Multimagic Squares

In this paper we give the first method for constructing n-multimagic squares (and hypercubes) for any n. We give an explicit formula in the case of squares and an effective existence proof in the higher dimensional case. Finally we prove that n-multimagic squares do not exist for certain orders. Introduction Magic squares have been studied over 4000 years. Recently some exciting new results have been found considering these squares. For instance the first method of constructing all most-perfect magic squares and their enumeration appeared in [3] (See also [14]). Also the non-existence of a 8 × 8 magic knights tour ([13]) has been verified by computers. Finally a very natural connection has been found between the properties and construction of Franklin’s squares and his magic circles ([10]). However, there are still many unsolved problems (e.g. [1], [12]) and as Clifford Pickover says “the field of magic square study is wide open” ([12], p. 26). In this paper we will concentrate on so called multimagic squares. Suppose that M is an m ×m matrix M consisting of natural numbers. Then M is called a magic square if the sum of all elements in each column, row and main diagonal gives the same number; the so-called magic number. Let M be the matrix obtained by raising each element of M to the d-th power. The matrix M is called an n-multimagic square (where n is a fixed positive natural number) if M is a magic square for d = 1, 2, . . . , n. The matrixM is called normal if its matrix elements consist of the consecutive integers 1, 2, . . . , m. Throughout this paper we always consider normal magic squares (of course if m > 1 and d > 1 then the matrix M is not normal). The first 2-multimagic square was published by Pfeffermann in 1891: it has order 8 (Figure 1, [11], [5]). In 1905 the first 3-multimagic square was constructed by Tarry: it has order 128. In 2001 both a 4and a 5-multimagic square were constructed by Boyer and Viricel respectively of order 512 and 1024 ([4], [5] where they also give a nice history on the subject). The record up to now was a 6-multimagic square of order 4096 constructed by Pan Fengchu in October 2003 ([8]). 1 56 34 8 57 18 47 9 31 33 20 54 48 7 29 59 10 26 43 13 23 64 38 4 49 19 5 35 30 53 12 46 60 15 25 63 2 41 24 50 40 6 55 17 11 36 58 32 45 61 16 42 52 27 1 39 22 44 62 28 37 14 51 21 3 Figure 1: G. Pfeffermann, 1891 In this paper we give a constructive procedure to make a large class of n-multimagic squares for each positive integer n ≥ 2. The problem of finding such squares is reduced to an easy linear algebra problem which is solved in general in section 4. A more explicit solution is described in section 3. This solution is used to give an explicit formula for n-multimagic squares for all n ≥ 3. In particular it gives the first 7-multimagic squares, of order 13 and 8-multimagic squares of order 17 etc. The method described for constructing n-multimagic squares can easily be extended to n-multimagic cubes and hypercubes. We refer to section 4 for all definitions and more details. 1 Preliminaries Throughout this paper R denotes a finite ring with q elements. Let R denote the set of elements in R which have a multiplicative inverse. Definition 1.1 For c ∈ R we call a bijection N : R → {0, 1, . . . , q − 1} of type c if N(a) +N(−a + c) = q − 1 for all a ∈ R. Lemma 1.2 i). If 2 ∈ R, then for every c ∈ R there exists a bijection N of type c. ii). If 2 / ∈ R, then for every c ∈ R there exists a bijection N of type c. Proof. For c ∈ R define φ = φc : R → R by φ(a) = −a + c for all a ∈ R. Then φ 2 is equal to the identity. So all orbits of φ have length 1 or 2. An element a is a fixed point of φ if and only if 2a = c. For a ∈ R denote its orbit under φ by O(a). i) Let 2 be a unit in R. Then φ has exactly one fixed point, namely a0 := 2 c. The orbit O(a0) of a0 has one element. Let O(a1), O(a2), . . . , O(as) be the other orbits. They all have two elements. In particular, we get s = (q− 1)/2. Define N(a0) = s, N(ai) = i− 1 and N(φ(ai)) = q − 1−N(ai) (= q − i), for i = 1, 2, . . . , s. Now N is as desired. ii) Let 2 not be a unit in R. Then for c ∈ R, φ = φc has no fixed points. Indeed, if 2 φ(a) = a, then we get 2a = c ∈ R. So 2 ∈ R and we have a contradiction. We can write R = ⋃s i=1 O(ai), where each O(ai) is an orbit with two elements. In particular, s = q/2. Define N(ai) = i− 1 and N(φ(ai)) = q− 1−N(ai) (= q− i) for all i. Then N is as desired. Let m be a positive integer. For each 1 ≤ j ≤ m we choose a bijection N(j) : R → {0, 1, . . . , q − 1} of type cj, for some cj ∈ R (this is possible by Lemma 1.2). Put c = (c1, . . . , cm) ∈ R m and define Nm : R m → {1, 2, . . . , q} by Nm(a1, . . . , am) = 1 + m ∑ j=1 qN(j)(aj). Since the coefficients of the q-adic expansion of any natural number are unique and each N(j) is a bijection, it follows that Nm is a bijection. Lemma 1.3 Nm(−a) = q m + 1−Nm(a + c), for all a = (a1, . . . , am) ∈ R . Proof. From definition 1.1 follows that Nm(−a) +Nm(a + c) = 2 + m ∑ j=1 q(N(j)(−aj) +N(j)(aj + cj)) = = 2 + (1 + q + . . .+ q)(q − 1) = 2 + (q − 1) = q + 1. To conclude this section we will give a result (proposition 1.5) which plays a crucial role in the next section. First some notations. Let n and s be positive integers. Suppose that L : R → R is an affine map, i.e., there exists an R-module homomorphism L0 : R n → R and a vector v ∈ R such that L(a) = L0(a) + v, a ∈ R . Lemma 1.4 If L : R → R is a surjective affine map, then #L(y) = q for all y ∈ R. Proof. Let y ∈ R. Since L is surjective there exists a0 ∈ R n such that L(a0) = y. It follows that L(y) = a0 + kerL0. Since L is surjective, so is L0. It follows that R/ kerL0 ≃ R , whence #kerL0 = q n−s and consequently #L(y) = q.