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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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Solution tried | Comment 4 of 10 |
Suppose that 2M -N +N define a interval of square numbers, 2N wide.
The distance between the extremes a^2 & b^2 can be formulate as n^2+2an, (n=b-a), as they are squares numbers.
So:
2M-N=a^2
2M+N=b^2
2N=b^2-a^2=n^2+2an

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

In other words
P=(n*a/8)*[1+(a/(a+n))]

See that in the above expression:
[When ratio a/n decreases, the remainder of (a/a+n) is convergent to 0 and P -> na/8 (but never reaching it). 
And:
[When ratio n/a decreases, the remainder of (a/a+n) is convergent to 1 and P -> na/4 (never reaching).
For each value of "a" there is no integer any more under a certain ratio in the P expression (and same for "n").

So for the higher P values a~n
With a=n
P=(a^2/8)*[(1+(a/2a)]=3a^2/16
Only fits for a=4 P=3 which is the same solution posted by Brian.


Edited on January 21, 2016, 1:43 am
  Posted by armando on 2016-01-19 16:40:48
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