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Getting Primed With Squares (Posted on 20080727) 

Determine all possible pair(s) of primes (M, N) such that each of M+N and MN is a perfect square.

Submitted by K Sengupta

Rating: 2.0000 (1 votes)


Solution:

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Jonathan Lindgren has provided a solution here.
I solved this problem in the following manner:
Suppose that each of M+N and MN are perfect squares. If so, then (M+N, MN) = (x^2, y^2), for some integers x and y.
Thus, 2N = x^2  y^2 ........(i)
Hence, each of x any y must have the same parity. Therefore, each of (x+y) and (xy) must be even, so that the rhs of (i) is always divisible by 4.
Accordingly, N is divisible by 2. Since N is prime, this is possible only when N=2.
Therefore, x^2  y^2 = 4, giving: (x^2, y^2) = (4, 0) as the only possiblility, so that:
M+2 = 4, giving: M = 2.
Consequently, (M, N) = (2,2) is the only possible solution to the given problem.

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Subject 
Author 
Date 
 re: Solution  K Sengupta  20080810 06:29:20 
 Solution  Jonathan Lindgren  20080727 17:06:59 


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