In this paper, we present the problem of counting magic squares and we focus
on the case of multiplicative magic squares of order 4. We give the exact
number of normal multiplicative magic squares of order 4 with an original and
complete proof, pointing out the role of the action of the symmetric group.
Moreover, we provide a new representation for magic squares of order 4. Such
representation allows the construction of magic squares in a very simple way,
using essentially only five particular 4X4 matrices

Barbero, Stefano

Cerruti, Umberto

Murru, Nadir

English

arXiv

Stefano Barbero

Umberto Cerruti

Nadir Murru

PORTO@iris (Publications Open Repository TOrino - Politecnico di Torino)

04 August 2020POLITECNICO DI TORINORepository ISTITUZIONALESquaring the magic squares of order 4 / Barbero, Stefano; Cerruti, Umberto; Murru, Nadir. - In: JOURNAL OFALGEBRA, NUMBER THEORY: ADVANCES AND APPLICATIONS. - 7:1(2012), pp. 31-46.OriginalSquaring the magic squares of order 4Publisher:PublishedDOI:Terms of use:openAccessPublisher copyright(Article begins on next page)This article is made available under terms and conditions as specified in the  corresponding bibliographic description inthe repositoryAvailability:This version is available at: 11583/2725055 since: 2019-02-12T21:13:23ZScientific Advances PublishersSquaring the magic squares of order 4Stefano Barbero, Umberto Cerruti, Nadir MurruDepartment of MathematicsUniversity of TurinAbstractIn this paper, we present the problem of counting magic squaresand we focus on the case of multiplicative magic squares of order 4.We give the exact number of normal multiplicative magic squares oforder 4 with an original and complete proof, pointing out the roleof the action of the symmetric group. Moreover, we provide a newrepresentation for magic squares of order 4. Such representation allowsthe construction of magic squares in a very simple way, using essentiallyonly five particular 4× 4 matrices .1 IntroductionA magic square is defined as an n × n matrix of integers, where the sumof the numbers in each line (i.e., in each row, in each column and in eachdiagonal) is the same. Magic squares have a very rich history (see, e.g.,the beautiful book of Descombes [6]) and they have many generalizations.There are a lot of contributes to their theory from several different fields(see, e.g., the classical book of Andrews [1]).A very difficult job is counting how many magic squares there are for a givenorder and a given line sum, but if we relax the requirements and we ask forthe so–called semi–magic squares (i.e., magic squares without the conditionon the diagonals), the counting is easier. For order 3, Bona [2] has foundthat the number of semi–magic squares is(r + 44)+(r + 34)+(r + 24),where r is the line sum (let us observe that in [2] semi–magic squares arecalled magic squares). For a general theory of counting semi–magic squaresyou can see [3], chapter 9.1The classic magic squares, which are called normal, are n×n matrices whoseentries consist of the numbers 0, 1, ..., n2− 1 and in each line the sum of thenumbers is the same. Their number is known for the orders 3, 4 and 5. Thecase of order 4 is especially interesting. There are exactly 880 normal magicsquares of order 4 (up to symmetries of the square, i.e., in total they are880×8 = 7040). They were enumerated for the first time in 1693 by Freniclede Bessy. The Frenicle method was analytically expanded and completedby Bondi and Ollerenshaw [4]. Of course, one can think to different kindsof magic squares, only changing the operation. Then a multiplicative magicsquare is an n× n matrix of integers, where the product of the numbers ineach line is the same. This is not a new