Wilson's theorem

The theorem was first stated by Ibn al-Haytham {{circa|1000 AD}}.{{MacTutor Biography|id=Al-Haytham|title=Abu Ali al-Hasan ibn al-Haytham}} Edward Waring announced the theorem in 1770 without proving it, crediting his student John Wilson for the discovery.Edward Waring, Meditationes Algebraicae (Cambridge, England: 1770), page 218 (in Latin). In the third (1782) edition of Waring's Meditationes Algebraicae, Wilson's theorem appears as problem 5 on [https://books.google.com/books?id=1MNbAAAAQAAJ&pg=PA380 page 380]. On that page, Waring states: "Hanc maxime elegantem primorum numerorum proprietatem invenit vir clarissimus, rerumque mathematicarum peritissimus Joannes Wilson Armiger." (A man most illustrious and most skilled in mathematics, Squire John Wilson, found this most elegant property of prime numbers.) Lagrange gave the first proof in 1771.Joseph Louis Lagrange, [https://books.google.com/books?id=_-U_AAAAYAAJ&pg=PA125 "Demonstration d'un théorème nouveau concernant les nombres premiers"] (Proof of a new theorem concerning prime numbers), Nouveaux Mémoires de l'Académie Royale des Sciences et Belles-Lettres (Berlin), vol. 2, pages 125–137 (1771). There is evidence that Leibniz was also aware of the result a century earlier, but never published it.Giovanni Vacca (1899) "Sui manoscritti inediti di Leibniz" (On unpublished manuscripts of Leibniz),

Bollettino di bibliografia e storia delle scienze matematiche ... (Bulletin of the bibliography and history of mathematics), vol. 2, pages 113–116; see [https://books.google.com/books?id=vqwSAQAAMAAJ&pg=PA114 page 114] (in Italian). Vacca quotes from Leibniz's mathematical manuscripts kept at the Royal Public Library in Hanover (Germany), vol. 3 B, bundle 11, page 10:

Original : Inoltre egli intravide anche il teorema di Wilson, come risulta dall'enunciato seguente:
"Productus continuorum usque ad numerum qui antepraecedit datum divisus per datum relinquit 1 (vel complementum ad unum?) si datus sit primitivus. Si datus sit derivativus relinquet numerum qui cum dato habeat communem mensuram unitate majorem."
Egli non giunse pero a dimostrarlo.

Translation : In addition, he [Leibniz] also glimpsed Wilson's theorem, as shown in the following statement:
"The product of all integers preceding the given integer, when divided by the given integer, leaves 1 (or the complement of 1?) if the given integer be prime. If the given integer be composite, it leaves a number which has a common factor with the given integer [which is] greater than one."
However, he didn't succeed in proving it.

For each of the values of n from 2 to 30, the following table shows the number (n − 1)! and the remainder when (n − 1)! is divided by n. (In the notation of modular arithmetic, the remainder when m is divided by n is written m mod n.)

The background color is blue for prime values of n, gold for composite values.

As a biconditional (if and only if) statement, the proof has two halves: to show that equality does not hold when $n$ is composite, and to show that it does hold when $n$ is prime.

Suppose that $n$ is composite. Therefore, it is divisible by some prime number $q$ where . Because $q$ divides $n$, there is an integer $k$ such that $n\; =\; qk$. Suppose for the sake of contradiction that $(n-1)!$ were congruent to $-1$ modulo $\{n\}$. Then $(n-1)!$ would also be congruent to $-1$ modulo $\{q\}$: indeed, if $(n-1)!\; \backslash equiv\; -1\; \backslash pmod\{n\}$ then $(n-1)!\; =\; nm\; -\; 1\; =\; (qk)m\; -\; 1\; =\; q(km)\; -\; 1$ for some integer $m$, and consequently $(n-1)!$ is one less than a multiple of $q$. On the other hand, since $2\; \backslash leq\; q\; \backslash leq\; n\; -\; 1$, one of the factors in the expanded product $(n\; -\; 1)!\; =\; (n\; -\; 1)\; \backslash times\; (n\; -\; 2)\; \backslash times\; \backslash cdots\; \backslash times\; 2\; \backslash times\; 1$ is $q$. Therefore $(n\; -\; 1)!\; \backslash equiv\; 0\; \backslash pmod\{q\}$. This is a contradiction; therefore it is not possible that $(n\; -\; 1)!\; \backslash equiv\; -1\backslash pmod\{n\}$ when $n$ is composite.

