Excellent analysis, Michael! And you will be credited of this nice 10x10 pandiag. multiplic. square in my future website update. Our 6x6 and 10x10 examples have a magic product 2^a * 3^b. I mentioned in my previous email some other squares. Here is for example my 8x8 square having a magic product 2^a * 3^b * 5^c * 7^d, allowing the use of smaller integers than a 8x8 square using only 2^a * 3^b integers: 1 3780 4 945 24 630 6 2520 216 70 54 280 9 420 36 105 5 756 20 189 120 126 30 504 1080 14 270 56 45 84 180 21 315 12 1260 3 7560 2 1890 8 840 18 210 72 35 108 140 27 63 60 252 15 1512 10 378 40 168 90 42 360 7 540 28 135 And it is also a most-perfect pandiagonal square, with same product for all 2x2 sub-squares. Unfortunately it seems difficult to construct 6x6 and 10x10 squares with smaller integers allowed by this other technique used for 8x8. Am I right? But it is possible for 9x9: my example (with 3x3 sub-squares) is a 2^a * 3^b * 5^c * 7^d construction. Christian. -----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 Michael Kleber Envoyé : vendredi 21 avril 2006 20:58 À : math-fun Objet : Re: [math-fun] 6x6 pandiag. multiplic. squares ARE NOT impossible! Christian Boyer wrote:

Here is an example!

7776 3 3888 96 243 48 2 46656 4 1458 64 2916 2592 9 1296 288 81 144 486 192 972 6 15552 12 32 729 16 23328 1 11664 162 576 324 18 5184 36

All the rows, columns, diagonals AND broken diagonals give the same product:[...] And best, it is a "most-perfect" pandiagonal magic square, à la Kathleen Ollerenshaw. All 2x2 sub-squares have always the same product:[...]

This is a very pretty construction! I think in the same way you can generate pandiagonal multiplictive magic squares of any size not a prime power (and various sub-square sizes of most-perfect-ness). Here's my interpretation of your example: --------- a) Take the 2x3 matrix [ 5 0 4 ] [ 1 6 2 ] This has relatively prime side lengths, distinct positive integer entries, a constant row sum, and a (different, of course!) constant column sum. Call this building-block matrix B. b) Form a 6x6 matrix by tiling with 6 copies of B. Now you have a square with constant row and col sum (equal to 18, the sum of the entries in B) and the same constant sum along all broken diagonals, because each diagonal traces a (2,3)-torus knot on B. Also, any 2x2 subsquare is exactly 2 columns of B, so will sum to 2/3 the other sums. Call this tiled matrix T. c) Make a multiplicative magic square M = 2^T 3^transpose(T). (This is pointwise exponentiation and multiplication.) --------- It is vital that the sides of the building block B are relatively prime (this ensures the pandiagonality and the fact that M's entries are all distinct). Oh, and the most-magic-ness is that all squares of side length the smaller side of B will have the same product. So to make a 10x10 square with all those desirable properties, you could start with the 2x5 rectangle [ 0 11 2 9 8 ] [ 12 1 10 3 4 ] and perform the same construction, to get 1, 725594112, 9, 80621568, 6561, 4096, 177147, 36864, 19683, 26873856 1088391168, 6, 120932352, 54, 165888, 1062882, 6144, 118098, 55296, 162 4, 181398528, 36, 20155392, 26244, 1024, 708588, 9216, 78732, 6718464 272097792, 24, 30233088, 216, 41472, 4251528, 1536, 472392, 13824, 648 256, 2834352, 2304, 314928, 1679616, 16, 45349632, 144, 5038848, 104976 531441, 12288, 59049, 110592, 81, 2176782336, 3, 241864704, 27, 331776 2048, 354294, 18432, 39366, 13436928, 2, 362797056, 18, 40310784, 13122 2125764, 3072, 236196, 27648, 324, 544195584, 12, 60466176, 108, 82944 512, 1417176, 4608, 157464, 3359232, 8, 90699264, 72, 10077696, 52488 136048896, 48, 15116544, 432, 20736, 8503056, 768, 944784, 6912, 1296 Every row/col/diag multiplies to 2^60 3^60 = 48873677980689257489322752273774603865660850176 and every 2x2 square multiplies to 2^24 3^24 = 4738381338321616896. For a 20x20 example you'd need a 4x5 B and every 4x4 block in the final square would have the same product. For an example of size pqr, like 30, you could start with a B of sizes 2x15, 3x10, or 5x6, and you'd get different most-magic subsquare sizes for each. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

