The least square (Posted on 2005-11-05)

	Submitted by pcbouhid
Rating: 4.0000 (3 votes)
Solution:	(Hide)
Let (15a + 16b) = r^2 and (16a - 15b) = s^2, where r, s belongs to N. We obtain: r^4 + s^4 = (15a + 16b)^2 + (16a - 15b)^2 = (15^2 + 16^2)*(a^2 + b^2) = 481*(a^2 + b^2). 481 = 13 x 37. We now use the fact that -1 is not a fourth power either mod 13 or mod 37 (to see why this holds, note that the congruence -1 (congr) x^4 (mod 13) for some x (of N) leads via Fermatīs theorem to (-1)^3 congr 1 (mod 13), which is false. Likewise, -1 congr x^4 (mod 39) leads to (-1)^9 congr 1 (mod 37), which is false too. Since r^4 + s^4 congr 0 (mod 13), either r congr s congr 0 (mod 13), or r and s not congr 0 (mod 13). The latter possibility cannot occur because -1 is not a fourth power mod 13; therefore r congr 0 and s congr 0 (mod 13), and similarly, r congr 0 and s congr 0 (mod 37). Therefore, r and s are multiples of 481, and so r >= 481, s >= 481. It is easy to check that r = s = 481 is realizable. We obtain a = 481*31 and b = 481, that leads to r^2 (or s^2) = 481^2.

	Subject	Author	Date
	re: pc	pcbouhid	2005-11-16 08:15:07
	pc	xdog	2005-11-16 07:16:10
	re: No Subject - to xdog	pcbouhid	2005-11-15 14:13:21
	No Subject	xdog	2005-11-15 09:25:46
	re(2): Solution	pcbouhid	2005-11-15 05:25:38
	re: Solution	Mindy Rodriguez	2005-11-14 21:57:46
	re: Solution	Mindy Rodriguez	2005-11-09 21:26:17
	Solution	xdog	2005-11-09 15:16:04
	re: Solution - not sure if least	Mindy Rodriguez	2005-11-07 21:48:58
	No Subject	Joe	2005-11-06 17:29:59
	Solution - not sure if least	goFish	2005-11-06 16:52:53