Just cleaning up old email, so this is a very, very belated response, which may have been taken care of by someone long ago. Copied to seqfans, in case someone wants to make a sequence or two out of it. Solutions of x(x+1)/2 + y(y+1)/2 = z! are solutions of (2x+1)^2 + (2y+1)^2 = 8(z!) + 2 and the first few values of z for which there are solutions can easily be ascertained: (x,y,z) = (0,1,0), (0,1,1), (1,1,2), (0,3,3), (2,2,3), (2,6,4), (0,15,5), (5,14,5), (45,89,7), (89,269,8), (210,825,9), (760,2610,10), (1770,2030,10), none for z = 11, 12, one for z = 13 (see Ed's solution below), none for z = 14, two for z = 15 (see below), one for z = 16, two for z = 17, none for z = 18, 19, 20, two for z = 21, none for z = 22, 23, two for z = 24, none for z = 25, 26, eight for z = 27, one for z = 28, two for z = 29, none for z = 30, 31, four for z = 32, 33, sixteen for z = 34, none for z = 35, 36, ..., 41, two for z = 42, none for z = 43, 44, ..., 48, sixteen for z = 49, none for z = 50, 51, 52, 53, one for z = 54, none for z = 55, 56, ..., 65, two for z = 66, none for z = 67, sixteen for z = 68, ... (E&OE, and PARI is slowing down a bit now: AND it would take rather longer to find the actual solutions!) R. On Tue, 20 Jan 2009, Ed Pegg Jr wrote:
A very belated response.
Not squares, but related. Using Triangular numbers. 1 3 6 10 15 21
T[3] = 3! T[3] + T[6] = 4! T[14]+T[5] = T[15] = 5! T[45] + T[89] = 7! T[210] + T[825] = 9! T[1770] + T[2030] = 10! T[71504] + T[85680] = 13! T[213384] + T[1603064] = T[299894] + T[1589154] = 15!
I don't see an easy way to extend these. The density of triangular numbers seems to be sufficient for extended solutions.
--Ed Pegg Jr
Date: Mon, 3 Jul 2006 11:44:20 -0600 (MDT) From: Richard Guy <rkg@cpsc.ucalgary.ca> Reply-To: math-fun <math-fun@mailman.xmission.com> To: Number Theory List <NMBRTHRY@listserv.nodak.edu>, Math Fun <math-fun@mailman.xmission.com> Subject: [math-fun] Factorial n
Presumably 0! = 1! = 0^2 + 1^2. 2! = 1^2 + 1^2 6! = 12^2 + 24^2
are the only integer solutions of
n! = x^2 + y^2
but is there a proof? [Later: for n > 6, n! will always contain a prime factor of shape 4k - 1 raised to an odd power, in fact raised to the first power, by a suitable form of the prime number theorem.] R. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun