No, because m=1, n=1, k=1, is a counter-example. To show that, we substitute


That 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