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Wilson’s Theorem
• Difficulty Level : Medium
• Last Updated : 19 Nov, 2016

Wilson’s theorem states that a natural number p > 1 is a prime number if and only if

```
(p - 1) ! ≡  -1   mod p
OR  (p - 1) ! ≡  (p-1) mod p
```

Examples:

```p  = 5
(p-1)! = 24
24 % 5  = 4

p  = 7
(p-1)! = 6! = 720
720 % 7  = 6
```

How does it work?
1) We can quickly check result for p = 2 or p = 3.

2) For p > 3: If p is composite, then its positive divisors are among the integers 1, 2, 3, 4, … , p-1 and it is clear that gcd((p-1)!,p) > 1, so we can not have (p-1)! = -1 (mod p).

3) Now let us see how it is exactly -1 when p is a prime. If p is a prime, then all numbers in [1, p-1] are relatively prime to p. And for every number x in range [2, p-2], there must exist a pair y such that (x*y)%p = 1. So

```    [1 * 2 * 3 * ... (p-1)]%p
=  [1 * 1 * 1 ... (p-1)] // Group all x and y in [2..p-2]
// such that (x*y)%p = 1
= (p-1)```

How can it be useful?
Consider the problem of computing factorial under modulo of a prime number which is close to input number, i.e., we want to find value of “n! % p” such that n < p, p is a prime and n is close to p. For example (25! % 29). From Wilson's theorem, we know that 28! is -1. So we basically need to find [ (-1) * inverse(28, 29) * inverse(27, 29) * inverse(26) ] % 29. The inverse function inverse(x, p) returns inverse of x under modulo p (See this for details).

See this for more applications of Wilson’s Theorem.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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