idea, for example multiplicativesemi–magic squares of order 3 are studied in [7]. There are several differ-ent methods to construct multiplicative magic squares. The most obviousmethod is to transform an additive magic square A = (aij) into M = (baij ),for any base b. More intriguing ways are explained in [6] (multiplicativemagic squares are usually constructed by using orthogonal latin squares orgeometric progressions).In [8], the problem of finding the exact number of multiplicative magicsquares of order 4 has been posed. In order to solve such question, in [9]a correspondence between additive and multiplicative magic squares is pro-posed. The authors claimed that such function is an isomorphism eventhough it is only an injective morphism, as pointed out in [10]. Since suchrevision is only a note, the above problem is not completely clarified and thecorrect number of multiplicative magic squares is given without a rigorousproof. In the next section, we clarify the above questions, proposing an orig-inal and complete proof. Moreover, in the last section a novel representationfor some magic squares is proposed, allowing an easy way to construct them.2 Counting multiplicative magic squares of order4A normal multiplicative magic square n×n is created using n primes p1, ..., pn.All the divisors of k = p1 · · · pn must appear once time in the matrix and ineach line the product is the same. The magic constant is k2 and it is well–known that in a normal multiplicative magic square, in each line, any primecompares with power one and exactly two times. For example, a normal2multiplicative magic square of order 4 is1 p3 · p4 p1 · p2 p1 · p2 · p3 · p4p1 · p3 · p4 p1 · p2 · p3 p4 p2p1 · p2 · p4 p1 p2 · p3 · p4 p3p2 · p3 p2 · p4 p1 · p3 p1 · p4 .In the following, we use Mk,n to indicate the set of normal multiplicativemagic squares n × n with magic constant k2. An hard problem to solveis to find the order of Mk,4, as posed in [8]. In order to answer to thisquestion, Mk,4 is made in correspondence with the set of normal additivemagic squares 4× 4 [9].A normal additive magic square is a matrix n× n, whose entries consist ofthe numbers 0, 1, ..., n2 − 1 and in each line the sum of the numbers is thesame. The magic constant is n(n2−1)2 . From now on, we name An the set ofnormal additive magic squares n× n.It is well–known that the order of A4 is 7040, or 880 up to symmetries ofthe square [4]. Can we find a similar result for Mk,4?In [9] the following correspondence is defined:f : Mk,n → A2n/2 .If f is bijective, |Mk,4| is determined. The function f maps M = (mij) ∈Mk,n into A = (aij) ∈ A2n/2 . The element mij is made in correspondencewith aij , where aij is the base 10 number , between 0 and n2−1, whose base2 expression is the n–string associated to mij . This string is constructed asfollows: arrange the n prime numbers in ascending order; if pk is a factor ofmij place a 1 in the kth position of the string ; if not, place a 0.Such a function f is surely injective, but unfortunately it is not surjectiveand the order of Mk,4 is not |A4| = 7040. Indeed, we can consider the groupS4 which permutes the four primes involved in Mk,4.Theorem 1. The order of S4 divides the order of Mk,4.Proof. Surely when a permutation ρ ∈ S4 acts on M ∈ Mk,4, then ρ(M) isstill an element in Mk,4. Furthermore, if ρ is not the identity, it does not fixany element of Mk,4, i.e., any orbit has 24 elements. Therefore by BurnsideLemma |S4| must be a divisor of |Mk,4|.As immediate consequence of the previous Theorem, 7040 can not be theorder of Mk,4, since 24 does not divide 7040. The correct answer is given inthe following theorem.3Theorem 2. The order of Mk,4 is 4224.Proof. If we consider A ∈ A4 and