The discussion centers on proving the equation (p-n+1) . (p-2)! + n–1 = 0 (mod p) using Wilson's Theorem. Participants confirm that (p-2)! simplifies to 1 when p is prime, leading to the conclusion that the equation holds true under the specified conditions. The proof involves manipulating the equation to show that both terms equal zero modulo p. Additionally, one participant mentions solving the problem using induction.
PREREQUISITESMathematicians, students studying number theory, and anyone interested in modular arithmetic and proofs involving Wilson's Theorem.
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