The smallest possible tetramagic square

What is the smallest tetramagic (4-multimagic) square possible? We don't know, we only know that its order is perhaps 24, or between 27 and 243.

The smallest known tetramagic square is order-243, found in February 2004 by Pan Fengchu.

We will prove here that tetramagic squares of orders ≤ 23, or of orders 25 and 26, cannot exist.

11th-order tetramagic square? Or smaller order?

Since there is no 11th or lower order trimagic squares, an 11th-order or lower tetramagic square cannot exist.

Moreover, even a partial construction of an 11th-order tetramagic square is impossible, since there is no 11th-order tetramagic series.

12th-order tetramagic square?

There are exactly 106 series, so that could be enough to build a 12th-order tetramagic square, since 26 series (12 rows + 12 columns + 2 diagonals) "well chosen" could be sufficient. But they are not numerous enough. For example, the number 1 should be present in the square, and among the 106 series, there are only 5 series using 1:

• S1:    1       23      44      47      48      50      75      96      112     115     120     139
• S2:    1       26      33      44      52      68      69      85      112     120     125     135
• S3:    1       29      30      45      54      58      71      97      110     112     125     138
• S4:    1       29      37      38      50      54      90      96      97       110     133     135
• S5:    1       30      34      38      50      65      69      103    110     112     117     141

Among these 5 series, the square needs two of them, one for the row + one for the column, intersecting on the number 1, no other number should be in common. This is impossible, any possible couple of these series has other numbers in common:

 S1 S2 S3 S4 S5 S1 X S2 44, 112, 120 X S3 112 112, 125 X S4 50, 96 135 29, 54, 110 X S5 50, 112 69, 112 30, 110, 112 38, 50, 110 X

So, a 12th-order tetramagic square cannot exist.

13th-order tetramagic square?

The magic sum S4 is 2152397897 = 9 mod 16. Because (2x+1)^4 = 1 mod 16 and (2x)^4 = 0 mod 16, each series has 9 odd numbers. So the 13 rows of the square would have 13x9 = 117 odd integers, even though in a 13th-order square we must place only 85 odd integers.

So, a 13th-order tetramagic square cannot exist.

14th or 15th-order tetramagic square?

There is no tetramagic series of such orders.

So, a 14th or 15th-order tetramagic square cannot exist.

16th-order tetramagic square?

There are 235275 series, and we may think that they could be sufficient, but we will prove that they can't be organized in a tetramagic square using modulo 9 reasoning. The magic sums of a 16th-order tetramagic square (S1 is not needed in the proof) are:

• S2 = 351576 = 0 mod 9
• S3 = 67634176 = 4 mod 9
• S4 = 13878462600 = 0 mod 9

•  n n2 mod 9 n3 mod 9 n4 mod 9 0 0 0 0 1 1 1 1 2 4 8 7 3 0 0 0 4 7 1 4 5 7 8 4 6 0 0 0 7 4 1 7 8 1 8 1

In a tetramagic series of 16 integers, calling ai the number of its integers which are equal to i mod 9, and using the above table, we have this system of equations:

• a0 + a1 + a2 + a3 + a4 + a5 + a6 + a7 + a8 = 16   (with 0  ≤ ai  ≤ 16)
• a1 + 4a2 + 7a4 + 7a5 + 4a7 +   a8 = 0 mod 9   (using S2)
• a1 + 8a2 +   a4 + 8a5 +   a7 + 8a8 = 4 mod 9   (using S3)
• a1 + 7a2 + 4a4 + 4a5 + 7a7 +   a8 = 0 mod 9   (using S4)

Analyzing all the possible solutions of this system, they are always:

• a0 + a3 + a6 = 7
• a1 + a4 + a7 = 2
• a2 + a5 + a8 = 7

Moving from mod 9 to mod 3, this also means that each tetramagic series has its 16 integers always split that way:

• 7 integers are 0 mod 3
• 2 integers are 1 mod 3
• 7 integers are 2 mod 3

With such series, it is obviously impossible to get 16 rows using all integers from 1 to 256.

So, a 16th-order tetramagic square cannot exist.

17th, 18th or 19th-order tetramagic square?

An 18th-order tetramagic square cannot exist simply because there is no trimagic (and of course tetramagic) series of any 4k+2 order.

On the 17th and 19th-orders, Michael Quist sent me these short proofs in May 2008 using (2x+1)^4 = 1 mod 16 and (2x)^4 = 0 mod 16:

Though there may be tetramagic series of order 17, there cannot be a tetramagic square. Since S(4,17)%16 = 1, each tetramagic series of order 17 has either 1 or 17 odd elements. In 1...17^2 there are (17^2-1)/2 = 145 odd numbers to account for. This can be done only by having exactly 8 entirely odd rows, and 9 rows with a single odd element each. The same reasoning applies to the columns: there must be 8 entirely odd columns, and 9 columns with only a single odd entry. But since there are 8 entirely odd rows, each column must have at least 8 odd elements; there can't be any with just one odd entry.

It is even simpler to show that there can't be tetramagic squares of order 19. Here S(4,19)%16 = 7, so each tetramagic series of order 19 has exactly 7 odd elements. In any collection of 19 rows, then, there are only 7*19 = 133 odd numbers, which is far short of the (19^2-1)/2 = 180 found in 1...19^2.

So, a 17th, 18th or 19th-order tetramagic square cannot exist.

20th, 21st, 22nd or 23rd-order tetramagic square?

A 22nd-order tetramagic square cannot exist simply because there is no trimagic (and of course tetramagic) series of any 4k+2 order.

In February-March 2013, Lee Morgenstern proved that 20th, 21st and 23rd-order tetramagic squares cannot exist. See his proofs (also including other proofs of orders ≥ 12)

24th-order tetramagic square?

May perhaps exist! Who will try?

25th or 26th-order tetramagic square?

A 26th-order tetramagic square cannot exist simply because there is no trimagic (and of course tetramagic) series of any 4k+2 order.

In April 2013, Lee Morgenstern proved that 25th-order tetramagic squares cannot exist. See his proof.

27th-order tetramagic square?

May perhaps exist! Who will try?