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Maximum Value Muse (Posted on 2016-01-14) Difficulty: 3 of 5
Each of M and N is a positive integer such that:
P = (N/4)*√((2M – N)/(2M + N)) is a prime number.

Determine the maximum possible value of P and prove that no higher value of P is possible.

No Solution Yet Submitted by K Sengupta    
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A better try | Comment 5 of 10 |
My last post here is unconclusive. So I try again.

Suppose 2M -N +N define a interval of square numbers, 2N wide.
The extremes will be a^2 and b^2 and
the distance between the extremes will be n^2+2an, (n=b-a).

So: 2M-N=a^2 and 2M+N=b^2
2N=b^2-a^2=n^2+2an

Then:
P=(N/4)*(sq a^2/b^2)=((n^2+2a*n)/8)*(a/a+n)

P=(n*a/8)*[1+(a/(a+n))] 
This is a product of three factors, n, a, and [  ]/8. P should be a divisor of one of the three factors. But the third one is always between 2/8 and 1/8, so P is not a divisor here.

If P is divisor of a: a=kP, then kP*(n+2kP)/8(a+n))=1 an it results that P= n*(8-kn)/(2kkn-8) which is positive only for interval (4-kn-8) For K=1 n=6 a=3 and P=3

If P is a divisor of n: n=kP there is a quite similar formula with no positive value for P.


So the only solution is P=3 and no other solution is posible.
(I edited again to add k factor)
Edited on March 1, 2016, 5:21 pm


Edited on March 2, 2016, 3:05 am
  Posted by armando on 2016-03-01 16:41:56

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