Question

Show that if n is a composite integers with n is not 4, then (n-1) ! =0 mod n
Please, help

Answer 1

I don't know how to do it but:
n! = n (n-1)! = 0 mod n
n = a. b for some a, b in N,
and \(a, b\cancel \equiv 0\)
hence only one way to get it hold is \((n-1)! \equiv 0 (mod n)\)
Am I crazy?

Answer 2

dat should work.

another way to prove is that n=a.b for some a,b in N and a,b<n
so (n-1)! = 1...a..b..(n-1) for n>4
then (n-1)! = 0 mod n will follow

Answer 3

Nope, I don't think so
a.b = n , hence a.b > n-1
both them cannot be on (n-1) !

Answer 4

hahahaa sorry - its a comma , not period .
alright a, b in N and n>a, n>b
so for n>4, (n-1)!=1...a...b..(n-1)

happy now? :)

Answer 5

no :)

Answer 6

\(n = a*b\rightarrow a < n\rightarrow a \leq n-1\rightarrow a| (n-1)!\)
Same to b
\(n = a*b\rightarrow b < n\rightarrow b \leq n-1\rightarrow b| (n-1)!\)
But nothing to guarantee that a.b | (n-1)!
like 4 | 8
8|8
but 32 not | 8

Answer 7

ahhhh....so difficult to put this in writing....im not saying a.b < n or a.b < (n-1) u are right: it is NOT

but cuz' n=a.b and
n>a and n>b AND
a and b are not equal to (n-1)
(something about consecutive numbers cannot have a common factor except 1) SO
n-1>a and n-1>b
it follows that (n-1)!=1...a..b..(n-1)

Answer 8

Let try other way:
Let \( n = p1^{a_1} p2^{a_2}........pr^{a_r}\)
each of pi < n , hence \(p_i^{a_i} \leq n-1\)
Hence by Euclid,
\((n-1)! \equiv 0 (mod ~~p_i^{a_i}\))
And \((p_i^{a_i}, p_j ^{a_j}) =1\)
Hence by Wilson's theorem
\((n-1)! \equiv 0 (mod~~lcm ~~(p_i^{a_i}))\) And the last notation is n

Answer 9

both of ur proves should work :)

but the thing that bugs me is that somewhere there should be a restriction that n is not equal to 4 as given.

Answer 10

If n = 4, then only one composition is 2*2 , then n -1 = 3 and 3! =6
and then 6 not | 4, so that \(6 \cancel \equiv 0 (mod 4)\)

Answer 11

just weird that its not intrinsic to ur proves but they will work :)