Problem 4
For a real number x, let floor(x) be the greatest integer not exceeding x. Prove that floor((n-1)! divided by n(n+1)) is even for every positive integer n.
Step 4 of 5: Use Wilson when n+1 is prime
Detailed analysis
Let q=n+1 be an odd prime, so n=q-1 is even and composite (the small case q=3 was already handled). Since n divides (n-1)!+n and Wilson gives n!=(q-1)! congruent to -1 modulo q, we obtain (n-1)!+n divisible by both n and q, hence by n(n+1). Put K=((n-1)!+n)/(n(n+1)). For n≥6, the factorial has strictly more powers of 2 than n, so ((n-1)!+n)/n is odd; q is odd, hence K is odd. The quotient in the problem is K-1/(n+1), whose floor is K-1, even.