In his http://www.maa.org/editorial/mathgames/mathgames_11_07_05.html column, our friend Ed Pegg Jr. wrote: "The latin squares method, used for 4×4 and 5×5 squares, does not work for the 6×6 square due to the famous "36-officers problem" of Euler. Because of this, a 6×6 pandiagonal multiplicative magic square may be impossible. Can anyone prove that, or find an example?" 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: 101,559,956,668,416 And best, it is a "most-perfect" pandiagonal magic square, à la Kathleen Ollerenshaw. All 2x2 sub-squares have always the same product: 2,176,782,336 My feeling was the same than Ed, I thought that such a square was impossible... I wrote the same "Impossible?" than Ed in my recapitulative table at www.multimagie.com/English/Multiplicative.htm A new update will be necessary! And I have also first examples of pandiagonal multiplicative squares of orders 8, 9, 12, 15, 16. Best regards. Christian.