Prove that n!-1 is a composite number when n>3 and n+2 is a prime.
When n+2=p is a prime the numbers 1..n-1 mod p are a cyclic multiplicative group.
Thus, there is some number j 1<j<n such that
for for all k 1<=k<n there is some integer m j^m mod n+2 = k
Exactly two of the elements are thier own inverse:
p-1 (congruent to -1 mod p and -1*-1=1)
n! is the product of all of those elements that have an inverse. So, arrange the factors into the pairs k, inverse(k) and you can see that n! is congruent to 1 mod p. Thus, n!-1 is congruent to 0 mod p and so n!-1 is divisible by n+2. For all n>3 n!-1 > n+2 so n!-1 is thus composite.
Posted by Joel
on 2007-01-25 02:45:12