The number 1 is not prime. Why?

Alonso Del Arte
Aug 2 · 10 min read
Photo by Matthijs van Heerikhuize on Unsplash

Sometimes in a science fiction story, the extraterrestrials need to let us know that they’re intelligent life. What message can they send that we would understand? They might send the first few positive prime numbers.

Two blips. Pause. Three blips. Pause. Five blips. Pause. Seven blips, pause. Then 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, maybe send a few more besides those.

The prime numbers are good because the pattern is unlikely to be a random natural occurrence.

The extraterrestrials might include the number 1 in that list. They shouldn’t include 1 in that list, but we shouldn’t ignore their message if they do include it. After all, human mathematicians regarded 1 as prime until fairly recently, and not all science fiction writers have caught up yet (e.g., Carl Sagan, the author of Contact).

The ancient Greeks defined prime numbers as positive integers that are divisible only by 1 and themselves. For example, 7 is prime because it is divisible by 1 and by 7, but not by 2, 3, 4, 5 nor 6.

Compare the number 8, which is divisible by 1 and 8 and also by 2 and by 4. By this definition, 1 seems to be a prime number. It’s divisible by 1 and it’s definitely divisible by itself.

The old Greek definition is a good enough definition if you’re only concerned about positive integers. But if you also want to consider negative integers, this definition becomes inadequate.

A number like −12 is clearly not prime. But what about a number like −13? We can amend the definition to say that an integer n is prime if it’s only divisible by −n, −1, 1 and n.

By this amended definition, 1 still seems to be prime. But there should be this nagging feeling that 1 might be too different from the “other” primes to actually be prime itself.

Even so, for the earliest mathematicians who accepted negative numbers, there wasn’t much incentive to question the presumed primality of 1.

Then Fermat’s so-called “last” “theorem” (more properly called a conjecture prior to 1994) spurred a lot of interest in algebraic integers. The algebraic integers are an extension of the integers we use on a daily basis.

Each of those familiar integers is the solution to a simple equation of the form xn = 0. For example, x = 15 is the solution to x − 15 = 0. That seems plainly obvious, you don’t need a blackboard to solve that equation. It’s hardly a revelation. But I’m going somewhere with this.

First though, we need a bit of notation and terminology. By “integers” without an adjective, I mean the “familiar” or “plain” or “ordinary” integers. The set of all such integers is often notated with the symbol Z (or sometimes with a much more stylized letter Z), after the German word “Zahlen” for “numbers.”

Here’s a more interesting equation: x³ − 9x² + 21x − 15 = 0. If you can solve that one in your head, I’m impressed. I just ask Wolfram Alpha, it tells me the equation has three solutions, one of which is 3 plus the cubic root of 2 plus the cubic root of 4, roughly 5.8473221 (we will be using Wolfram Alpha to spare ourselves some of the more intensive number crunching).

That number (but not its approximation) is an algebraic integer of degree 3 (because of the x³ in the equation). The coefficient for x² is 9, and the coefficient for x is 21. The coefficient for x³ is 1, but that’s almost never written explicitly.

An algebraic integer is a solution to an equation like x³ − 9x² + 21x − 15 = 0, in which all the coefficients are integers from Z and the leading coefficient (the coefficient for the x with the highest exponent) is 1.

Note however that rational numbers other than those of Z are algebraic numbers but not algebraic integers.

For example, x = 1/2 is an algebraic number but not an algebraic integer, as it’s the solution to 2x − 1 = 0, but there is no equation for which the leading x doesn’t have a coefficient other than 1.

The study of algebraic integers, and the quest to resolve Fermat’s conjecture, led to the realization that not all number domains have unique factorization, a property we take for granted in Z. It was a painful realization for the earliest mathematicians who thought they had solved Fermat’s puzzle.

Just because Z has unique factorization doesn’t mean other domains also have unique factorization. In Z, each nonzero integer can be factorized into primes in exactly one way.

For example, in Z, the number −60 can be uniquely factorized as −1 × 2² × 3 × 5. This factorization can be expressed in different ways, like 3 × 2 × 5 × 2 × −1, but the underlying factorization is the same.

Now consider the domain of numbers of the form a + b√2, where a and b are ordinary integers from Z. Obviously −60 is still −1 × 2 × 2 × 3 × 5 in this other domain of numbers. But we can break this factorization down further.

The number √2 is an algebraic integer, as it is the solution to x² − 2 = 0, and it is a number in this domain of numbers of the form a + b√2, with a = 0 and b = 1. Since √2 × √2 = 2, we see that −60 = −1 × (√2)⁴ × 3 × 5.

The numbers 3 and 5 are still primes in this other domain. You should be having misgivings about −1 being prime now if you didn’t before.

However… I seem to have found an alternate factorization for 2 in this domain, as (2 − √2)(2 + √2) = 2. Verify that both 2 − √2 and 2 +√2 are solutions to x² − 4x + 2 = 0.

But what happens if we try dividing either of those numbers by √2? That gives us the numbers −1 + √2 and 1 + √2. The former is a solution to x² + 2x − 1 = 0, the latter is a solution to x² − 2x − 1 = 0.

The numbers −1 + √2 and 1 + √2 are not primes. We can divide one by the other, divide the result by either number, and iterate this process indefinitely, obtaining more algebraic integers that are solutions to x² + ax − 1 = 0, where a is an integer from Z that gets arbitrarily far away from 0 in either direction, while the trailing coefficient firmly remains either −1 or 1.

Likewise, we can multiply any number in this domain that is a solution to x² + ax ± 2 = 0 by a number in this domain that is a solution to x² + ax − 1 = 0 and obtain another number in this domain that is a solution to x² + ax ± 2 = 0. The sign of the trailing coefficient will alternate.

Numbers in this domain that are solutions to x² + ax − 1 = 0 are called units. Indeed any algebraic integer for which the equation has a trailing coefficient of either −1 or 1 is a unit (“the equation” is more commonly written without the “= 0” part and is called a “characteristic polynomial” or a “minimal polynomial”).

This means that both −1 and 1 are units. The pertinent equations are x + 1 = 0 and x − 1 = 0. In both of those, a = 0, so ax = 0 also, and so we leave it out of the polynomial.

Multiplying an algebraic integer x for which the equation has trailing coefficient N by a unit gives an algebraic integer for which the equation has trailing coefficient that is either −N or N (the trailing coefficient is generally called the “norm,” hence the the letter N).

So 2 = (2 − √2)(2 + √2) = (−1)(4 − 3√2)(4 + 3√2) = (10 − 7√2)(10 + 7√2) = (−1)(24 − 17√2)(24 + 17√2) and so on and so forth, these aren’t distinct factorizations of 2 in the domain of numbers of the form a + b√2.

Likewise in Z, the expression −15 = −3 × 5 = 3 × −5 does not give two distinct factorizations of −15.

Multiplying by units does not create alternative factorizations. This is as true in unique factorization domains (UFDs) like Z and the the domain of numbers of the form a + b√2 as it is in domains that are not UFDs, like the domain of numbers of the form a + b√10.

The presence or absence of unique factorization in a given domain of numbers does not change the fact that 1 is not a prime number in any of those domains.

Consider the number 31. It’s prime in Z. But in the domain of numbers of the form a + b√10, it can be factorized as (−1)(3 − 2√10)(3 + 2√10). You can verify this on any calculator with a square root key, though you might have to cut it some slack for precision.

The numbers 3 − 2√10 and 3 + 2√10 are prime numbers. In this domain of numbers, they’re divisible only by numbers with the norm −31, −1, 1 or 31. Note that in this domain the norm of 31 is its square, 961.

So, if the norm of a number in a given domain is a prime number in Z, then we know that that number is a prime in its domain.

Knowing that 3 + √10 has minimal polynomial x² − 6x − 1, you can find other seemingly different factorizations. But you would just be multiplying by units.

The number 3 is prime in Z, but something kind of screwy happens to it in the domain of numbers of the form a + b√10.

On your calculator, verify that (2 − √10)(2 + √10) = −6 (you might get something like −6.00000001 or −5.99999999 instead, depending on your calculator’s precision).

Clearly −6 is not prime in Z, as it is divisible by 2 and by 3. It must not be prime in the domain of numbers of the form a + b√10 either. Nor can 2 − √10 nor 2 + √10 be prime either, since their norms are not prime in Z.

And yet neither 2 − √10 nor 2 + √10 are divisible by 2 or by 3. We get polynomials like 2x² − 4x − 3, telling us that such divisions are algebraic numbers but not algebraic integers.

Well then, 2 − √10 and 2 + √10 must be divisible by some other numbers of the form a + b√10 with b ≠ 0 and norm −3, −2, 2 or 3. To find those numbers, we need to solve the equations a² − 10b² = ±2 or ±3.

Turns out those equations have no solutions in integers from Z, a fact that can easily be verified by looking for a perfect square integer with least significant digit 2, 3, 7 or 8.

So, by the definition of primality we’ve been using, 2 − √10 and 2 + √10 would seem to be prime numbers, but their norm is not prime in Z.

If we’re going to study prime numbers in number domains that are not UFDs, we need a stronger definition of primality.

We don’t need to discard the earlier definition, we simply use it for a different term, “irreducible.” So the numbers 2 − √10 and 2 + √10 are irreducible but neither of them is prime by the stronger definition of primality.

Here is that stronger definition of primality: a number p that is potentially prime in a given domain is indeed prime if, for any combination of numbers a and b from that same domain such that a × b is divisible by p, either a or b is also divisible by p.

We saw earlier that 2, 3 and 5 are prime factors of −60 in Z. They’re still prime under the stronger definition. Not much changes for unique factorization domains like Z.

If we express −60 as ab, such as −4 × 15 we will find for every possible combination of a and b that 2 is a divisor of either a or b, sometimes both. The same holds true for 3 and 5.

It doesn’t hold true for numbers like 15, for example. Given −60 = 5 × −12, we see that neither 5 nor −12 is divisible by 15. Obviously 15 is not in prime in Z, unlike its divisors 3 and 5.

This suggests that there are numbers in a given domain that are not divisible by a particular prime p. For example, in Z, the numbers …, −8, −6, −4, −2, 0, 2, 4, 6, 8, … are divisible by the prime 2, but the numbers …, −7, −5, −3, −1, 1, 3, 5, 7, … are not divisible by 2.

Where does this leave units like 1 and 3 + √10? Well, if u is a unit in a given domain, then both a and b from that same domain are divisible by u, so a × b is obviously also divisible by u, always.

In this way, units are different from prime numbers and from composite numbers. The set of multiples of a prime number or of a composite number in a given domain is a proper subset of the domain.

But the set of multiples of a unit is the entire domain.

That is an extremely important concept in number theory.

Clearly units are not primes. And 1 is a unit. Ergo, it’s not prime. Regardless of whether the domain under consideration is a unique factorization domain or not.

Removing 1 from the list of primes must not be thought of as a demotion, like knocking Pluto down from planet to planetoid. Rather, it is acknowledging a special property that puts it in a very special and select group of numbers.

I hope you find this all as interesting as I do. But I also acknowledge that if we ever make contact with extraterrestrial life, whether they consider 1 to be a prime number or not would be one of the least important details.

Physically, they might be very different from us. Science fiction has postulated how they might differ physically from us. Science fiction has not postulated much on how their conception of mathematics might be different from ours.

But the history of our mathematical development might provide some suggestions as to how other life forms in the universe might think about numbers, if they’re enough like us.

Star Gazers

“If you want to master something, teach it.”