Example of order 8 bimagic square
Following bimagic square is made by "Mahoujin" (Magic square) which is written by Kiyomi Oomori method.
| 2 | 13 | 24 | 27 | 35 | 48 | 53 | 58 |
| 23 | 28 | 1 | 14 | 54 | 57 | 36 | 47 |
| 37 | 42 | 51 | 64 | 8 | 11 | 18 | 29 |
| 52 | 63 | 38 | 41 | 17 | 30 | 7 | 12 |
| 16 | 3 | 26 | 21 | 45 | 34 | 59 | 56 |
| 25 | 22 | 15 | 4 | 60 | 55 | 46 | 33 |
| 43 | 40 | 61 | 50 | 10 | 5 | 32 | 19 |
| 62 | 49 | 44 | 39 | 31 | 20 | 9 | 6 |
|
S = 260
|
Squared each cell
| 4 | 169 | 576 | 729 | 1225 | 2304 | 2809 | 3364 |
| 529 | 784 | 1 | 196 | 2916 | 3249 | 1296 | 2209 |
| 1369 | 1764 | 2601 | 4096 | 64 | 121 | 324 | 841 |
| 2704 | 3969 | 1444 | 1681 | 289 | 900 | 49 | 144 |
| 256 | 9 | 676 | 441 | 2025 | 1156 | 3481 | 3136 |
| 625 | 484 | 225 | 16 | 3600 | 3025 | 2116 | 1089 |
| 1849 | 1600 | 3721 | 2500 | 100 | 25 | 1024 | 361 |
| 3844 | 2401 | 1936 | 1521 | 961 | 400 | 81 | 36 |
|
S = 11180
|
About order-8 bimagic square, construction method is known.
How about order-7 and smaller?
The answer is no.
I confirmed it, I don't have any information that about order-7 and smaller bimagic square cannot constract at January 2000.
Multimagic square records
This site describe mainly bimagic, following multimagic records.
Example: Trimagic square is a magic square that is still magic when each cell of the square is squared or cubed.
Multimagic square records
| Power |
Order |
Minimum? |
Inventor(Date) |
| 1 |
3 |
Minimum |
China (B.C.) |
| 2 |
8 |
Minimum |
G. Pfeffermann (1890) |
| 3 |
12 |
Minimum |
Walter Trump (2002) |
| 4 |
243 |
? |
Gao Zhiyuan (2004?) |
| 5 |
729 |
? |
Li Wen (2003) |
| 6 |
4096 |
? |
Pan Fengchu (2003) |
| 7 |
Unknown |
? |
(Do you have any information?) |
Order 3 : Impossible
Order-3 magic square is only one, it cannot keep magic when each cell of the square.
Preparation of order 4 to 7:
Order-4 to order-7 bimagic square cannot make, I check them by the conditions of both base magic square and squared one.
Preparation : THe constant of base magic square and the magic square of each cell squared.
THe constant of magic square is sums of all numbers divide order number.
The constant of magic square of order-n (called S1)
(1+2+ … +n2) / n = n (n2+1) / 2
The constant of squared magic square of order-n (called S2)
(12+22+ … +(n2)2) / n = n (n2+1) (2n2+1) / 6
Order 4 : Impossible
[ S1=34, S2=374 ]
(16,E1,E2,E3) means elements which include maximum number 16.
The conditions of magic square provide the following equations.
16 + E1 + E2 + E3 = 34
E1 + E2 + E3 = 18 ... (1-1)
The conditions of squared magic square provide the following equations.
162 + E12 + E22 + E32 = 374
E12 + E22 + E32 = 118 ... (1-2)
Any E1, E2, E3 set don't satisfy both (1-1) and (1-2), we cannot make lines which include 16.
Then, we cannot make order 4 bimagic square.
Order 5 : Impossible
[ S1=65, S2=1105 ]
(25,E1,E2,E3,E4) means the element of line which include maximum number 25.
The conditions of magic square provide the following equations.
25 + E1 + E2 + E3 + E4 = 65
E1 + E2 + E3 + E4 = 40 ...(2-1)
The conditions of squared magic square provide the following equations.
252 + E12 + E22 + E32 + E42 = 1105
E12 + E22 + E32 + E42 = 480 ...(2-2)
Only (16,12,8,4) satisfied both (2-1) and (2-2).
To make bimagic, we need at lease two lines which stisfy above condition, we cannot make order-5 bimagic square.
Order 6 : Impossible
[ S1=111, S2=2701 ]
(36,E1,E2,E3,E4,E5) means the element of line which include maximum number 36.
E1 > E2 > E3 > E4 > E5
I find same method as order-5, 18 sets of line satisfy the contitions of magic square and squared magic square.
Following is my confirm method:
Maximum E1 of (36,E1,E2,E3,E4,E5) is 29.
Maximum E1 of (35,E1,E2,E3,E4,E5) is 31.
Following are those maximum number and second number list.
Max Second
36 29
35 31
34 32
33 32
Regarding to above conditions, 36, 35, 34 and 33 are not exist on same line.
Following is example of 36 and 35.
P1 and P2 are the number which is included 36's line and 35's line.
Bimagic square finding method is following.
1. Find pairs of different numbers set (E1, E2, E3, E4, E5) from 36's line to 33's line. I call one of pair "(36, E1, E2, E3, E4, E5)" 36v (vertical), other one 36h (horizontal).
2. Find Max36-pairs and Max35-pairs which satisfy following conditions.
No number exist both 36v and 35v
No number exist both 36h and 35h
Only one number exist both 36v and 35h
Only one number exist both 36h and 35v
3. If the pair line is exist, find Max34-pairs which satisfy following conditions.
No number exist both 34v and (36v or 35v)
No number exist both 34h and (36h or 35h)
Only one number exist both 34v and 36h
Only one number exist both 34v and 35h
Only one number exist both 34h and 36v
Only one number exist both 34h and 35v
4. If the pair line is exist, find Max33-pairs which satisfy following conditions.
No number exist both 33v and (36v or 35v or 34v)
No number exist both 33h and (36h or 35h or 34h)
Only one number exist both 33v and 36h
Only one number exist both 33v and 35h
Only one number exist both 33v and 34h
Only one number exist both 33h and 36v
Only one number exist both 33h and 35v
Only one number exist both 33h and 34v
I checked those Max36-pair to Max33-pair are impossible using computer.
Then, we cannot make order-6 bimagic square.
Order 7 : Impossible
[ S1=175, S2=5775 ]
Method is same as order-6, but relationship of maximum number and second number is not enough to check order-7 bimagic square.
Max Second
49 45
48 46
47 46
To check order-7 bimagic square, maximum number of second (to seventh) pairs using a never using number.
The list of maximum number of each line, number of lines and number of virtical-horizontal pairs are following.
Max Lines Virtical-horizontal pair
49 209 7238
48 261 12226
47 209 7343
46 299 17662
45 229 9539
44 233 10286
43 138 3326
42 131 2704
41 69 466
40 49 234
39 9 2
38 7 1
I checked above combinations, it is impossible to construct bimagic square.
An order 7 bimagic square cannot exist.
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