The puzzle reserves a tricky final because the formula
P = (N/4)*√((2M-N)/(2M + N))
involves only 2M and N as variables. It turns out that there is an infinite number of pairs (2M, N) resulting in P prime.
So the formula is satisfied by all the pairs (2M, N) = (5K, 4K) con K=3p e p any prime.
But then 2M is always odd, so that M is not an integer.
And then the only other solution is M=20, N=24, P=3.
Edited on March 3, 2016, 11:40 am
Posted by armando
on 2016-03-03 04:46:03