To this day there are mathematics aficionados who consider 1 to be a prime number, and there are also people alive today who were taught this as children. Until 1899, professional mathematicians considered 1 a prime number. The reason they stopped doing so is sometimes made to appear to be a matter of convenience, as Timothy Gowers says in A Very Short Introduction to Mathematics.
The tables below list arguments for both sides, presented more or less like a trial in court, with a lawyer for the defense, a lawyer for the prosecution and a hopefully impartial judge. If I don't know a particular argument on one side to counter the other side, I will leave the cell blank. But if you have an counter-argument for that particular cell, or for a completely new argument, or for a stronger version of an argument already listed, please let me know. E-mail me at firstname.lastname@example.org.
|A prime number is only divisible by 1 and itself. 1 is divisible by 1 and itself. This is a simple definition. Ergo, 1 is prime.
||A prime number has two distinct factors, 1 and itself. 1 has only one factor, itself. Ergo, 1 is not a prime.|
|If 1 is not a prime, and it's certainly not a composite, then what the hell is it?
||1 is unity, or if that sounds too New Agey for you, 1 is the unit of the integers.|
|Given a prime number p, we find that φ(p) + σ(p) = 2p (φ is Euler's totient function and σ is the sum of divisors function). This is true for p = 1. (Thanks to Anthony Morris for this argument).
||But with prime numbers, φ(p) = p - 1 and σ(p) = p + 1, resulting in the inequality φ(p) < σ(p). With 1, this is an equality. (Thanks to Alonso Delarte for this counter-argument).|
|A lucky number of Euler x is a number that give primes in the formula n2 - n + x for all -1 < n < x. We can agree that most of these lucky numbers of Euler are primes. One of these numbers is 1.
||Maybe in Sloane's 1973 Handbook of Integer Sequences, when 1's were added to sequences that didn't already have them. If you look up A014556 today, it should read "2, 3, 5, 11, 17, 41." Rejecting the trivial cases, neither 0 nor 1 counts as one of Euler's lucky numbers. (Thanks to Arthur Rubin for this counter-argument).|
||Furthermore, given x = 1, the only n allowed is n = 0, which plugged into the equation just gives back the initial 1. The argument thus fails to convince anyone who doesn't already consider 1 prime in the first place, and is therefore a circular and logically faulty argument. (Thanks to Tom Creed for clarifying this point.)|
|There are only nine Heegner numbers. Eight of them are -2, -3, -7, -11, -19, -43, -67, -163 (Mathworld gives them as negative, but the OEIS gives them as positive even if you enter them as negative in the search box). These eight numbers listed so far are all prime (or primes multiplied by -1 if you don't consider negative numbers as having primality). -1 is also a Heegner number. (Thanks to Peter for pointing out the logical fallacy of the first version of this argument).
||That may be the case, but it does not necessarily follow that that first element must be prime just because the eight others are.|
|In order to make a magic square using only prime numbers, you have to accept 1 as a prime number. Take for example this lovely magic square with magic constant 111:
||You're wrong, you can make a magic square using only prime numbers and not using 1. For example, this Chen prime magic square with magic constant 177:
(Thanks to Robert Happelberg for this argument and counter-argument).|
|When reckoning Zeisel numbers, p0 = 1.
|Also, in Wilson's primeth sequence (Sloane's A007097), a(0) = 1.
||You have given two arguments in a row in which 1 is at index 0, but not a single one where 1 is at index 1. Accepting 1 as a prime would render Wilson's primeth sequence merely a copy of the sequence of prime numbers with 1 added at the beginning.|
|To make a numerical representation where place value x (with the place value of the least significant digit being numbered either 0 or 1) is the xth prime, you have to accept 1 as a prime and assign it to the least significant digit place value. Otherwise you'd be unable to represent semiprimes in an unambiguous manner. (Thanks to Wilfredo Lopez for this argument.)
|Accepting 1 as a prime number would invalidate the fundamental theorem of arithmetic, that each integer has a unique factorization. By adding any number of multiplications by 1 we would be able to create many different factorizations, e.g., 30 = 2 × 3 × 5 = 1 × 2 × 3 × 5 = 1 × 1 × 1 × ... × 1 × 2 × 3 × 5, etc. Ergo, 1 is not a prime.
||It would not. The fundamental theorem accepts the commutative property of multiplication, so no one argues that 30 = 2 × 3 × 5 is any different than 30 = 5 × 2 × 3. Then why can't it make room for the multiplicative identity? Ergo, 1 is a prime.|
|What about exponents? For example, 108 = 22 × 33. If we allow 1 as a prime, then that would allow not just 108 = 11 × 22 × 33, but open the door on 108 = 147 × 22 × 33, 108 = 11729 × 22 × 33, etc.
||It would be quite simple to just add to the fundamental theorem of arithmetic that the lowest exponent possible is used (or tacit), just as rearrangements (redistributions) are already ignored. (Thanks to Graeme McRae for this argument and counter-argument).|
|When reckoning the sieve of Erathostenes, we circle a prime and cross off its subsequent multiples. Moving forward, we find the next number that hasn't been crossed off, circle it, and cross off its subsequent multiples. We start this procedure with 2. If we started with 1, we'd have to cross off all the other positive integers! (This argument, independently thought of by several different people, appears on page 130 of the 1996 edition of Conway & Guy's book The Book of Numbers.)
||The sieve is a process, not a definition. We could just as easily say that the first step of the sieve is to circle 1, move on to the next step of circling 2 and cross off its subsequent multiples, circle 3, cross off... etc. (Thanks to Hal Westhead for this counter-argument).|
|For a prime number p, there is no solution to the equation nq = p, where q is another prime number and n is any other positive integer; this is to say, no prime is the product of another prime and an integer. If we accepted 1 as a prime, it would give 1 as a solution to that equation and in the process rule out all integers from being prime, including 1 itself! (Thanks to Hal Westhead for this argument).
|n! + p is never a prime for p < n, because obviously it will be a multiple of p, just as n! is. But n! + 1, even though it certainly is a multiple of 1, can be a prime. There's even a special name for primes like these, factorial primes. (This argument still works if we replace the + signs with - signs).
|In his C++ program to find primes, James Wolovchek hard-coded 1 as a prime number.
||James Wolovchek's programming skills are not on trial here. And in any case, he also hard-coded 2 and 3 as primes.|
|Given q = p2, it is obvious that √q = p. Perhaps not quite so obvious, q - φ(q) = p. This is true of the primes 2, 3, 5, 7, ... but it's not true of 1, since 1 - φ(1) = 0 and √1 = 1.
|Generalizing the previous argument to higher powers, we find that φ(px) = px - 1(p - 1) so long as x > 0. This works with p = 2, 3, 5, 7, ... but it doesn't work with p = 1 just like it doesn't work with composite numbers. (Thanks to Alonso Delarte for this argument).
||φ(1) = 1 just so that the function can be multiplicative. One way of calculating φ(x) is to list the fractions from 1/x to x/x, reduce all the fractions you can (e.g., reduce 12/21 to 4/7) then count how many fractions still have x as their denominator. For x > 1 this gives the expected results, but suggests that φ(1) = 0. Then your formula for φ(px) does work with p = 1. (Thanks to Anthony Morris for this somewhat tortuous counter-argument).|
|No untouchable number is one more than a prime. As it happens, 2 is an untouchable number.
||Objection, circular argument! To accept that 2 is an untouchable number not of the form p + 1 requires us to accept that 1 is not a prime number, which is precisely what the argument was supposed to do in the first place. (Thanks to Michael Blackburn for pointing this out).|
|Can the prosecution restate the argument without recourse to circularity or other logical fallacies?|
|Yes, we can, your Honor. This just goes back to what we saying earlier about σ(p) = p + 1. Therefore a number of the form p + 1 is "touched" by p. Since 1 has only one divisor, itself, σ(1) = 1. That's why 2 is therefore untouchable. (Thanks to Lisa Havelock for this non-circular rephrasing of the argument).
|There are more Smarandache-Wellin primes if you don't consider 1 to be prime. (Thanks to Wilfredo Lopez for this argument.)
Both sides presented strong arguments, and a few weak ones. Given these arguments presented before me, it is my verdict that 1 is not a prime number, but only because deeming it prime introduces more inconveniences than conveniences.