Getting Squared With 1 And 16 (Posted on 2007-09-01)

	Submitted by K Sengupta
Rating: 3.6667 (3 votes)
Solution:	(Hide)
We observe from the given equation that q cannot be negative, for that will yield negative values for RHS of the given equation, which is a contradiction. Again, if (p, q) is a solution, then (-p, q) is also a valid solution………(#). q=0 gives p = +/- 1, while p=0 gives q^4 = 1+ 16q, which do not yield integer solutions. In terms of (#), we can consider both p and q to be positive. From the given equation, we obtain: EITHER, p^2 = q^2 + V(1+16q) OR, p^2 = q^2 - V(1+16q) In the former situation, p> q. Now, (q+1)^2 > q^2 + V(1+16q) iff (2q+1)^2 > 1+ 16q Or, q> 3. But, for q> 3, it follows that : q^2< p^2< (q+1)^2, so that p is not an integer. This is a contradiction. For q = 1, 2, the expression (1+16q) is not a perfect square, while q = 3 gives p = 4 In the latter situation, p< q. Now, (q-1)^2 > q^2 - V(1+16q) iff: (2q-1)^2 > 1- 16q, so that q > 5 But, for q> 5, it follows that (q-1)^2< p^2< q^2, so that p is not an integer. This is a contradiction. For q = 1, 2, 4 we observe that the expression (1+16q) is not a perfect square, while q = 3 gives p^2 = 2, a contradiction. However, q = 5 in the latter expression gives p = 4 Thus, in terms of (#), we obtain: (p, q) = (+/- 1, 0); (+/-4, 3); (+/-4, 5) as all possible solutions to the given problem.

	Subject	Author	Date
	Solution	Brian Smith	2007-09-05 10:22:23
	Some observations.	xdog	2007-09-04 17:06:12
	re: some general solutions	brianjn	2007-09-02 10:53:42
	some general solutions	Daniel	2007-09-02 05:37:51
	Some values	brianjn	2007-09-01 23:42:28