The smallest possible trimagic square

What is the smallest trimagic (3-multimagic) square possible? The smallest known trimagic square is order-12, found in June 2002 by Walter Trump.

We will prove here that a trimagic square of order lower than 12 cannot exist.

7th-order trimagic square? Or smaller order?

Since there is no 7th or lower order bimagic squares, a 7th-order or lower trimagic square cannot exist.

Moreover, even a partial construction of a 7th-order trimagic square is impossible, since there is no 7th-order trimagic series.

8th-order trimagic square?

If we study with a computer the possible 8th-order trimagic squares, we are surprised to find so many trimagic series, that is to say series of different integers from 1 to 64, having for magic sums S1 = 260, S2 = 11 180, and S3 = 540 800.

There are exactly 121 series, so that could be enough to build a 8th-order trimagic square, since 18 series (8 rows + 8 columns + 2 diagonals) "well chosen" could be sufficient. Let try to find 8 completely different series, that means using the whole 64 integers. We can find very numerous (5 719) groups of 6 different trimagic series, like for example the group built from the G1, G14, G33, G54, G84 and G103 series:

• G1 = 64, 51, 38, 33, 32, 27, 14, 1
• G14 = 63, 53, 36, 34, 31, 29, 12, 2
• G33 = 62, 52, 41, 39, 26, 24, 13, 3
• G54 = 61, 50, 48, 35, 30, 17, 15, 4
• G84 = 59, 54, 43, 42, 23, 22, 11, 6
• G103 = 57, 56, 45, 40, 25, 20, 9, 8

But unfortunately we cannot go further, it is impossible to find a group of 8 trimagic series (or 7, that's the same thing).

So, a 8th-order trimagic square cannot exist.

Download the 121 trimagic series of order-8, and the 5 719 groups of 6 series, zipped Excel file 176Kb.

9th-order trimagic square?

For the supposed 9th-order trimagic square, let try to place the number 81. Only 3 different series of different integers exist, from 1 to 81, having S1 = 369, S2 = 20 049, S3 = 1 225 449, and containing 81:

• G1 = 81, 67, 51, 49, 41, 35, 31, 11, 3
• G2 = 81, 63, 55, 49, 47, 35, 23, 9, 7
• G3 = 81, 59, 55, 53, 51, 31, 21, 11, 7

Like each number, 81 has to be present in one row and one column of the square, so we must have 2 different series having in common only the 81 number (and we should even have 3 different series if 81 is in a diagonal, and 4 series if 81 is in the central cell). But:

• G1 and G2 have 81, 49, and 35 in common
• G1 and G3 have 81, 31, and 11 in common
• G2 and G3 have 81, 55, and 7 in common

So, a 9th-order trimagic square cannot exist.

10th-order trimagic square?

For the supposed 10th-order trimagic square, the situation is more simple. The magic sums are: S1 = 505, S2 = 33 835, S3 = 2 550 250. It is impossible to construct an integers series having S3 even, even though S1 and S2 are odd.

So, a 10th-order trimagic square cannot exist.

11th-order trimagic square?

Here is a nice proof of the non-existence of 11th-order trimagic square done by Walter Trump (Germany) in May 2002.

The magic sum : S1 = 671, S2 = 54 351, S3 = 4 952 651

Like his complete demonstration shows from considerations about S1, S2 and S3 modulo 4 and 8 (S1 = 3 mod 4, S2 = 3 mod 4, and S3 = 7 mod 8), a 11th-order trimagic series is forced to have 7 odd integers. So, the 11 rows of the squares would have 7x11 = 77 odd integers, even though in a 11th-order square we must place only 61 odd integers.

So, a 11th-order trimagic square cannot exist.

We have proved that a trimagic square of order lower than 12 cannot exist. It is impossible to do best that the 12th-order trimagic square of Walter Trump!