Perfect magic squares - Inlaid square (1)

Inlaid square (1)

	Perfect magic squares

Inlaid square (1)

How to produce an inlaid square?

On page Bordered suares I present an method of construction to produce odd respectively even
bordered magic squares. More complicated are inlaid squares; see for example on the website of
Harvey Heinz: www.magic-squares.net/magicsquare.htm#Orders 3, 5, 7, 9 Inlaid

Someone asked me how to produce a complicated inlaid square. There is no algorithm to produce
each imaginable inlaid square. It challenged me to produce the below mentioned inlaid square.

The challenge is: how to produce a 12x12 magc square existing of four 6x6 magic squares with in
each 6x6 magic square an 4x4 (panmagic) inlaid square. To meet the challenge I followed the steps
below:

· The easiest step is to produce the four 4x4 panmagic inlaid squares. Use a random chosen 8x8
most perfect (Franklin pan)magic square (see explanation most perfect magic squares), add 40
to each digit and split up the 8x8 square in four 4x4 (inlaid) squares.

Most magic 8x8 square + 40 = four 4x4 inlaid squares

1	54	12	63	3	56	10	61	41	94	52	103	43	96	50	101
16	59	5	50	14	57	7	52	56	99	45	90	54	97	47	92
53	2	64	11	55	4	62	9	93	42	104	51	95	44	102	49
60	15	49	6	58	13	51	8	100	55	89	46	98	53	91	48
17	38	28	47	19	40	26	45	57	78	68	87	59	80	66	85
32	43	21	34	30	41	23	36	72	83	61	74	70	81	63	76
37	18	48	27	39	20	46	25	77	58	88	67	79	60	86	65
44	31	33	22	42	29	35	24	84	71	73	62	82	69	75	64

· To produce the four borders you need (4 x 20 =) 80 digits. Take the digits 1 up to 40 and 105 up
to 144 and translate the digits 105 up to 144 into -/- 1 up to -/- 40.

· See method of Construction to produce even bordered squares. Each side of the border consists of
3 positive and 3 negative digits and the sum of the 6 digits is 0. For the four times four corners you
need 16 digits, that is 8 digits positive/negative, double. Given that the average digit is (the lowest
digit plus the highest digit devided by two: [1+40]/2 =) 20,5, the sum of the 8 double digits must
be (8 x 20,5 = )164. The sum of 3 digits must be (3 x 20,5 =) 61,5, that is alternate 61 or 62.
I puzzled and got finally the table below:

+	15	20	26	61	16	21	25	62	17	22	23	62	18	19	24	61	164
+	7	28	26	61	5	32	25	62	8	31	23	62	1	36	24	61
-/-	15	9	37	61	16	6	40	62	17	10	35	62	18	4	39	61
-/-	13	14	34	61	3	29	30	62	2	27	33	62	11	12	38	61

· Use the table to produce the borders of the four 6x6 squares (fill in the digits from the table, fill in
the opposite digits and translate the negative digits -/- 1 up to -/- 40 into 105 up to 144).

15	20	-13	-14	-34	26	16	21	-3	-29	-30	25	17	22	-2	-27	-33	23	18	19	-11	-12	-38	24
					28						32						31						36
					7						5						8						1
					-37						-40						-35						-39
					-9						-6						-10						-4
					-15						-16						-17						-18


15	20	-13	-14	-34	26	16	21	-3	-29	-30	25	17	22	-2	-27	-33	23	18	19	-11	-12	-38	24
-28					28	-32					32	-31					31	-36					36
-7					7	-5					5	-8					8	-1					1
37					-37	40					-40	35					-35	39					-39
9					-9	6					-6	10					-10	4					-4
-26	-20	13	14	34	-15	-25	-21	3	29	30	-16	-23	-22	2	27	33	-17	-24	-19	11	12	38	-18


15	20	132	131	111	26	16	21	142	116	115	25	17	22	143	118	112	23	18	19	134	133	107	24
117					28	113					32	114					31	109					36
138					7	140					5	137					8	144					1
37					108	40					105	35					110	39					106
9					136	6					139	10					135	4					141
119	125	13	14	34	130	120	124	3	29	30	129	122	123	2	27	33	128	121	126	11	12	38	127

· Put the borders and the 4x4 inlaid squares together.

12x12 square = four 6x6 squares with 4x4 inlaid

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

The magic sum of the four 4x4 panmagic inlaid squares is each time 290. The magic sum of the four 6x6
magic squares is each time 435. The magic sum of the 12x12 magic square is 870.

Notify that the 12x12 magic square consists of four proportional 6x6 magic squares, and that is why (as
extra magic feature) half of the rows/columns/diagonals of the 12x12 magic square give 435 (= ½ of the
magic sum of 870).

And now the finishing touch!!!

We can enlarge the above produced 12x12 inlaid square to a 14x14 inlaid square.

[1] Add 26 to each digit.

[2] Make a border of 52 digits (1 up to 26 and 171 up to 196) around the 12x12 inlaid square.

For method to produce the border, see: www.perfectmagicsquares.com/Bordered_squares.html.

The sum of the digits 1 up to 26 is 351. If you add 33, than you get 384, that is 4x96. To get 33
take for example the digits 16 and 17 double. Solve the puzzle and you get for example this table:

16	17	1	26	2	25	9	96
16	4	24	5	22	6	19	96
17	3	23	7	21	10	15	96
8	11	12	13	14	18	20	96

Use the table to make the border (notify that the 26 highest digits, 171 t/m 196, are translated into
-/- 1 up to -/- 26):

16	1	26	2	25	9	-8	-11	-12	-13	-14	-18	-20	17
													3
													23
													7
													21
													10
													15
													-4
													-24
													-5
													-22
													-6
													-19
													-16


16	1	26	2	25	9	-8	-11	-12	-13	-14	-18	-20	17
-3													3
-23													23
-7													7
-21													21
-10													10
-15													15
4													-4
24													-24
5													-5
22													-22
6													-6
19													-19
-17	-1	-26	-2	-25	-9	8	11	12	13	14	18	20	-16


16	1	26	2	25	9	189	186	185	184	183	179	177	17
194													3
174													23
190													7
176													21
187													10
182					<