In fact, more is true. With the sole exception of the case $n\; =\; 4$, where $3!\; =\; 6\; \backslash equiv\; 2\; \backslash pmod\{4\}$, if $n$ is composite then $(n\; -\; 1)!$ is congruent to 0 modulo $n$. The proof can be divided into two cases: First, if $n$ can be factored as the product of two unequal numbers, $n\; =\; ab$, where , then both $a$ and $b$ will appear as factors in the product $(n\; -\; 1)!\; =\; (n\; -\; 1)\backslash times\; (n\; -\; 2)\; \backslash times\; \backslash cdots\; \backslash times\; 2\; \backslash times\; 1$ and so $(n\; -\; 1)!$ is divisible by $ab\; =\; n$. If $n$ has no such factorization, then it must be the square of some prime $q$ larger than 2. But then , so both $q$ and $2q$ will be factors of $(n-1)!$, and so $n$ divides $(n-1)!$ in this case, as well.

The first two proofs below use the fact that the residue classes modulo a prime number form a finite field (specifically, a prime field).{{cite book|last=Landau|first=Edmund|orig-date=1927|date=1966|title=Elementary Number Theory|edition=2nd|chapter=Part One, Chapter V: Congruences, Theorem 77|url=https://archive.org/details/elementarynumber0000edmu_u9d2/page/50|location=New York|publisher=Chelsea Publishing Company|pages=51-52|lccn=66002147|oclc=1420155|ol=5976039M|access-date=2025-02-06}}

The result is trivial when $p\; =\; 2$, so assume $p$ is an odd prime, $p\; \backslash geq\; 3$. Since the residue classes modulo $p$ form a field, every non-zero residue $a$ has a unique multiplicative inverse $a^\{-1\}$. Euclid's lemma implies{{efn|Because if $a\; \backslash equiv\; a^\{-1\}\; \backslash pmod\{p\}$ then $a^2\; -1\; \backslash equiv\; 0\; \backslash pmod\{p\}$, and if the prime $p$ divides $a^2\; -\; 1\; =\; (a\; -\; 1)(a\; +\; 1)$, then by Euclid's lemma it divides either $a\; -\; 1$ or $a\; +\; 1$.}} that the only values of $a$ for which $a\; \backslash equiv\; a^\{-1\}\backslash pmod\{p\}$ are $a\; \backslash equiv\; \backslash pm\; 1\; \backslash pmod\{p\}$. Therefore, with the exception of $\backslash pm\; 1$, the factors in the expanded form of $(p\; -\; 1)!$ can be arranged in disjoint pairs such that product of each pair is congruent to 1 modulo $p$. This proves Wilson's theorem.

For example, for $p\; =\; 11$, one has

$$10!\; =\; [(1\backslash cdot10)]\backslash cdot[(2\backslash cdot6)(3\backslash cdot4)(5\backslash cdot9)(7\backslash cdot8)]\; \backslash equiv\; [-1]\backslash cdot[1\backslash cdot1\backslash cdot1\backslash cdot1]\; \backslash equiv\; -1\; \backslash pmod\{11\}.$$

Again, the result is trivial for p = 2, so suppose p is an odd prime, {{nowrap|1=p ≥ 3}}. Consider the polynomial

:$g(x)=(x-1)(x-2)\; \backslash cdots\; (x-(p-1)).$

g has degree {{nowrap|1=p − 1}}, leading term {{nowrap|1=x^{p − 1}}}, and constant term {{nowrap|1=(p − 1)!}}. Its {{nowrap|1=p − 1}} roots are 1, 2, ..., {{nowrap|1=p − 1}}.

Now consider

:$h(x)=x^\{p-1\}-1.$

h also has degree {{nowrap|1=p − 1}} and leading term {{nowrap|1=x^{p − 1}}}. Modulo p, Fermat's little theorem says it also has the same {{nowrap|1=p − 1}} roots, 1, 2, ..., {{nowrap|1=p − 1}}.

Finally, consider

:$f(x)=g(x)-h(x).$

f has degree at most p − 2 (since the leading terms cancel), and modulo p also has the {{nowrap|1=p − 1}} roots 1, 2, ..., {{nowrap|1=p − 1}}. But Lagrange's theorem says it cannot have more than p − 2 roots. Therefore, f must be identically zero (mod p), so its constant term is {{nowrap|1=(p − 1)! + 1 ≡ 0 (mod p)}}. This is Wilson's theorem.

It is possible to deduce Wilson's theorem from a particular application of the Sylow theorems. Let p be a prime. It is immediate to deduce that the symmetric group $S\_p$ has exactly $(p-1)!$ elements of order p, namely the p-cycles $C\_p$. On the other hand, each Sylow p-subgroup in $S\_p$ is a copy of $C\_p$. Hence it follows that the number of Sylow p-subgroups is $n\_p=(p-2)!$. The third Sylow theorem implies

:$(p-2)!\; \backslash equiv\; 1\; \backslash pmod\; p.$

Multiplying both sides by {{nowrap|1=(p − 1)}} gives

:$(p-1)!\; \backslash equiv\; p-1\; \backslash equiv\; -1\; \backslash pmod\; p,$

that is, the result.

