Basic key method (ultra magic)

How to construct as perfect as possible magic squares which are an odd multiple of 4?First we analyze how the basic key method works for the construction of the 16x16 magic square. Than we investigate the possibilities to use a basic key to construct a 12x12 magic square.

Why does the basic key of the 16x16 magic square work?

How is it possible that the basic key leads to most perfect (Franklin pan)magic squares for each order is a multiple of 8.

1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16 = 136.

(1st) sum = sum = 34 = 1/4 x 136

1	2	16	15
16	15	1	2

So, in the 16x16 magic square each 1/4 row/column gives 1/4 of the magic sum.

(2nd) sum = sum = 34 = 1/4 x 136

1	2	16	15
16	15	1	2

So, in the 16x16 magic square each 1/4 diagonal gives 1/4 of the magic sum.

(3rd) Between opposite combinations must be an odd number of cells (opposite combinations must not be arranged next to each other).

1	2	16	15
16	15	1	2

So the first grid can be reflected to construct the second grid and if you combine the first grid with the second grid you get all the numbers from 1 up to 256 in the magic square.

How to get the right basic key to construct a 12x12 magic square?

For magic squares which are an odd multiple of 4 no basic key exists to comply to all three above mentioned conditions. You have to make a choice.

[Option a]

(1st) sum = sum = 26 = 1/3 x 78

1	10	11	4	8	6	5	7	12	3	2	9
12	3	2	9	5	7	8	6	1	10	11	4

So each 1/3 row/column gives 1/3 of the magic sum (but 1/2 row/column gives not 1/2 of the magic sum).

(2nd) sum = sum = 39 = 1/2 x 78

1	10	11	4	8	6
12	3	2	9	5	7

So each 1/2 diagonal gives 1/2 of the magic sum (and each [[parallel] bended diagonal gives the magic sum = Franklin feauture).

(3rd) Between each opposite combination is and odd number of cells, so the first grid can be reflected to construct the second grid and if you combine the first grid with the second grid you get all the numbers from 1 up to 144 in the magic square.

1x number from first grid

1	10	11	4	8	6	5	7	12	3	2	9
12	3	2	9	5	7	8	6	1	10	11	4
1	10	11	4	8	6	5	7	12	3	2	9
12	3	2	9	5	7	8	6	1	10	11	4
1	10	11	4	8	6	5	7	12	3	2	9
12	3	2	9	5	7	8	6	1	10	11	4
1	10	11	4	8	6	5	7	12	3	2	9
12	3	2	9	5	7	8	6	1	10	11	4
1	10	11	4	8	6	5	7	12	3	2	9
12	3	2	9	5	7	8	6	1	10	11	4
1	10	11	4	8	6	5	7	12	3	2	9
12	3	2	9	5	7	8	6	1	10	11	4

+ 12 x (number -/- 1) from second grid

9	4	9	4	9	4	9	4	9	4	9	4
2	11	2	11	2	11	2	11	2	11	2	11
3	10	3	10	3	10	3	10	3	10	3	10
12	1	12	1	12	1	12	1	12	1	12	1
7	6	7	6	7	6	7	6	7	6	7	6
5	8	5	8	5	8	5	8	5	8	5	8
6	7	6	7	6	7	6	7	6	7	6	7
8	5	8	5	8	5	8	5	8	5	8	5
4	9	4	9	4	9	4	9	4	9	4	9
11	2	11	2	11	2	11	2	11	2	11	2
10	3	10	3	10	3	10	3	10	3	10	3
1	12	1	12	1	12	1	12	1	12	1	12

= 12x12 magic square (option a)

97	46	107	40	104	42	101	43	108	39	98	45
24	123	14	129	17	127	20	126	13	130	23	124
25	118	35	112	32	114	29	115	36	111	26	117
144	3	134	9	137	7	140	6	133	10	143	4
73	70	83	64	80	66	77	67	84	63	74	69
60	87	50	93	53	91	56	90	49	94	59	88
61	82	71	76	68	78	65	79	72	75	62	81
96	51	86	57	89	55	92	54	85	58	95	52
37	106	47	100	44	102	41	103	48	99	38	105
132	15	122	21	125	19	128	18	121	22	131	16
109	34	119	28	116	30	113	31	120	27	110	33
12	135	2	141	5	139	8	138	1	142	11	136

[option b]

(1st) sum = sum = 39 = 1/2 x 78

1	11	10	2	7	8	12	4	3	9	6	5
12	2	3	11	6	5	1	9	10	4	7	8

So each 1/2 row/column gives 1/2 of the magic sum.

