perplexus.info :: Just Math : Maximum Value Muse

Maximum Value Muse (Posted on 2016-01-14)

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.

See The Solution Submitted by K Sengupta

Rating: 5.0000 (1 votes)

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Square Thoughts

| Comment 2 of 10 |

The fraction (2M-N)/(2M+N) must reduce to the square of a fraction, otherwise P will not be rational, let alone integral.

Multiplying the numerator and denominator by (2M+N):

(2M-N)/(2M+N) = [(2M)^2-N^2]/[(2M+N)^2]

The denominator is square, therefore the numerator must be a square to satisfy the problem. Then for each pair (M,N) there must be an X such that X^2 + N^2 = (2M)^2

From the Pythagorean triple (24,32,40): M=20, N=24 yields:

P = 24/4*sqrt[(40-24)/(40+24)] = 6*sqrt[1/4] = 3

This shows there are M,N,P which exist to satisfy the problem.

Posted by Brian Smith on 2016-01-15 09:46:53

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