\sqrt{-1} = i$, an imaginary but rational number. Of course lots of composite numbers also have irrational square roots. But this particular quality makes $-1$ look much more like certain composite numbers than like any prime number.

Also think about the powers of prime numbers. If $p$ is a positive prime number, then $\sqrt{p} < p < p^2$. But with $1$, we have $\sqrt{1} = 1 = 1^2$. The roots and powers of a positive prime number extend to the infinities, while the roots and powers of $1$ are fixed upon a single point on the real number line. What about the roots and powers of $-1$? They form a little four-"star" "constellation" centered upon $0$.

Once you start studying the notion of algebraic integers, you will come across a very useful concept which is the norm function. For example, $7i$ has a norm of $49$. $i$ has a norm of $1$, and so it is considered a unit. Multiplying an arbitrary algebraic integer from a given algebraic integer domain by a unit in that domain results in a number that has the same norm [or absolute value of the norm] as the arbitrary algebraic integer. Guess what multiplying by $-1$ does to the norm of an arbitrary real, imaginary or complex integer.

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One way to get at the answer here is with ideals, specifically, principal ideals. Given $a \in \mathbb{Z}$, the ideal $\langle a \rangle$ consists of all multiples of $a$ in $\mathbb{Z}$. For example, $\langle 2 \rangle$ consists of all even integers: $$\ldots, -8, -6, -4, -2, 0, 2, 4, 6, 8, \ldots$$

If $p$ is a positive prime number, then $\langle p \rangle$ is what is called a "maximal ideal": you can't find a proper ideal $\langle a \rangle$ that is not all of $\mathbb{Z}$ which contains $\langle p \rangle$. If $n$ is composite, then $\langle n \rangle$ is not maximal, e.g., $\langle 4 \rangle$ is contained in $\langle 2 \rangle$.

It turns out that $\langle -2 \rangle$ is the same thing as $\langle 2 \rangle$. Therefore it makes sense to consider $-2, -3, -5, -7, -11, -13, -17, -19, \ldots$ to be primes just like their positive counterparts. If $p$ is prime, so is $-p$.

Now for the moment of truth: is $\langle -1 \rangle$ a maximal ideal? That ideal turns out to be all of $\mathbb{Z}$, which means that no ideal properly contained within $\mathbb{Z}$ can properly contain $\langle -1 \rangle$.

This is yet another way in which $-1$ fundamentally differs from the negative primes.

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I can see why someone would think that $-1$ is prime. Consider for example the prime factorization of $-4$. That would be $(-1) \times 2^2$. Or could it be $-2 \times 2$ or $2 \times -2$? Since multiplication is commutative, those latter two factorizations are not seen as a threat to the sanctimoniousness of unique factorization.

But shouldn't the primality of an eternal number be determined by its own intrinsic properties, and not the vagaries of the thought process of some short-lived animal? And yet it's not entirely possible to divorce our thinking of numbers from the way in which we represent them. We might learn something if we more fully embrace how our thinking about numbers is influenced by our representation of them.

If $b$ is a composite positive integer, then there exists divisibility tests in base $b$ for numbers with absolute value other than $b$, and these divisibility tests depend solely on the least significant digit. For example, with our familiar $b = 10$, there are divisibility tests for $2$ and $5$ which need only inspect the least significant digit to deliver their true or false result.

If $b$ is a composite negative integer, then there also exist divisibility tests in base $b$ for numbers with absolute value other than $|b|$, and these divisibility tests also depend solely on the least significant digit. For example, $-4$ would be represented as $16$, and the divisibility tests for $2$ and $5$ would tell us this number is divisible by $2$ but not by $5$.

With prime $b$, the only available LSD divisibility tests are for $b$ and $-b$. One could argue that there is an LSD divisibility test with $b = -1$. For example, $-3$ would be represented as $101010$. But it could also be represented as $10101011$. The divisibility test for $-1$ in base $-1$ depends not on the least significant digit, but on the absence or presence of the negunal point. But there's not much point for a negunal point since it's impossible to represent numbers like $\frac{1}{2}$ in negunary.

Given all the many different properties positive and negative primes share which $-1$ and $1$ lack, it is only reasonable to conclude that $-1$ and $1$ are not prime numbers.

