For the triangular number problem, multiply the equation by 8 and add 2. Using T(x) = x(x+1)/2, T(a) + T(b) = N! becomes (2a+1)^2 + (2b+1)^2 = 8 N! + 2. The RHS should sometimes meet the criteria for being the sum of two squares, often multiple times. The criteria is that there is no 4K+3 prime divisor to an odd power. If you have the factorization, or know the number is twice a prime, you can calculate the sum-of-two-square representations, and then the triangles. Wrt RKG's original problem, N! = A^2 + B^2: N! has a gaggle of prime divisors that occur to the first power: Any prime P in the range N/2 < P <= N will divide N! to exactly the first power. If any prime in this range is a 4K+3 prime, it prevents splitting N! into two squares. I assume there's a quantitative version of PNT for 4K+3 primes which will give a calculable bound on N. Maybe there's a clever way to adopt the proof of Bertrand's postulate (there's always a prime between J and 2J) for 4K+3 primes. The final tidying step is to provide a sequence of 4K+3 primes, each less than twice the previous, to cover the bottom end of the range. 7, 11, 19, 31, 59, ... This suggests some related puzzles: Three squares; and whats the mod2,4,8 pattern of the odd-part of N!? Rich ________________________________________ From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com [math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Ed Pegg Jr [ed@mathpuzzle.com] Sent: Tuesday, January 20, 2009 8:14 AM To: math-fun Subject: [math-fun] Triangular+Triangular = Factorial 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? R. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun