Prove that for each positive integer n there exist n consecutive positive integers none of which is a prime or a prime power.
Solution
Consider (N!)2+2, (N!)2+3, ... , (N!)2+N. (N!)2+r is divisible by r, but ((N!)2+r)/r = N! (N!/r) + 1, which is greater than one, but relatively prime to r since N! (N!/r) is divisible by r. For each r we may take a prime pr dividing r, so (N!)2+r is divisible by pr, but is not a power of pr. Hence it is not a prime or a prime power. Taking N = n+1 gives n consecutive numbers as required.
Solutions are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
12 Nov 1998