No, because m=1, n=1, k=1, is a counter-example. To show that, we substituteThat only holds when p=1, but 1 is not a prime. It does not work for the first prime 2 And as p increases through larger and larger primes, the left side increases approaching 1, but the right side decreases approaching 0. So no larger integer for p can possibly be a solution, let alone a larger prime number. Edwin