we write its entries in base 4, then in eachline the sum of the digits in the first position multiplied by 4 and added tothe sum of the digits in the second position must yield 30. The only possiblecombinations are30 = 7× 4 + 2 = 6× 4 + 6 = 5× 4 + 10.So if A has a line in which the digits in the second position have sum 2, weare in the only two situations·0 · 0 · 1 · 1·0 · 0 · 0 · 2where · indicates the first position of the number in base 4. Consideringthe representation in base 2 of these strings, we have one of the followingsituations· · 00 · ·00 · ·01 · ·01· · 00 · ·00 · ·00 · ·10 .Now if we try to create M ∈Mk,4, using the inverse of f , if A has a line whichpresents one of these situations we can not have a normal multiplicativemagic square. In fact, if we are in the first situation, for example, then theprime p3 will not appear in this line. Similarly if a line of A has digits inthe second position with sum 10, the possible cases are·3 · 3 · 3 · 1·2 · 2 · 3 · 3which correspond in base 2 to the strings· · 11 · ·11 · ·11 · ·01· · 10 · ·10 · ·11 · ·11 .4Once more, if we try, we fail to obtain M ∈Mk,4 .Finally, if A has a line whose digits in the second position have sum 6, wehave the possibilities·0 · 1 · 2 · 3·0 · 0 · 3 · 3·1 · 1 · 2 · 2·0 · 2 · 2 · 2·1 · 1 · 1 · 3 .With similar arguments as the ones used before, it is easy to see that the lasttwo situations do not allow to obtain a normal multiplicative magic square.Therefore the squares A ∈ A4 from which we obtain a normal multiplicativemagic square are only those with lines whose entries in base 4 have digits inthe first and second position composed by strings·0 · 1 · 2 · 3·0 · 0 · 3 · 3·1 · 1 · 2 · 2 .In [4], all the squares in A4 are classified. The squares in A4 generatedusing only these strings, i.e., which correspond to squares in Mk,4, are onlythose in category one ([4], p. 510). They are exactly 528 unless of symmetriesof the square. Therefore our arguments and the injectivity of f allow us toconclude that the order of Mk,4 is 528× 8 = 4224.3 A new representation for magic squaresIn this section we see a representation for normal additive magic squareswhich correspond to normal multiplicative ones.In [5] a lower bound for the distance between the maximal and minimalelement in a multiplicative magic square is given. In order to do that, therelation between additive and multiplicative magic squares is highlighted,recalling that a multiplicative magic square can be factorized as∏i pAii ,where Ai’s are additive magic squares. Moreover, focusing on magic squaresof order 4, in [5] the Hilbert basis (composed by 20 magic squares) for suchmagic squares is explicited. However, such representation and the relation5between additive and multiplicative magic squares are not really manageablein order to construct additive and multiplicative magic squares of order 4.Here, the proposed representation allows to determine all (and not only)the normal additive magic squares of order 4, which corresponds to normalmultiplicative magic squares. In this way, they can all be easily constructedessentially using only 5 basic matrices.We consider M = (mij) ∈ Mk,4, by means of f we have the correspondentA = (aij) ∈ A4 and we consider its entries in base 2. Now we decompose Ainto four matrices A1, A2, A3, A4, whose entries are only 0 or 1, so that theentries in position ij of the matrices A1, A2, A3, A4 form the string aij .Example 1. From the normal multiplicative magic squarep1p2p3 p3p4 1 p1p2p4p2p4 p1 p1p3p4 p2p3p1p4 p2 p2p3p4 p1p3p3 p1p2p3p4 p1p2 p4 ,using f , we obtain the normal additive magic square1110 0011 0000 11010101 1000 1011 01101001 0100 0111 10100010 1111 1100 0001 =14 3 0 135 8 11 69 4 7 102 15 12 1and it can be decomposed into81 0 0 10 1 1 01 0 0 10 1 1 0+ 41 0 0 11 0 0 10 1 1 00 1 1 0+ 21 1 0 00 0 1 10 0 1 11 1 0 0+0 1 0 11 0 1 01 0 1 00 1 0 1 .Since A is derived from a matrix in Mk,4, by means of f , the matricesA1, A2, A3, A4 are all and only those having in each line exactly two ones.We call forms these matrices which allow us to construct any magic squarein A4 corresponding to magic squares in Mk,4. These forms are exactly 16.Theorem 3. There are 16 different matrices 4×4, with entries 0 or 1, suchthat in each line there are exactly two ones.Proof. The strings that we can use to generate these forms are only six:1100, 1010, 1001, 0110, 0101, 0011 .6If we choose as a diagonal a string with different extremes, then we haveonly two possible different forms. For example:1 · · ·· 1 · ·· · 0 ·· · · 0→1 · · 0· 1 · ·· · 0 ·· · · 0→1 0 1 00 1 0 10 1 0 11 0 1 01 · · ·· 1 · ·· · 0 ·· · · 0→1 · · 1· 1 · ·· · 0 ·· · · 0→1 0 0 10 1 1 01 0 0 10 1 1 0 .On the other hand, if we choose as a diagonal a string with same extremes,then we have four possible forms. Since we have four strings with differentextremes and two strings with same extremes, our forms are 4 · 2 + 2 · 4 =16 .From these 16 forms we can individuate five fundamental forms:A0 =0 0 1 10 1 0 11 0 1 01 1 0 0B0 =0 0 1 11 1 0 00 0 1 11 1 0 0C0 =0 0 1 11 1 0 01 1 0 00 0 1 1D0 =0 1 0 11 0 1 01 0 1 00 1 0 1E0 =0 1 0 11 1 0 00 0 1 11 0 1 0 .All the remaining can be found acting on these five forms with the group ofsymmetries of the square D8. Combining any four forms, as we have donein the previous example, the resulting matrix is always an additive magicsquare, not necessarily normal, and it corresponds to a multiplicative one.Theorem 4. f A1, A2, A3, A4 are fundamental forms, or matrices obtainedfrom fundamental forms through the action of D8, then8A1 + 4A2 + 2A3 +A4is always an additive magic square which corresponds to a multiplicativemagic square.7Given a magic square 8A1+4A2+2A3+A4 constructed using our forms,we can consider all the permutations of the forms A1, A2, A3, A4. After apermutation, we have still a magic square. Thus we can use the group S4over our forms.Example 2. We consider80 0 1 11 1 0 00 0 1 11 1 0 0+ 41 0 1 01 0 1 00 1 0 10 1 0 1+ 20 1 1 00 1 1 01 0 0 11 0 0 1+1 1 0 00 0 1 10 0 1 11 1 0 0which corresponds to 5 3 14 812 10 7 12 4 9 1511 13 0 6 .Permuting the positions of the forms, we obtain81 1 0 00 0 1 10 0 1 11 1 0 0+ 40 0 1 11 1 0 00 0 1 11 1 0 0+ 21 0 1 01 0 1 00 1 0 10 1 0 1+0 1 1 00 1 1 01 0 0 11 0 0 1which is equal to 10 9 7 46 5 11 81 2 12 1513 14 0 3 .We have seen that using our forms we obtain always an additive magicsquare but it is not necessarily normal. Finally, let us see how to utilize theforms in order to generate all and only the normal additive magic squarescorresponding to normal multiplicative magic squares. The orbit of thefundamental forms with respect to the action of D8 areA = {A0, A1}B = {B0, B1, B2, B3}C = {C0, C1, C2, C3}D = {D0, D1, D2, D3}E = {E0, E1}.8We easily observe thatA0 +A1 = E0 + E1 = U =1 1 1 11 1 1 11 1 1 11 1 1 1 .Furthermore, for any form in the orbits B, C, D there is another form, inthe same orbit, such that their sum is U.We call class the set (A,B,C,D), whose elements are all the magicsquare obtained combining the forms belonging to the orbits A, B, C, D(e.g., 8A0 + 4B1 + 2C2 + D3 or 8B2 + 4C1 + 2D0 + A0 are elements of(A,B,C,D)). In the next theorem we show all the classes which providenormal additive magic squares.Remark 1. We obtain normal additive magic squares, which correspondsto all the normal multiplicative magic squares, only from the classes(A,C,D,E)(B,B,C,C)(B,B,C,D)(B,B,D,D)(B,C,C,D)(B,C,D,D)(C,C,D,D) .We have to made clear