Smallest magic squares of triangular numbers (and of polygonal numbers)
Expanded version of the solution published in The American Mathematical Monthly, Vol 114, N. 8, October 2007, p. 745-746
by Christian Boyer. V2.0, June 2008, with additions (magic squares of pentagonal numbers, by Lee Morgenstern)


An old problem proposed in 1941, solved in 2007

In 1941, Royal Vale Heath proposed this short problem in The American Mathematical Monthly [4], when H. S. M. Coxeter was in charge of the “Problems and Solutions” column:
                
What is the smallest value of n for which the n² triangular numbers 0, 1, 3, 6, 10, …, n²(n²-1)/2 can be arranged to form a magic square?
This problem remained unsolved. Here is the solution found in April 2007… 66 years later:
                
n = 6.

The samples were not easy to find, but are easy for the reader to check, as a factorization problem. Contents of this expanded solution:


Triangular and polygonal numbers

This figure will help us to understand what triangular numbers are, and more generally polygonal numbers.


Relationship to bimagic squares

In his partial solution [5] published in the Monthly 1942, R. V. Heath remarked that a bimagic square (magic square which is still magic after the original entries are all squared [2a]) can be directly used to construct a magic square of triangular numbers:

“Clearly, the magic property will still be retained if each of the original numbers is subtracted from its square. The resulting numbers are all even, and their halves are the triangular numbers”

Using a bimagic square of order 8 found in the well-known book [1, p. 212] by W.W. Rouse Ball and initially constructed by H. Schots in 1931 [8, p.357], Heath [5] built a magic square of 64 triangular numbers and with it showed that n£8, but the smallest possible n remained unknown, as he said:

“But it remains possible that a smaller set of triangular numbers might form a magic square without the corresponding natural numbers forming a magic square. Moreover, it has never been satisfactorily proved that there is no doubly-magic (=bimagic) square of order 7”

By an exhaustive search, we proved with Walter Trump in 2002 [2b] that a bimagic square of order smaller or equal to 7 does not exist: this means that Heath’s trick in using bimagic squares cannot be used for orders n<8. Here is a study of the problem for the smallest orders.


Order n=3, impossible

Is it possible to construct a magic square using the 9 triangular numbers 0, 1, 3, 6, 10, 15, 21, 28, 36?

No! The total sum of these numbers being 120, such a square would have a magic sum 120/3=40. The central number of any 3x3 magic square being one third of its magic sum, and 40 being not divisible by 3, since n=3, it is impossible.

Order n=4, impossible

A magic square using the 16 triangular numbers 0, 1, 3, …, 120 would have a magic sum equal to 170. There are only 10 series of 4 triangular numbers giving this sum:

        0         1         78       91
        0         10       55       105
        1         21       28       120
        1         28       36       105
        1         36       55       78
        3         10       66       91
        3         21       55       91
        6         28       45       91
        15       28       36       91
        21       28       55       66

In a magic square, each number needs to use at least two or three series: one for the row, one for the column, and one more if the number is located on a diagonal. Because the number 6 -for example- is present in only one series, a magic square of order n=4 is impossible.

Order n=5, impossible

A magic square using the 25 triangular numbers 0, 1, 3, …, 300 would have a magic sum equal to 520. There are 118 series giving this sum. Combining the series, there are 148 possible ways to get 5 series using the 25 triangular numbers. This means that it is possible to have 5 magic rows. An exhaustive search, however, shows that it is impossible to arrange these rows and make all the columns magic. The best possible arrangements are 5 magic rows and 3 magic columns, for example:

Order n=6, possible! Solution of the problem.

A magic square using the 36 triangular numbers 0, 1, 3, …, 630 would have a magic sum equal to 1295. There are 1921 series giving this sum.

Good news: it is possible to arrange these series to form magic squares! Here is an example, solution of our problem.

Order n=7, also possible

The order n=7 allows also magic squares of the first triangular numbers.


Numbers from 1 instead of 0

If we prefer to use consecutive polygonal numbers starting from 1 instead of 0 (see below the da Silva’s challenge), a similar reasoning shows that the minimum order is again 6. The 6 series of order 4 and the 91 series of order 5 are not sufficient to construct a magic square. Here are examples of order 6 and 7 starting from 1.


Squares of polygonal numbers, p ≤ 10

We can generalize Heath’s Problem E496 to other polygonal numbers.

Reminder: the i-th p-gonal number is equal to

With p=3, we get triangular numbers. With p=4, we get square numbers. With p=5, we get pentagonal numbers. And so on…

Any bimagic square can be used to construct magic squares of k2i²+k1i+k0 numbers: using the same bimagic square of order n=8 as R.V. Heath, Charles W. Trigg published squares of polygonal numbers for p=3, 5, 6, 7, 8 in [9], and for p=9, 10 in [10]. But is it possible to construct squares of orders n<8? Yes!

An interesting remark: a magic square of polygonal numbers can be turned into a magic square of squares by multiplying each term by 8(p – 2) then adding (p - 4)² to each term, because:


An unsolved problem: the smallest magic square of distinct triangular numbers

All the above examples use the first consecutive polygonal numbers. But what is the smallest order n if we allow any polygonal numbers, consecutive or not, but distinct?

The first 4x4 magic square of squares, using 16 distinct squares, was constructed by Euler, in a letter sent to Lagrange in 1770 [3]. I found the first 5x5 magic square of squares in 2004 [2c] [3] [6]. Now I am please to give the first 4x4 and 5x5 magic square of triangular numbers:

It is still unknown if a 3x3 magic square of squares is possible [2d] [3] [6] [7], but what about a 3x3 magic square of triangular numbers? As remarked by John P. Robertson (author of [7]), in a private communication of April 2007:

“If there is a 3x3 magic square of squares, then all the entries are odd, and so congruent to 1 modulo 8.  Because if T is a triangular number then 8T + 1 is a square, and if S is an odd square then (S - 1)/8 is a triangular number, the question of whether there is a 3x3 magic square of squares is equivalent to the question of whether there is a 3x3 magic square of triangular numbers.”

Open problem. Who will construct a 3x3 magic square of distinct triangular numbers, or its equivalent 3x3 magic square of squares? Or who will prove that it is impossible?


Another unsolved problem: the smallest magic square of distinct pentagonal numbers

We have seen that we do not have the answer to the problem of the smallest magic square of triangular or of square numbers: there are 4x4 magic squares, but we still don’t know if 3x3 squares are possible.

But after triangular numbers (p=3) and square numbers (p=4), what about pentagonal numbers (p=5)? We find that 6x6 squares are possible, as shown in the figure below, but it should be possible to construct 5x5 squares or smaller ones.

Open problem. What is the smallest possible magic square of distinct pentagonal numbers: 3x3, 4x4, 5x5 or 6x6?

In November 2007, Lee Morgenstern worked on this very difficult problem. He constructed the first known 4x4 and 5x5 magic squares of distinct pentagonal numbers. Congratulations!

Lee constructed also this 3x3 semi-magic square:

All this means that the above open problem becomes now:

Open problem. Who will construct a 3x3 magic square of distinct pentagonal numbers? Or who will prove that it is impossible?

As seen above, a magic square of polygonal numbers can be turned into a magic square of squares: an example of a 3x3 magic square of pentagonal numbers would also solve the 3x3 magic square of squares problem.


Acknowledgements

Particular thanks to Sebastião A. da Silva, Brazil, who challenged me to find solutions of order n < 8 in March 2007. Without knowing the references [4] and [5] to Problem E496 by Heath and [9] and [10] to articles by Charles W. Trigg, Sebastião had independently found the relationship with bimagic squares and sent me this solution of order 8, constructed using the G. Pfeffermann’s first bimagic square built in 1890 [2a].

But Sebastião was unsuccessful in finding examples of smaller orders. He offered a bottle of Brazilian Curaçao if I succeeded in answering his question, asking in French:

               “Est-il possible de construire un carré triangulaire quand il n´existe pas un bimagique du même ordre ?”
               (“Is it possible to construct a triangular square when there is no bimagic of the same order?”)

A bottle? Very interesting! It’s because I worked on his challenge that I looked for mathematical references and found that the same problem had been proposed already a long time ago in the Monthly –without the reward of a bottle- and which had remained unsolved. The only difference is the starting triangular number: Sebastião started from 1, while Heath started from 0. Both cases are solved now. Because I won his challenge, and because it seems unfortunately difficult to send a bottle through the airmail post, Sebastião sent me in May 2007 this nice gift instead of a bottle. Thanks Sebastião for your interesting challenge and for your gift. We will drink together another bottle when you come to Paris, or when I go to Rio!

Thanks to Doug Hensley, Dept. of Mathematics, Texas A&M University, and Douglas B. West, Dept. of Mathematics, University of Illinois at Urbana-Champaign, both in charge of the “Problems and Solutions” column of The American Mathematical Monthly. They were the first to read the solution n=6, after Sebastião, and immediately accepted to publish it, 66 years after the problem was proposed in the same column.

Thanks to George P. H. Styan, McGill University, Montreal, Canada, who sent me a copy of the paper [9], wich I was unable to find it in the main mathematical libraries of Paris. And thanks also to him and to Evelyn Matheson Styan, his wife, for correcting my English in this expanded solution of Problem E 496!


References


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