Office Applications and Entertainment, Magic Squares

15.0    Special Magic Squares, Bimagic Squares, Part 3

15.3    Bimagic Squares (8 x 8)

15.3.14 Benson/Jacobi

The Pan Diagonal/Complete Bimagic Square shown below has originally been constructed by Benson and Jacobi.
The main - and semi diagonals are trimagic.

47	28	6	49	23	36	62	9
8	51	45	26	64	11	21	34
53	2	32	43	13	58	40	19
30	41	55	4	38	17	15	60
42	29	3	56	18	37	59	16
1	54	44	31	57	14	20	39
52	7	25	46	12	63	33	22
27	48	50	5	35	24	10	61

The square has following additional properties:

• The half rows sum to half the Magic Sum;
• All 4 x 4 squares (64 ea) sum to two times the Magic Sum (4 x 4 compact);
• Twelve of the 2 x 2 squares sum to half the Magic Sum;
• The corner points of all 5 x 5 sub squares (16 ea) sum to half the Magic Sum;
• The corner points of sixteen 3 x 3 sub squares sum to half the Magic Sum;
• Twenty four 2 x 4 rectangles and twenty four 4 x 2 rectangles sum to the Magic Sum;
• The horizontal main bent diagonals (4 ea) are bimagic;
• Six pairs of horizontal and two pairs of vertical ZigZag lines sum to the Magic Sum.

The properties mentioned above are illustrated in Spreadsheet Solution CnstrSngl8m and result,
after deduction, in following set of linear equations:

a(61) = 0.5  * s1 - a(62) - a(63) - a(64)
a(57) = 0.5  * s1 - a(58) - a(59) - a(60)
a(54) = 0.5  * s1 - a(55) - a(62) - a(63)
a(53) =           - a(56) + a(62) + a(63)
a(52) =-0.5  * s1 + a(55) + a(58) + a(62) + a(63) + a(64)
a(51) = 0.5  * s1 + a(56) - a(58) - a(59) - a(60) - a(62)
a(50) =           - a(56) + a(60) + a(62)
a(49) = 0.5  * s1 - a(55) + a(59) - a(62) - a(63) - a(64)
a(47) =             a(48) - a(58) - a(60) + a(62) + a(63)
a(46) =           - a(48) + a(58) + a(60)
a(45) = 0.5  * s1 - a(48) - a(62) - a(63)
a(44) =             a(48) - a(58) - a(59) - a(60) + a(62) + a(63) + a(64)
a(43) =             a(48) - a(60) + a(63)
a(42) =           - a(48) + a(58) + a(59) + a(60) - a(63)
a(41) = 0.5  * s1 - a(48) + a(60) - a(62) - a(63) - a(64)
a(40) =           - a(48) - a(56) + a(58) + a(60) + a(62)
a(39) =        s1 - a(48) - a(55) - 2 * a(62) - 2 * a(63) - a(64)
a(38) =-0.5  * s1 + a(48) + a(55) + a(62) + a(63) + a(64)
a(37) =             a(48) + a(56) - a(58) - a(60) + a(63)
a(36) = 0.5  * s1 - a(48) - a(55) + a(58) + a(59) + a(60) - a(62) - 2 * a(63) - a(64)
a(35) = 0.5  * s1 - a(48) - a(56) + a(60) - a(63) - a(64)
a(34) =            a(48) + a(56) - a(58) - a(59) - a(60) + a(63) + a(64)
a(33) =-0.5  * s1 + a(48) + a(55) - a(60) + a(62) + 2 * a(63) + a(64)

The solutions can be obtained by:

• selecting suitable bimagic bottom lines, containing 2 half lines but no pairs
• guessing a(48), a(55) and a(56) and filling out these guesses in the equations shown above

which makes a guessing routine relatively fast.

An optimized guessing routine (MgcSqr15314), produced - based on preselected bimagic series (9450 ea) - 3584 Bimagic Squares (896 Unique). A set of unique squares is shown in Attachment 15.3.14.

Verification:
For each of the 896 Unique Bimagic Squares mentioned above, an aspect could be found in the collection of “Unique Bimagic Squares, Complete with bimagic semi-diagonals” (10496 ea) as made available by Walter Trump.

15.3.15 Summary

The obtained results regarding the miscellaneous types of order 8 Bimagic Squares as deducted and discussed in previous sections are summarized in following table:

Next section will provide some examples of classical construction methods for Bimagic Squares of order 9.