Question

show that \((p-2)!\) is divisible by \(p-1\) for all primes \(p\gt 5\)

Answer 1

same as showing that (p-1)! is divisible by p
hehe ok trying to restore Wilson proof

Answer 2

(p-1)! is never divisible by p by wilson thm because

(p-1)! = -1 (mod p)

Answer 3

\(\large \color{black}{\begin{align} (p-2)! \pmod {(p-1)}\hspace{.33em}\\~\\
\implies (p-2)(p-1)p! \pmod {(p-1)}\hspace{.33em}\\~\\
\implies (p-2)(p-1)p! \pmod {(p-1)}\equiv 0\hspace{.33em}\\~\\
\end{align}}\)

Answer 4

p = 3
p-2 = 1
1! = 1
1 is not divisible by 2

Is there an appropriate minimum p?

Answer 5

yes sorry p > 5 @tkhunny

Answer 6

let me update the question

Answer 7

oh hehe misunderstood

Answer 8

also mathmath's method wont work, we can't multiply a number..

Answer 9

ok
(p-2)! =1*2*3*.....*(p-3)*(p-2)
humm (thinking)

Answer 10

6 is not prime

Answer 11

p is prime :\

Answer 12

updated the question again sry ;p

Answer 13

ok u want primes

Answer 14

ok got it
p-1 is even composite thus
p-1=2*n such that n<p-2
hence
(p-2 )! |n
and
(p-2)!|2 since p>5 there would be extra 2's in 4 so it wont matter if n is even or not

Answer 15

wew! so cute

Answer 16

can i have a bit more rigorous proof

Answer 17

ok
p is odd thus p-1 is even
p-1=2*n (such that n <p-2)

so \( n \in (1,2,3,.....,p-3) \\ n| (1*2*3*.....*p-2)=(p-2)!\)

now when n is odd
n>2 thus n can be any number btw 3 and p-2 (so n wont be 2)
hence 2*n|(p-2)!

when n is even
p-1>=8
thus
n btw 4 and p-2 (so n wont be 2)
hence
2*n|(p-2)!

Answer 18

hmm :\ sounds clear to me idk what ur looking for

Answer 19

Okay nice that will do

Answer 20

hmm

Answer 21

do u have some other idea ?