Proof positive: squares of integers (Posted on 2007-04-07)

	Submitted by K Sengupta
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3p^2 + p = 4q^2 + q Or, p-q = 4q^2 - 3p^2 Now: (p-q)(1+4q+4p) = p-q + 4p^2 - 4q^2 = 4q^2 - 3p^2 + 4p^2 - 4q^2 = p^2....(i) Similarly, (p-q)(1+ 3q +3p) = q^2.....(ii) Multiplying (i) by (ii), we obtain: (p-q)^2(3p+3q+1)(4p+4q+1) = (pq)^2.....(iii), so that: (p-q)^2(3p+3q+1)(4p+4q+1) is a perfect square. Now, 3(4p+4q - 1) - 4(3p+3q-1) = 1, so that (3p+3q-1) and (4p+4q-1) must be relatively prime to each other. Accordingly, each of (3p+3q+1)and (4p+4q+1) must be perfect squares and this in conjunction with (iii) implies that (p-q) must correspond to a perfect square. Therefore, each of p-q, 3p+3q+1 and 4p+4q+1 are squares of integers.

	Subject	Author	Date
	Third of three proved	Brian Smith	2007-04-09 14:32:52
	Two of three proved	Brian Smith	2007-04-09 11:34:14
	Hint	K Sengupta	2007-04-09 05:55:15