Christian Boyer wrote:

Here is for example my 8x8 square having a magic product 2^a * 3^b * 5^c * 7^d, allowing the use of smaller integers than a 8x8 square using only 2^a * 3^b integers:

1 3780 4 945 24 630 6 2520 216 70 54 280 9 420 36 105 5 756 20 189 120 126 30 504 1080 14 270 56 45 84 180 21 315 12 1260 3 7560 2 1890 8 840 18 210 72 35 108 140 27 63 60 252 15 1512 10 378 40 168 90 42 360 7 540 28 135

Here's what I can make of this. The end result is that I can improve on this square a little -- decrease both the magic product and the maximum entry -- but only in a sort of shallow way. As usual, look at the matrix through "log_p"-colored glasses for the various p which appear. If you write down the highest powers of 2, 3, 5, 7 in each entry -- that is, take the prime factorization, and spread the exponents of each p out into separate matrices -- here is the result: 02203113 03031212 01010101 01010101 31130220 30302121 01010101 01010101 02203113 03031212 10101010 01010101 31130220 30302121 10101010 01010101 02203113 21213030 10101010 10101010 31130220 12120303 10101010 10101010 02203113 21213030 01010101 10101010 31130220 12120303 01010101 10101010 The powers of 5 and 7 make clear what's going on here: you've overlayed a bunch of pandiagonal magic bit planes with the property that every 2x2 box contains precisely two 1's, and you've managed to arrange things so that you get all distinct entries at the end. The matrices for powers of 2 and 3 are showing exactly the same thing, but those each have two bits worth of information. We should separate them out into the 2-plane and the 4-plane. Likewise the 3s should be separated into 3s and 9s bit-planes. 02203113 00001111 01101001 03031212 01011010 01010101 31130220 11110000 10010110 30302121 10100101 10101010 02203113 00001111 01101001 03031212 01011010 01010101 31130220 11110000 10010110 30302121 10100101 10101010 02203113 00001111 01101001 21213030 01011010 10101010 31130220 11110000 10010110 12120303 10100101 01010101 02203113 00001111 01101001 21213030 01011010 10101010 31130220 11110000 10010110 12120303 10100101 01010101 So all in all, you have six bit planes, all of which exhibit very regular patterns, and your matrix is the result of considering them as the exponents of 2,4, 3,9, 5, 7. This now gives us an easy way to lower the magic product and max entry in your square: your use of 9 as the base for one plane is inefficient, since a smaller prime power, 8, is still available. Taking your matrix and replacing every multiple of 9 with 8/9 of itself, you get 1 3360 4 840 24 560 6 2240 192 70 48 280 8 420 32 105 5 672 20 168 120 112 30 448 960 14 240 56 40 84 160 21 280 12 1120 3 6720 2 1680 8 840 16 210 64 35 96 140 24 56 60 224 15 1344 10 336 40 168 80 42 320 7 480 28 120 with max entry 6720 (= 8/9 of your 7560) and magic product 2039281090560000 (=(8/9)^4 of yours, since each bit plane had four 1's in each row). So there is a natural path for generalizing this construction, where you first need to understand all magic bit planes, and how many of them you need to get all entries distinct (as bit strings), then you pack them into exponents as tightly as possible. In this case you've managed to find 6 bit planes which do the job, and this is clearly best possible, because your matrix has 8^2 = 2^6 = 64 entries. It makes one hope that you can always attain this natural bound -- now that would be a nice construction! (Does your 9x9 matrix similarly break apart into four trit-planes, with exponents 2,3,5,7?) Note that this means your multiplicative magic could also be expressed as a regular magic square, thinking of the same six bit-planes as the binary expansion of numbers 0-63. But this would hide some of the niceness; your magic works if you add the numbers in base two column-by-column, while I suppose a normal 8x8 magic square might fail this test and only be magic after you do the carrying in the base 2 addition. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.

Franklin T. Adams-Watters wrote:

This does not work. You have duplicated values: e.g., the 8 in the middle of the 2nd line and the end of the 5th line.

Oof, that was dumb. Quite right, I'd need to use 16 as the base, which is clearly worse than 9... --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.