In practice, Wilson's theorem is useless as a primality test because computing (n − 1)! modulo n for large n is computationally complex.Lagrange, p. 132: "cette méthode devient extrémement laborieuse, & presque impracticable"

Using Wilson's Theorem, for any odd prime {{nowrap|1=p = 2m + 1}}, we can rearrange the left hand side of

$$1\backslash cdot\; 2\backslash cdots\; (p-1)\backslash \; \backslash equiv\backslash \; -1\backslash \; \backslash pmod\{p\}$$

to obtain the equality

$$1\backslash cdot(p-1)\backslash cdot\; 2\backslash cdot\; (p-2)\backslash cdots\; m\backslash cdot\; (p-m)\backslash \; \backslash equiv\backslash \; 1\backslash cdot\; (-1)\backslash cdot\; 2\backslash cdot\; (-2)\backslash cdots\; m\backslash cdot\; (-m)\backslash \; \backslash equiv\backslash \; -1\; \backslash pmod\{p\}.$$

This becomes

$$\backslash prod\_\{j=1\}^m\backslash \; j^2\backslash \; \backslash equiv(-1)^\{m+1\}\; \backslash pmod\{p\}$$

or

$$(m!)^2\; \backslash equiv(-1)^\{m+1\}\; \backslash pmod\{p\}.$$

We can use this fact to prove part of a famous result: for any prime p such that p ≡ 1 (mod 4), the number (−1) is a square (quadratic residue) mod p. For this, suppose p = 4k + 1 for some integer k. Then we can take m = 2k above, and we conclude that (m!)² is congruent to (−1) (mod p).

Wilson's theorem has been used to construct formulas for primes, but they are too slow to have practical value.

Wilson's theorem allows one to define the p-adic gamma function.

Gauss provedGauss, DA, art. 78{{cite journal

| last1 = Cosgrave | first1 = John B.

| last2 = Dilcher | first2 = Karl

| journal = Integers

| mr = 2472057

| article-number = A39

| title = Extensions of the Gauss–Wilson theorem

| url = https://emis.de/journals/INTEGERS/papers/i39/i39.Abstract.html

| volume = 8

| year = 2008}} that

\prod_{k = 1 \atop \gcd(k,m)=1}^m \!\!k \ \equiv

\begin{cases}

-1 \pmod{m} & \text{if } m=4,\;p^\alpha,\;2p^\alpha \\

\;\;\,1 \pmod{m} & \text{otherwise}

\end{cases}

where p represents an odd prime and $\backslash alpha$ a positive integer. That is, the product of the positive integers less than {{mvar|m}} and relatively prime to {{mvar|m}} is one less than a multiple of {{mvar|m}} when {{mvar|m}} is equal to 4, or a power of an odd prime, or twice a power of an odd prime; otherwise, the product is one more than a multiple of {{mvar|m}}. The values of m for which the product is −1 are precisely the ones where there is a primitive root modulo m.

• Wilson prime
• Table of congruences
• Agoh–Giuga conjecture

{{noteslist}}

{{Reflist}}

{{refbegin}}

The Disquisitiones Arithmeticae has been translated from Gauss's Ciceronian Latin into English and German. The German edition includes all of his papers on number theory: all the proofs of quadratic reciprocity, the determination of the sign of the Gauss sum, the investigations into biquadratic reciprocity, and unpublished notes.

• English translation: {{cite book

| last1 = Gauss

| first1 = Carl Friedrich

| last2 = Clarke

| first2 = Arthur A.

| title = Disquisitiones Arithemeticae

| edition = 2nd corrected

| publisher = Springer

| location = New York

| year = 1986

| isbn = 0-387-96254-9

}}

• German translation: {{cite book

| last1 = Gauss

| first1 = Carl Friedrich

| last2 = Maser

| first2 = H.

| title = Untersuchungen über hohere Arithmetik (Disquisitiones Arithemeticae & other papers on number theory)

| edition = 2nd

| publisher = Chelsea

| location = New York

| year = 1965

| isbn = 0-8284-0191-8

}}

• {{cite book

| last = Landau

| first = Edmund

| title = Elementary Number Theory

| publisher = Chelsea

| location = New York

| year = 1966

}}.

• {{Cite book

| last = Ore

| first = Øystein

| author-link = Øystein Ore

| title = Number Theory and its History

| publisher = Dover

| year = 1988

| pages = [https://archive.org/details/numbertheoryitsh0000orey/page/259 259–271]

| isbn = 0-486-65620-9

| url = https://archive.org/details/numbertheoryitsh0000orey/page/259

}}

{{refend}}

• {{springer|title=Wilson theorem|id=p/w098010}}
• {{Mathworld|urlname=WilsonsTheorem|title=Wilson's Theorem}}
• Mizar system proof: http://mizar.org/version/current/html/nat_5.html#T22
• {{Cite web |last=Ohana |first=Andrew |title=A Generalization of Wilson's Theorem |url=https://sites.math.washington.edu/~morrow/336_09/papers/Andrew.pdf}}

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