papacy said:if p is prime and n is natural number, show that (p-n+1) . (p-2)! + n–1 =0 (mod p)
i think i have to show that (p-n+1) . (p-2)! = 0 (mod p) and n – 1 =0 (mod p)
using Wilson theorem
(p-1)!=-1(p)
DonAntonio said:Well, of course [itex]\,n-1\neq 0\pmod p\,[/itex] almost always. Working modulo p in the following:
$$(p-2)!=\frac{(p-1)!}{p-1}=\frac{-1}{-1}=1\Longrightarrow (p-n+1)(p-2)!+n-1=(-n+1)\cdot 1+n-1=-n+1+n-1=0$$
DonAntonio
Wilson's Theorem is a mathematical theorem that states that for any prime number p, (p-1)! is congruent to -1 (mod p). In other words, if p is a prime number, then (p-1)! + 1 is divisible by p.
The formula is (p-n+1) . (p-2)! + n–1 =0 (mod p).
Wilson's Theorem is often used in conjunction with the formula (p-n+1) . (p-2)! + n–1 =0 (mod p) to solve for the value of p. By substituting (p-1)! with -1 (mod p) in the formula, we can simplify it to p-n+1 + n–1 =0 (mod p), which can be further simplified to p=0 (mod p). This means that p is a factor of p, which is true for all prime numbers.
Solving (p-n+1) . (p-2)! + n–1 =0 (mod p) allows us to find the value of p, which can be useful in various mathematical problems and proofs. It also demonstrates the relationship between Wilson's Theorem and the formula, highlighting the power and applicability of modular arithmetic.
Yes, there are a few limitations. First, Wilson's Theorem only applies to prime numbers, so it cannot be used for composite numbers. Second, the formula may not always yield a unique solution for p, as there may be multiple combinations of n and p that satisfy the equation. Lastly, the formula can be computationally intensive for large values of p, making it impractical for certain applications.