Minimality of K = 216 for 3x3 mulgic square: Proof sketch: K must be a cube, with at least 9 divisors. 1,8,27,64,125 have 1,4,4,7,4 divisors. Done. -----Original Message----- From: Schroeppel, Richard Sent: Thursday, September 22, 2005 10:07 AM To: 'math-fun' Cc: Schroeppel, Richard; 'rcs@cs.arizona.edu' Subject: RE: [math-fun] Multiplicative Magic Squares There's a standard proof that the center of a 3x3 magic square is K/3, where K is the row sum. (Add up the four lines through the center, subtract the whole square.) Of course this works for multiplication too, so the magic product is always the cube of the center. There's a collection of similar results for a 4x4 square, such as "the sum of the four corners is K" etc. Although the arguments offered so far are plausible, I think a bit more proof is in order before claiming minimality of the various squares. Maybe a program? There aren't that many potential products < 5040 with 16 divisors, and we can also require the exponents in the prime factorization form a descending staircase. Allowing 0 as an entry doesn't seem helpful. I suspect that allowing negatives will turn out to be equivalent to the positive case with duplications allowed (but not triplications & up). But I don't see the proof that minus signs can always be attached to a positive square in a consistent way. Rich rcs@cs.arizona.edu -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of Christian Boyer Sent: Thursday, September 22, 2005 9:02 AM To: 'math-fun' Subject: RE: [math-fun] Multiplicative Magic Squares Of course replace "magic sum" by "magic product", twice in my previous message... -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de Christian Boyer Envoyé : jeudi 22 septembre 2005 16:30 À : 'math-fun' Objet : RE: [math-fun] Multiplicative Magic Squares In his old book on recreative mathematics, Boris Kordiemski described exactly the Michael's method (using two latin squares) for 4x4 multiplicative squares. He also described a method for 3x3 multiplicative squares: a b² a²b a²b² ab 1 b a² ab² With a=2 and b=3, you get 2 9 12 36 6 1 3 4 18 Magic sum = 216 (*). It is the smallest possible multiplicative square. In 1667 (a long time ago...), Arnauld gave a 3x3 multiplicative square. Perhaps the oldest published multiplicative square? Using powers of 2, its magic sum was bigger : 4096. Christian. (*) 216 is also the smallest cube being also sum of three cubes. 216 = 6^3 = 3^3 + 4^3 + 5^3 _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun