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. V3.0, February 2017, with additions (semi-magic squares of triangular numbers, by Wesolowski/"Anonuser"/Ridders)
previous version: 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 , 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.

First polygonal numbers Relationship to bimagic squares

In his partial solution  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  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:

 0 3 105 276 136 66 1 253 190 10 210 45 6 28 231 91 300 36 15 78 153 171 120 21 55

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.

 0 406 120 528 105 136 1 300 435 378 171 10 66 276 496 15 91 351 595 78 153 28 210 231 3 190 55 21 465 561 630 45 36 325 253 6

Order n=7, also possible

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

 0 378 1176 210 595 6 435 3 351 45 465 703 1128 105 946 171 561 820 190 21 91 741 528 36 325 120 15 1035 1081 300 55 496 780 10 78 28 666 66 231 276 630 903 1 406 861 253 136 990 153

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.

 28 666 78 1 528 105 45 276 351 3 406 325 66 378 136 171 190 465 496 21 153 630 15 91 210 10 435 595 36 120 561 55 253 6 231 300

 36 406 276 3 528 946 780 45 903 351 6 1225 10 435 561 861 496 741 105 21 190 990 120 630 1 66 703 465 253 136 666 1081 153 91 595 55 378 231 15 820 1176 300 1035 171 325 1128 78 28 210

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

((p - 2)i² - (p - 4)i)/2

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 , and for p=9, 10 in . But is it possible to construct squares of orders n<8? Yes!

• The case p=3, triangular numbers. The smallest solution is n=6, as analyzed above.
• The case p=4, square numbers. The smallest solution is slightly larger: n=7. This question was solved in 2005: I constructed a magic square of squares of order 7 using the first squares 0², 1², …, 48². The magic square is on my web site [2c] and is also published in the MAA MathTrek column of Ivars Peterson .
• The cases p=5, 6, 7, 8, 9, 10. The smallest solution is again n=7 for all these p. It might be boring to give all my samples, but here is one with p=5, a magic square of pentagonal numbers of order 7.

•  1617 3015 35 0 1162 715 1520 2882 330 12 5 210 3290 1335 1926 2752 2501 247 117 376 145 70 176 2380 1 3432 925 1080 51 532 2262 2625 1717 287 590 1426 782 22 2035 1001 651 2147 92 477 852 3151 425 1820 1247

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:

8(p – 2) [((p - 2)i² - (p - 4)i)/2] + (p - 4)²   =   [2(p - 2)i² - (p - 4)]²

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 . I found the first 5x5 magic square of squares in 2004 [2c]  . Now I am please to give the first 4x4 and 5x5 magic square of triangular numbers:

 66 465 780 91 1 630 105 666 300 171 496 435 1035 136 21 210

 351 0 210 91 171 36 136 153 378 120 105 406 15 231 66 325 253 10 45 190 6 28 435 78 276

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

In December 2015, Arkadiusz Wesolowski, Poland, proposed this problem on mersenneforum.org, but asking for only one magic diagonal. And in January 2016, Carlos Rivera, Mexico, proposed the same problem on primepuzzles.net. "Anonuser" gave eight examples on mersenneforum, his smallest being:

 25894806 18547095 4846941 589155 13825911 34873776 22804881 16915836 9568125

In March 2016, Arkadiusz Wesolowski published the 22 first solutions in OEIS (sequence A271020), S < 4000000000. And in April 2016, Marc Ridders, Netherlands, found 5 more solutions, his largest having S=6491506641.

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.

 1426 1520 1080 176 0 376 1335 5 782 2147 22 287 1 1820 651 1926 145 35 92 51 925 247 2262 1001 1247 852 715 70 532 1162 477 330 425 12 1617 1717

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!

 4030 1001 145 2262 117 1426 1247 376 3290 0 70 176 2501 2882 1926 4187 35 715 477 925 782 3151 1162 425 2035 5 145 3876 51 2262 1426 2147 22 1335 2625 70 1335 782 1001 3151 1247 1080 3725 651 852 651 3577 590 1520 1
 3725 1908012 659022 20475 12650 1969401 578151 31032 760060 115787 500837 1214550 722107 83426 455126 1330575 300832 543305 1431305 315792 247051 495650 1557032 291501 1526617 24130 70 1040417 1609426 42757 925 938126

Lee constructed also this 3x3 semi-magic square:

 356972 651 54626 19780 275847 116622 35497 135751 241001

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  and  to Problem E496 by Heath and  and  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].

 1596 595 36 1653 171 1128 45 496 561 210 1485 1176 28 435 1770 55 351 946 91 276 2080 741 10 1225 190 15 630 465 1431 78 1081 1830 120 325 2016 3 861 300 1275 820 21 1540 153 66 666 1711 528 1035 1891 136 903 1378 378 1 780 253 990 1953 406 703 105 1326 231 6

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!

Pão de Açucar, Rio de Janaeiro, received from Sebastião A. da Silva 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 , 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

•  W. W. R. Ball and H. S. M. Coxeter, Mathematical Recreations and Essays, 13th edition, Dover Publications, 1987
•  C. Boyer, Multimagic squares and cubes web site, www.multimagie.com/indexengl.htm
•  C. Boyer, Some notes on the magic squares of squares problem, The Mathematical Intelligencer 27(2005), n°2, 52-64
•  R. V. Heath, Problem E 496, The American Mathematical Monthly 48(1941), 699
•  R. V. Heath, A magic square of triangular numbers - Solution E 496, The American Mathematical Monthly 49(1942), 476
•  I. Peterson, Magic squares of squares, MAA Online, MathTrek column, June 27, 2005, http://www.maa.org/mathland/mathtrek_06_27_05.html
•  J. P. Robertson, Magic squares of squares, Mathematics Magazine 69(1996), n°4, 289-293
•  H. Schots, Helsch vierkant, Bulletins de la Classe des Sciences – Académie Royale de Belgique, 5ème Série 17(1931), 339-361
•  C. W. Trigg, Magic squares with polygonal elements, School Science and Mathematics 71(1971), 195-197
•  C. W. Trigg, Magic squares with nonagonal and decagonal elements, Journal of Recreational Mathematics 5(1972), 203-204