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..n1 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)
p1 (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 20070125 02:45:12 