Let P be the largest prime less than p. Then P divides p!, hence P also divides a^2. But for prime P to divide a^2, it must also divide a, that is, P^2 divides p!, so that p must be >= 2P. However, we know that for any number x >= 1, there is a prime between x and 2x.
It follows that the sole solution is a=p=1.
Edited on February 24, 2008, 10:24 am
Posted by FrankM
on 2008-02-24 10:21:29