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:

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:

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:

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:

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

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

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:

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?


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