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:
1 (obviously)
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