Christian Boyer wrote:
Yes, Mike, hmmm, hmmm... I pretended to have not well understood the Michael Kleber's question ;-)
Yes, I had a funny feeling...
Before to explain the method, I will try also to answer at least to the more difficult question:
what is the minimum 5x5 multiplicative square?
I was reluctant to open that can of worms. But roughly the same idea which produced the 7! 4x4 magic square also produces a 9! 5x5: [ 10 24 4 42 9 ] [ 28 54 5 16 3 ] [ 8 2 21 36 30 ] [ 27 20 48 1 14 ] [ 6 7 18 15 32 ] magic product 9! = 362880 = 2^7.3^4.5.7 This time it's built on the permutation matrix [ 1 0 0 0 0 ] [ 0 0 0 1 0 ] [ 0 1 0 0 0 ] [ 0 0 0 0 1 ] [ 0 0 1 0 0 ] Jessica, aka "my wife the weaver", would probably be annoyed if I didn't call this the "satin" matrix, though readers here might prefer to think of it as the order in which you visit vertices of a pentagon when inscribing a star. Anyway, this permutation matrix is a 5x5 magic square with sum 1 -- one of 20 such; see A007016 for further counts. Better yet, it's what David Wilson just reminded us is called "diabolic": every broken diagonal likewise has sum 1. This means that we can take it and four more copies with rows shifted cyclically, and tile a 5x5 matrix, getting each row, column, and broken diagonal to have one entry from each of the five tiling matrices. Laying the transpose on top of that matrix, you get the beauty: [ Aa Bc Ce Db Ed ] [ Cb Dd Ea Ac Be ] [ Ec Ae Bb Cd Da ] [ Bd Ca Dc Ee Ab ] [ De Eb Ad Ba Cc ] The 25 entries in the matrix are all distinct, and it's diabolically magic, with product, if you're thinking multiplicatively, ABCDEabcde. Now we just need to assign those ten variables numerical values such that all 25 entries remain distinct. The 4x4 5040 square came from setting two variables to 1 and the other six to 2-7 in some fortuitious order, and we can do the parallel thing here: set (A,B,C,D)=(2,3,4,6) and (a,b,c,d)=(5,7,8,9), and E=e=1, to get [ 10 24 4 42 9 ] [ 28 54 5 16 3 ] [ 8 2 21 36 30 ] [ 27 20 48 1 14 ] [ 6 7 18 15 32 ] diabolical magic product 9! = 362880 = 2^7.3^4.5.7 Q: Is it always possible to partition the integers from 2 to 2n-1 into two sets of size n-1, such that the products of two elements, one from each set, all distinct integers >=2n? Seems like the answer is "yes with high probability", but can someone change that to just "yes"? Once again, I feel like one ought to be able to argue from elementary principles that this must be the minimum possible, but I'm unable to do so (though I do think the Hilbert basis for the 4x4 case should now make that argument easy). So we can wait for Christian's algorithm to get fast enough to check everything smaller, or hope to use David Wilson's logic to limit the smaller possible signatures, etc. All right, back to real work... --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.