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  • $\begingroup$ A+ for effort. I think you're getting at the issue fields $\mathbb{F}_p$. Last and least, there may be some minor errors in the negunary representation of $-3$. $\endgroup$ – Robert Soupe Mar 7 '15 at 3:28
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It's not because it's not positive, it's because it just isn't prime in the same way that $1$ isn't prime. Both $-1$ and $1$ are units. You can multiply and multiply by units as many times as you want, but in a lot of ways, you're just spinning your wheels in the air.

We agree that $2$ is prime, right? $$-1 \times 2 = -2$$ Is that prime? Maybe, maybe not, a separate question, I suppose. $$-1 \times -1 \times 2 = 2$$ We're back where we started. $$-1 \times -1 \times -1 \times 2 = -2$$ $$\ldots$$ $$-1 \times \ldots \times -1 \times 2 = \pm 2$$

Now try putting in more "$\times 2$"s instead of more "$\times -1$"s: $$-1 \times \ldots \times -1 \times 2 \times 2 = \pm 4$$ $$-1 \times \ldots \times -1 \times 2 \times 2 \times 2 = \pm 8$$ $$-1 \times \ldots \times -1 \times 2 \times 2 \times 2 \times 2 = \pm 16$$

The numbers may be positive or negative, but they're getting far away from $0$. Multiplying a nonzero number by $-1$ changes on what side of $0$ the number is, but not its distance from $0$.

By the way, I don't have to show that $4, 8, 16, 32, \ldots$ are composite, because units are not composite either. They're units.

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I'm not answering this for the bounty, so I'm free to invoke algebraic number theoretic concepts.

$R$ is some ring of algebraic integers, and we know that it has all the real integers, and we know that it has prime numbers, but we don't know if it's a unique factorization domain (UFD). $m$ is either a prime number in $R$ or it's a product of primes in $R$. And $n \neq 0$ is any number in $R$.

Then the norm of $n$ is different from the norm of $mn$. Multiplication by a single prime number, or by a power of a prime, or by a product of various powers of primes, causes the norm to change. Multiplication by $1$ and $-1$, and possibly other numbers, causes no change to the norm. These numbers are called units.

A concrete example: in $\mathbb{Z}[\sqrt{-5}]$, I'm told that the number $13$ is irreducible and prime in this domain. $2 + 2 \sqrt{-5}$ has a norm of $24$. Then $13(2 + 2 \sqrt{-5}) = 26 + 26 \sqrt{-5}$, which has a norm of $4056$.

Now let's try that with $-1$ instead of $13$. Then $(-1)(2 + 2 \sqrt{-5}) = -2 - 2 \sqrt{-5}$. Of course $2 + 2 \sqrt{-5}$ and $-2 - 2 \sqrt{-5}$ are different numbers. But they both have a norm of $24$. Check for yourself: $(-2 - 2 \sqrt{-5})(-2 + 2 \sqrt{-5}) = 24$.

Here I've taken it on faith that $13$ is prime in this ring, and it seems to behave like a prime. But I know for a fact that this ring is not a UFD, since $2$ and $3$ are both irreducible yet $3 \nmid (\pm 2 \pm 2 \sqrt{-5})$.

On the other hand, $-1$ behaves like a unit in $\mathbb{Z}[\sqrt{-5}]$ just as it does in our familiar $\mathbb{Z}$. I can say with confidence that $-1$ is always a unit, never a prime. $13$ may or may not be a prime (e.g., $2 + 3i$), but that's another story.

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  • $\begingroup$ I think I had already awarded the bounty when you posted your answer anyway. $\endgroup$ – Robert Soupe Mar 8 '15 at 21:22
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I'm going to play devil's advocate and tell you that $-1$ is in fact a prime number. The Greek word "πρώτος" is a root for words like "primary," which have to do with being first. So the prime numbers in the original Greek sense are primary numbers. They didn't think of $0$ as a number, nor of negative numbers, so they considered $1$ to be the first number. Since $1$ is divisible only by $1$ and itself, the other primary numbers are those numbers that are also divisible only by $1$ and themselves, and the composite numbers are secondary numbers.

Over the last few centuries, human mathematicians, especially the duller ones, became concerned with the use of prime factorization as unique signatures for numbers. The definition of prime number had to be changed, and so unique factorization was "saved." But unique factorization is a fact of existence, and not a human invention, and thus it doesn't need protection in the way that human laws (like the beloved Second Amendment) do.

If you wish to extend the definition of prime number to the negative integers in a manner consistent with the way the ancient Greeks thought about prime numbers, then $-1$ is in fact a prime number. But as far as today's human mathematicians are concerned (especially the duller ones), $-1$ is a unit, not a prime number. And it's perfectly possible to come to that conclusion in domains that don't have unique factorization. That's because $-1$ is a very special number, regardless of whether or not humans are smart enough to recognize it, and most of them aren't.

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  • $\begingroup$ That's not where the term "prime number" came from, though. "Prime" refers to the fact that they are fundamental numbers, that they can't be split up into more fundamental numbers. Kind of like the concept (at the time) of the atom - and as with the atom, they are actually able to be split up to more fundamental numbers, but only in special circumstances (gaussian integers, for instance). And in this context, 1 and -1 aren't prime because they aren't fundamental, but rather empty. They're the default states, much as how $0!=1$. $\endgroup$ – Glen O Mar 7 '15 at 3:04
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    $\begingroup$ I can't say either of you is right, but I'd like to point out the Spanish word "primero" and the Italian word "primo," which both mean first. On the other hand, "primo" in Spanish means nephew, I don't know what to make of that. $\endgroup$ – Robert Soupe Mar 8 '15 at 20:15
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    $\begingroup$ @RobertSoupe It doesn't mean "nephew", it means "cousin" - as in "first cousin". It also means "first" in other contexts. $\endgroup$ – Kyle Strand Jan 26 at 7:28
  • $\begingroup$ @KyleStrand Thanks for the correction. It's been years since I've actually needed to speak Spanish, all the Spanish speakers I come across seem to know English already. $\endgroup$ – Robert Soupe Jan 26 at 18:23
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Why did mathematicians choose ...

To answer this we need to to back to when that choice was made. "Prime number" can be found in the classical Greek mathematics. They had no notion of "negative number" at all. And (at least at first) the term "number" excluded $1$ simply by its linguistics. There is a "unit" ... and then there is a "number" made up of two or more units. Similarly, $a$ measures $b$ iff $b$ is a multiple of $a$ (where $a$ taken two or more times equals $b$).

Euclid's Elements, Book VII, Definition 2:
A number is a multitude composed of units.

Definition 5:
The greater number is a multiple of the less when it is measured by the less.

Definition 11:
A prime number is that which is measured by a unit alone.

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The simplest answer to the question is that "numbers divisible only by 1 and themselves" isn't the strict mathematical definition of prime numbers, but rather the simple description that is used to teach the concept of prime numbers.

A more exact definition (in the context of integers) is "numbers, $n$, for which any factorisation must contain exactly one value divisible by $n$".

So no matter how you factorise $3$, such as $1\times3$, $(-1)\times(-1)\times3$, or $(-1)\times(-3)$, there will be exactly one factor that is divisible by $3$.

In the case of $-1$, we have $(-1)\times(-1)\times(-1)$, all of which are divisible by $-1$, and thus it does not satisfy the condition (indeed, $(-1)\times1$ demonstrates the same issue, as $-1$ divides $1$). That factorisations can contain multiple values divisible by $-1$ makes $-1$ a "unit".

And of course, in the case of $4$, we can write $2\times2$, with neither of the $2$s being divisible by $4$. That there are factorisations of $4$ for which no factors are divisible by $4$ makes $4$ a "composite".

I do not know for certain if this definition works in general for the more general concepts of prime elements and units, but my instinct says that it should work generally for any domain in which the concept of divisibility is well-defined (note that this isn't strictly consistent with the definition of "unit" in all domains, as the normal definition of "unit" is "number with a multiplicative inverse", which means that $2$ (and all other reals except zero) is a unit in $(\mathbb{R},+,\times)$, whereas my definition makes $1$ (and all other reals except zero) "composite" in that domain).

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In general, a definition that describes relationships between numbers without describing the domain in which such relationships are valid is incomplete.

In other words, when you say that the "definition of prime is having only two factors," you are leaving out a crucial element: for which set(s) of numbers is this relationship between primes and factors valid?

The typical definition of "prime numbers" describes a relationship between natural numbers; since $-1$ is not a natural number, any definition that only applies to natural numbers has no bearing on whether $-1$ is prime.

Thus, in order to make the question "is $-1$ prime" meaningful, you must find an extension (or generalization) of the definition of "prime" to a domain that includes $-1$.

As pointed out in @Hugh's answer, perhaps the most natural extension is that of prime elements. In this extension, $-1$ is considered a unit and is therefore by definition not prime. Another extension is the Gaussian Primes, in which $-1 = -1 + 0i$ is non-prime because the coefficient of $i$ is $0$ and $|-1| = 1$ is not prime (according to the standard definition on the natural numbers). @BillDubuque, meanwhile, gives an example of a context in which it might be reasonable to consider $-1$ to be prime.