(2nd) The sum of the half winding rows is not 39, but the sum of the whole winding rows is 78.

So each diagonal gives the magic sum (but each 1/2 diagonal gives not 1/2 of the magic sum and each [parallel] bended diagonals gives not the magic sum).

1x number from first grid

1	11	10	2	7	8	12	4	3	9	6	5
12	2	3	11	6	5	1	9	10	4	7	8
1	11	10	2	7	8	12	4	3	9	6	5
12	2	3	11	6	5	1	9	10	4	7	8
1	11	10	2	7	8	12	4	3	9	6	5
12	2	3	11	6	5	1	9	10	4	7	8
1	11	10	2	7	8	12	4	3	9	6	5
12	2	3	11	6	5	1	9	10	4	7	8
1	11	10	2	7	8	12	4	3	9	6	5
12	2	3	11	6	5	1	9	10	4	7	8
1	11	10	2	7	8	12	4	3	9	6	5
12	2	3	11	6	5	1	9	10	4	7	8

+ 12x (number -/- 1) from second grid

5	8	5	8	5	8	5	8	5	8	5	8
6	7	6	7	6	7	6	7	6	7	6	7
9	4	9	4	9	4	9	4	9	4	9	4
3	10	3	10	3	10	3	10	3	10	3	10
4	9	4	9	4	9	4	9	4	9	4	9
12	1	12	1	12	1	12	1	12	1	12	1
8	5	8	5	8	5	8	5	8	5	8	5
7	6	7	6	7	6	7	6	7	6	7	6
2	11	2	11	2	11	2	11	2	11	2	11
10	3	10	3	10	3	10	3	10	3	10	3
11	2	11	2	11	2	11	2	11	2	11	2
1	12	1	12	1	12	1	12	1	12	1	12

= 12x12 magic square (option b)

49	95	58	86	55	92	60	88	51	93	54	89
72	74	63	83	66	77	61	81	70	76	67	80
97	47	106	38	103	44	108	40	99	45	102	41
36	110	27	119	30	113	25	117	34	112	31	116
37	107	46	98	43	104	48	100	39	105	42	101
144	2	135	11	138	5	133	9	142	4	139	8
85	59	94	50	91	56	96	52	87	57	90	53
84	62	75	71	78	65	73	69	82	64	79	68
13	131	22	122	19	128	24	124	15	129	18	125
120	26	111	35	114	29	109	33	118	28	115	32
121	23	130	14	127	20	132	16	123	21	126	17
12	134	3	143	6	137	1	141	10	136	7	140

Ultra panmagic 12x12 square

It is also possible to construct an ultra panmagic 12x12 square in which each 4x4 sub-square has the following symmetric structure:

Two numbers of the same colour sums to (the lowest + highest number of the magic 12x12 square = 1 + 144 =) 145.

*1x number*
12	2	2	12	5	6	6	5	10	4	4	10
1	11	11	1	8	7	7	8	3	9	9	3
12	2	2	12	5	6	6	5	10	4	4	10
1	11	11	1	8	7	7	8	3	9	9	3
12	2	2	12	5	6	6	5	10	4	4	10
1	11	11	1	8	7	7	8	3	9	9	3
12	2	2	12	5	6	6	5	10	4	4	10
1	11	11	1	8	7	7	8	3	9	9	3
12	2	2	12	5	6	6	5	10	4	4	10
1	11	11	1	8	7	7	8	3	9	9	3
12	2	2	12	5	6	6	5	10	4	4	10
1	11	11	1	8	7	7	8	3	9	9	3
*+ 12 x (number -/- 1)*
10	3	10	3	10	3	10	3	10	3	10	3
4	9	4	9	4	9	4	9	4	9	4	9
4	9	4	9	4	9	4	9	4	9	4	9
10	3	10	3	10	3	10	3	10	3	10	3
5	8	5	8	5	8	5	8	5	8	5	8
6	7	6	7	6	7	6	7	6	7	6	7
6	7	6	7	6	7	6	7	6	7	6	7
5	8	5	8	5	8	5	8	5	8	5	8
12	1	12	1	12	1	12	1	12	1	12	1
2	11	2	11	2	11	2	11	2	11	2	11
2	11	2	11	2	11	2	11	2	11	2	11
12	1	12	1	12	1	12	1	12	1	12	1
*= ultra panmagic 12x12 square*
120	26	110	36