New Method for Loubère, Méziriac and Méziriac Type Magic Squares (Part IB)

A stairs

The Siamese method which includes both the Loubère and Méziriac magic squares have the property that the center cell must always contain the middle number of the series of numbers used, i.e. a number which is equal to one half the sum of the first and last numbers of the series, or ½(n2 + 1). In addition, the sum of the horizontal rows, vertical columns and corner diagonals are equal to the magic sum S. Both squares also require an upward stepwise addition of consecutive numbers, i.e., 1,2,3...

Summary of Part IA

The previous page gave an account of a new method to generate 7×7 and 9×9 Méziriac type squares, a method previously unreported since Loubère and Méziriac concentrated only on those squares where the initial digit 1 were located on the periphery of the square (i.e., the cell at the center of row one) or on the cell adjacent to the central cell. There was no attempt to look into those squares where the digit 1 resided on cells other than these two positions. It was shown on that page that the 7×7 square was the first to contain three cells to the right of the central cell and that one could also place the digit 1 in the second cell to generate a new type of Méziriac square.

It was also shown that certain squares, particularly those where the size of the number n used to generate the square was divisible by three, produced less than the required number of squares. Thus, where the starting digit 1 is placed to the right of the central cell towards C determines what right hand shift to use when moving from one diagonal end to the start of the next diagonal. It was mentioned that there are four positions in which the digit 1 can be placed. An E with an arrow meant that a right cell shift of the value in the cell is required towards the periphery of the square. On the other hand, if it points inwards the movement is towards C:

Right Cell Shifts
C 2 468 10 1214 1618202224 26... E →

In addition, the caption for each square included both a letter and the two arrows showing the shift to the right and the accompanying shift to the left, e.g., K (6→ 5←) where the sum of the two numbers is n.

Two Possible Squares Derived from 11×11 and 13×13

Examination of the 11×11 (K) and the 13×13 (L) Méziriac type squares where the initial digit 1 is place three cells to the right of the central cell are shown below. Again the position of this cell implies that a shift of 6 cells to the right is to be used when moving from one diagonal to the next. Although the right hand shifts of these squares are 6 one may also visualize them as left hand shifts of 5 for (K) and 7 for (L) where the sum of the

left hand shift + right hand shift = n
K (6→ 5←)
79 103 63054 67 91115 1842 66
102 5 295377 90 11417 4165 78
4 28 527689 113 1640 6488 101
27 51 7599112 15 3963 87100 3
50 74 9811114 38 6286 1102 26
7397121133761 85109125 49
96 120 123660 84 10811 2448 72
119 22 355983 107 1023 4771 95
21 34 5882106 9 3346 7094 118
44 57 811058 32 4569 93117 20
56 80 104731 55 6892 11619 43
L (6→ 7←)
63 35 714812092 77 49 21162 134106 91
34 6 14711910476 48 20 161133 10590 62
5 146 1181037547 19 160 132117 8961 33
145 130 102744618 159 131 11688 6032 4
129 101 734517158 143 115 8759 313 144
100 72 4416157142 114 86 5830 2156 128
71431516914111385 57291 155127 99
42 14 16814011284 56 28 13154 12698 70
26 167 1391118355 27 12 153125 9769 41
166 138 110825439 11 152 12496 6840 25
137 109 81533810 151 123 9567 5224 165
108 80 65379150 122 94 6651 23164 136
79 64 368149121 93 78 5022 163135 107

So the upshot is that the full complement of m is possible for these squares when n is prime and it doesn't matter whether n is of type 4N+1 or 4N+3.

A Subset of Three 15×15 Squares

The next square, the 15×15 (M), shows (1) that filling in the right hand diagonal, (2) then placing the starting digit 1 six cells to the right of the central cell, (3) doing a stepwise construction, will eventually result in a cell number being superimposed over another. The same result is produced when placing the digit 1 thre cells to the right of the central cell (not shown). In both cases the 30 superimposes over 111 even though the former is a 12 cell right shift and the latter is a 6 cell right shift.

M (12→ 3←)
248 217 120
23 7 218 119
22 6 215 118
5214 117 21
4 213 116 20
3 212 115 19
211 114 18 2
225113 17 1
224 112 16 15
111 30 14 233
110 29 13 222
109 28 12 221
10827 11 220
107 26 10 219
106 25 9 218

Since n = 3×5 then those right shifts divisible by 5 will not produce a square. Thus, placing the digit 1 five cells to the right of the central cell will, according to the shift table below, result in a right cell shift of 10 cells to the right, whereby the number 15 superimposes over the cell value of 112. The upshot is that one more square is not constructible as seen in partial square Mp.

15×15 Right Cell Shifts
C 2 468 10 1214 E →
Mp (10→ 5←)
115 3 
114 2
113 1
112 15
111 14

Although, the square (N) with the digit 1 in position 4 cells from the center is magic (below), the square with the digit in 2 cells to the east of the central cell is not magic (square not shown here but the identical square rotated by 180o degrees along the main diagonal is shown at the end of Part IV). Thus according to the 15×15 right shift table above, only those right shifts of 2, 8 and 14 are possible while the 4 is non magic the 6, 10, and 12 being impossible.

N (8→ 7←)
137 169 20184072 104 121 153185 21724 5688 120
168 200 73971103 135 152 184216 2355 87119136
199 6 3870102134 151 183215 225486 118150167
5 37 69101133165 182 214 2153 85117 149166 198
36 68 100132164181 213 20 5284 116148 180197 4
67 99 131163195212 19 51 83115 147179 1963 35
98 130 16219421118 50 82 114146 178210 234 66
129161193225174981113 145177 2091 3365 97
160 192 224164880 112 144 176208 1532 6496 128
191 223 304779111 143 175 20714 3163 95127 159
222 29 4678110142 174 206 1345 6294 126158 190
28 60 77109141173 205 12 4461 93125 157189 221
59 76 108140172204 11 43 7592 124156 188220 27
90 107 13917120310 42 74 91123 155187 21926 58
106 138 170202941 73 105 122154 186218 2557 89

Finally, for the larger squares, we can mention that when n = 25, i.e., p1 × p2= 25 and p1 = p2 then those cells in the shift table divisible by five will be impossible to construct. For example, placing the digit 1 into the 5th cell from center will result in a right shift of 10, superimposing the digit 51 onto number 311 of the main diagonal. Similarly, for n = 3 × 7, placing the digit 1 in cell number 7 will result in a right shift of 14 and superimpose the digit 21 onto 220 of the main diagonal. The superimpositions, thereby, reduce the number of Méziriac type magic squares.

This completes this section (Part IB). To go to next section Part II. To return to Part IA. To return to homepage.


Copyright © 2020 by Eddie N Gutierrez. E-Mail: enaguti1949@gmail.com