Now, we can perhaps make further generalizations about all possible extensions of the definition of primes within the natural numbers. Some of the other answers provide some good intuition about this; for instance, it is observed that the units are typically not considered prime, because this simplifies the definition of prime factorization. Thus, we could answer that $-1$ is not considered prime in any domain in which $-1$ is a unit (though of course this requires us to define "unit" for each such domain).

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Rereading the existing answers after all these years, it seems like no one said anything about your assertion that $-1$ has two factors on account of $-1 \times 1$. Or maybe someone did address it but it got kind of buried.

So I will address it now. Limiting ourselves to $\mathbb Z^+$ (the positive integers), we see that prime numbers do indeed have two factors. One factor is 1, the other factor is the prime itself. That is, if $p$ is a positive prime, then $1 \times p$ is a factorization.

Of course if we wanted to have three or four or more factors we can just append "$1 \times$" as many times as we want: $1 \times 1 \times 1 \times \ldots \times p$. But when we strip out all the units (all 1's in this case), we're left with just $p$.

Likewise with a positive composite number $n$ that is the product of various primes $p_i$, not necessarily distinct, we can also prepend as many "$1 \times$" or append as many "$\times 1$" as we want, and as long as we don't mess with the $p_i$, we have the same number $n$.

So there is no need to whine about unique factorization. The fundamental theorem of arithmetic does not need any protection. It just is, regardless of whether we understand it or misunderstand it.

Things get just a tiny bit more complicated when we figure in negative numbers. Do you accept that numbers like $-2$ and $-47$ are prime? I hope you do, but if you don't, it's no problem. We can just say $-p = -1 \times p$, where $p$ is a positive prime.

We can't prepend "$(-1) \times$" or append "$\times (-1)$" arbitrarily. For example, $-1 \times -1 \times 2 = 2$, not $-2$. However... we can prepend "$(-1) \times$" or append "$\times (-1)$" and still have the same number as long as we make sure to do so in pairs and then put in an extra one so it's an odd one out.

Note that 1 is its own square root, its own cubic root, its own fourth root, etc. Likewise, although $-1$ is not its own square root, it is its own cubic root, its own fifth root, its own seventh root, etc. You can't say that about bona fide prime numbers.

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Prime numbers are (by definition) elements of the natural numbers. There is, however, a natural extension of prime numbers to all the integers, known as the prime elements in the set of integers. The definition of a prime element $p$ (in the integers) is an integer that is neither zero nor $\pm1$ and whenever $p$ divides $xy$ it must also divide $x$ and $y$ (where $x$ and $y$ are both integers).

Of course, this definition excludes $-1$ ipse dixit, so this definition may not be very satisfying. The reason for excluding $-1$ is the same as excluding $1$ from the prime numbers: both are done to ensure unique factorization at least cost.

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  • $\begingroup$ But what about a domain that doesn't have unique factorization? Can $-1$ possibly be prime in such a domain? Or is there a more profound reason applicable to any domain that may or may be UFD? $\endgroup$ – Robert Soupe Mar 5 '15 at 1:16
  • $\begingroup$ @RobertSoupe A nonzero nonunit element in any ring is defined to be prime if it satisfies the conclusion of Euclid's lemma. Since $-1$ is always a unit, it is never prime. $\endgroup$ – Pedro Tamaroff Mar 5 '15 at 1:59
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    $\begingroup$ @PedroTamaroff Alright, you know that. Does Hugh? $\endgroup$ – Robert Soupe Mar 7 '15 at 4:18
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In this answer we provide elementary definitions so that $-1$ is a prime number but $1$ is not. The motivation for this came from reviewing Bill Dubuque's work here as well as well as his answer and comments found on thread

Proof of Lemma: Every integer can be written as a product of primes

Definition: Let $a \in \mathbb Z$.
If there exists a $b \in \mathbb Z$ with $ab \ne a$ and $ab \ne b$, then $a$ is said to be a BB integer.

Definition: Let $n \in \mathbb Z$.
If there exist two BB numbers $a$ and $b$ such that
$\quad a \ne n$
$\quad b \ne n$
$\quad n = ab$
then $n$ is said to be a composite integer.

Definition: Let $p \in \mathbb Z$.
If $p$ is a BB integer and also not a composite integer, then $p$ is said to be a prime integer.

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