This is my last try to this puzzle. After Brian's note I am changing approach.
If the sq. root (2M-N/2M+N) is not a direct ratio of two square numbers, let multiply both -numerator and denominator- by (2M+N).
Then (4M^2-N^2) in the numerator should be a square number if we want the sq. root to be useful to get an integer in the P expression.
This calls for a pitagorean approach: 2M, N should be pitagorean (not primitive, because 2M is even). Then:
N=k(m^2-n^2) and the sq. root turns out very simple as: n/m.
So the P expression is now:
P=kn(m^2-n^2)/4m with m and n coprimes and even/odd pair.
But m is not a divisor of (n, m-n, m+n), so or k=4m or k=m and n=0 (mod4) or k=2m and n=2.
If k=4m, P is a product (or if n=1, is a sq. diff: see below)
If k=m and n=0 (mod4), P=n(m^2-n^2)/4 which is not a product when n=4. Then P=m^2-16, but this is a sq diff and can be express as a product (f.ex: a(a+8),with a=m-4).
Analog situation for k=2m.
Can conclude that when N=k(m^2-n^2) no way to a valid P prime.
But N is pitagorean, so it can also take the form: N=k(2mn)
The P expression become then: P=kmn(m-n)/2(m+n)
but m+n do not divide (m,n,or (m-n)).
So k should be 0 (mod m+n).
Then k=k'(m+n) and P=k'mn(m-n)/2 which for k'=1 m=3 n=1 gives P=3 (2M=40 N=24: so this is a possible solution);
For m=2, n=1 P=k'=k/3 (So P is prime when k/3 is prime).
Then it turn out that an infinite number of pairs (2M, N) result in P prime, with no limit in the value for P.
Verifying: 2M=5k, N=4k,
F. es: k=51, 2M=255 N=204 P=17
But as 2M=5k and k=3P, 2M=15P is always odd (except if P=2), so M is not an integer in this wide range of solutions.
P=2 meets also the contrains of the puzzle: M=15 N=24 P=2, but it is a lower solution than P=3.
The solution to this puzzle is then: M=20, N=24, P=3
Edited on March 4, 2016, 4:31 am
Posted by armando
on 2016-03-02 16:40:40