some details:1. when we choose a form in the orbit C, the forms available in the orbitD are only two and vice versa, except for the class (C,C,D,D);2. when we have a class with two forms from the same orbits, their summust not be U;3. we can not take two times the same form in the same class.Considering this remarks, we can count the magic squares obtainable fromthese classes and we check that they are exactly 4224.For the class (A,C,D,E) we can choose 2 forms from the orbit A, 4 fromthe orbit C, only 2 from the orbit D and 2 from E. Thus from this class wecan obtain 32 normal additive magic squares, unless of permutations. Thus9we have 32 · 24 = 768 normal additive magic squares from (A,C,D,E), andwe write|(A,C,D,E)| = 768.Similarly we find|(B,B,C,C)| = (4 · 2 · 4 · 2) · 6 = 384|(B,B,C,D)| = (4 · 2 · 4 · 2) · 12 = 768|(B,B,D,D)| = (4 · 2 · 4 · 2) · 6 = 384|(B,C,C,D)| = (4 · 4 · 2 · 2) · 12 = 768|(B,C,D,D)| = (4 · 2 · 4 · 2) · 12 = 768|(C,C,D,D)| = (4 · 2 · 4 · 2) · 6 = 384and768 + 384 + 768 + 384 + 768 + 768 + 384 = 4224.Remark 2. All the magic squares that can be represented through our no-tation can be classified and identified by the membership class.Remark 3. Such representation allows to construct in a simple way all thenormal multiplicative magic squares of order 4.We conclude this paper with a further example.Example 3. Let us consider the factorization 2 · 3 · 5 · 67 of 2010. We takea magic square in M2010,42010 5 67 63 134 10 10052 201 15 670335 30 402 1 =2 · 3 · 5 · 67 5 67 2 · 33 2 · 67 2 · 5 3 · 5 · 672 3 · 67 3 · 5 2 · 5 · 675 · 67 2 · 3 · 5 2 · 3 · 67 1 .its image through f is1111 0010 0001 11000100 1001 1010 01111000 0101 0110 10110011 1110 1101 0000 =15 2 1 124 9 10 78 5 6 113 14 13 0 .This square belongs to the class (C,C,D,D), in fact it can be decomposedas follows81 0 0 10 1 1 01 0 0 10 1 1 0+ 41 0 0 11 0 0 10 1 1 00 1 1 0+ 21 1 0 00 0 1 10 0 1 11 1 0 0+1 0 1 00 1 0 10 1 0 11 0 1 0 .10References[1] W. S. Andrews, Magic squares and cubes, Dover Publications, NewYork, 1960.[2] M. Bona, A new proof of the formula for the number of 3 × 3 magicsquares, Mathematics Magazine, Vol. 70 (1997), 201–203.[3] M. Bona, Introduction to enumerative combinatorics, McGraw Hill,2007.[4] Sir H. Bondi, Dame K. Ollerenshaw, Magic squares of order four, Phil.Trans. R. Soc. Lond. A, 306 (1982), 443–532.[5] J. Cilleruelo, F. Luca, On multiplicative magic squares, The ElectronicJournal of Combinatorics, Vol. 17, #N8, 2010.[6] R. Descombes, Les carre´s magiques, Vuibert, 2000.[7] D. Friedman, Multiplicative magic squares, Mathematics Magazine,Vol. 49, No. 5, (1976), 249–250.[8] Macalester College Problem of the Week, Nov. 30, 1994.[9] C. Libis, J. D. Phillips, M. Spall, How many magic squares are there?,Mathematics Magazine, Vol. 73 (2000), 57–58.[10] L. Sallows, C. Libis, J. D. Phillips, S. Golomb, News and letters, Math-ematics Magazine, Vol. 73, No. 4, 332–334, 2000.11

Squaring the magic squares of order 4

Abstract In this paper, we present the problem of counting magic squares and we focus on the case of multiplicative magic squares of order 4. We give the exact number of normal multiplicative magic squares of order 4 with an original and complete proof, pointing out the role of the action of the symmetric group. Moreover, we provide a new representation for magic squares of order 4. Such representation allows the construction of magic squares in a very simple way, using essentially only five particular 4 4 × matrices

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SQUARING THE MAGIC SQUARES OF ORDER 4

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Squaring the magic squares of order 4

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