Multimagic formulas

How do we know what the theoretical values should be of sums of rows, columns and diagonals of multimagic squares?

Start with the single magic square. It is known, and easy to demonstrate (except for an 8-year old child like Gauss who is supposed to have demonstrated at this age at school), that the sum of the first integer numbers from 1 to N is:

N(N+1)/2

A n-th order magic square contains all the numbers from 1 to n². So, the sum of the first integer numbers from 1 to n² is:

n²(n²+1)/2

Since this square has n rows, it is sufficient to divide by n in order to know the magic sum S1 of each row, column or diagonal, so:

• S1 = n(n²+1)/2

For the bimagic square, the sum of the squares of the first integers from 1 to N is:

N(N+1)(2N+1)/6

Replacing N by n², then dividing by n, we get the sum S2 of each row, column or diagonal, so:

• S2 = n(n²+1)(2n²+1)/6 = S1*(2n²+1)/3

For the trimagic square, the sum of the cubes:

• S3 = n * S1²

For the tetramagic, Fermat gave as early as 1636, in a letter sent to Roberval, the solution for the sum of the integers from 1 to N to the 4th-power. Here is Fermat's formula after replacing N by n², then dividing by n:

S4 = ((4n² + 2)*n*S1² - S2) / 5

Replacing (4n² +2) * S1 by 6S2, we can write an easier formula than Fermat:

• S4 = S2 * (6n * S1 - 1) / 5

For the pentamagic (sum of 5th-power):

• S5 = (3n*S2² - S3) / 2

If we prefer to develop Sp, here is the following generous table, which permits us to calculate the sums of the future hexa, hepta or octomagic squares:

 S1 = (1/2) n^3   + (1/2) n S2 = (1/3) n^5   + (1/2) n^3   +  (1/6) n S3 = (1/4) n^7   + (1/2) n^5   +  (1/4) n^3 S4 = (1/5) n^9   + (1/2) n^7   +  (1/3) n^5   - (1/30) n S5 = (1/6) n^11  + (1/2) n^9   + (5/12) n^7   - (1/12) n^3 S6 = (1/7) n^13  + (1/2) n^11  +  (1/2) n^9   -  (1/6) n^5  + (1/42) n S7 = (1/8) n^15  + (1/2) n^13  + (7/12) n^11  - (7/24) n^7  + (1/12) n^3 S8 = (1/9) n^17  + (1/2) n^15  +  (2/3) n^13  - (7/15) n^9  +  (2/9) n^5  - (1/30) n

In order to calculate Sp when the n-th order p-multimagic square contains the integers from 0 to n²-1 (instead of 1 to n²), it is sufficient to put a - sign (instead of the + red) in the 2nd column above. You may remark that the first column of Sp is equal to (1/(p+1)) n^(2p+1). This feature has been known since the middle of the XVIIth century, thanks to Pascal who wrote a Treatise on the subject. Jacques Bernoulli later took an interest in this question of the sum of integer powers. After his Ars conjectandi published posthumously at the beginning of the XVIIIth century, the numbers used in this development have been called Bernoulli numbers (1/6, -1/30, 1/42, -1/30, 5/66, ).

Here are the formulas for some n-orders of p-multimagic squares, containing numbers from 1 to n². Here we find the sums of Pfeffermann's 8th-order bimagic square, Pfeffermann's 9th-order bimagic square, and William Benson's 32-th order trimagic square:

 Sp 3rd-order 4th-order 5th-order 6th-order 7th-order 8th-order 9th-order ... 32th-order S1 15 34 65 111 175 260 369 ... 16400 S2 95 374 1105 2701 5775 11180 20049 ... 11201200 S3 675 4624 21125 73926 214375 540800 1225449 ... 8606720000

And here are the formulas of some n-orders of p-multimagic squares, containing numbers from 0 to n²-1. Here we find the sums of our record 512th-order tetramagic and 1024th-order pentamagic squares:

 Sp 32nd-order 512th-order 1024th-order S1 16368 67108608 536870400 S2 11168432 11728056920832 375299432076800 S3 8573165568 2305825417061203968 295147342229667840000 S4 7019705733392 483565716171561366524160 247587417561640996243120640 S5 5987221633671168 105636341097042573844228866048 216345083469423421673932062720000

Bibliography

For more details about the sum of powers, see the excellent chapter 14 « Sommation des puissances numériques », book II of the Théorie des Nombres by Edouard Lucas, Librairie Blanchard, Paris.