Getting Primed With Squares (Posted on 2008-07-27)

	Submitted by K Sengupta
Rating: 3.5000 (2 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.

	Subject	Author	Date
	re: Solution	K Sengupta	2008-08-10 06:29:20
	Solution	Jonathan Lindgren	2008-07-27 17:06:59