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Getting Primed With Squares (Posted on 2008-07-27) Difficulty: 2 of 5
Determine all possible pair(s) of primes (M, N) such that each of M+N and M-N is a perfect square.

  Submitted by K Sengupta    
Rating: 2.0000 (1 votes)
Solution: (Hide)
Jonathan Lindgren has provided a solution here.

I solved this problem in the following manner:

Suppose that each of M+N and M-N are perfect squares. If so, then (M+N, M-N) = (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 (x-y) 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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: SolutionK Sengupta2008-08-10 06:29:20
SolutionJonathan Lindgren2008-07-27 17:06:59
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