MAGIC SQUARES OF TWO DEGREES.
(
Magic Squares Problem.)
While reading a French mathematical work I happened to come across, the
following statement: "A very remarkable magic square of 8, in two
degrees, has been constructed by M. Pfeffermann. In other words, he has
managed to dispose the sixty-four first numbers on the squares of a
chessboard in such a way that the sum of the numbers in every line,
every column, and in each of the two diagonals, shall be the same; and
more, that if one substitutes for all the numbers their squares, the
square still remains magic." I at once set to work to solve this
problem, and, although it proved a very hard nut, one was rewarded by
the discovery of some curious and beautiful laws that govern it. The
reader may like to try his hand at the puzzle.
Answer:
The following is the square that I constructed. As it stands the
constant is 260. If for every number you substitute, in its allotted
place, its square, then the constant will be 11,180. Readers can write
out for themselves the second degree square.
[Illustration:
7 53 | 41 27 | 2 52 | 48 30
12 58 | 38 24 | 13 63 | 35 17
------+-------+-------+------
51 1 | 29 47 | 54 8 | 28 42
64 14 | 18 36 | 57 11 | 23 37
------+-------+-------+------
25 43 | 55 5 | 32 46 | 50 4
22 40 | 60 10 | 19 33 | 61 15
------+-------+-------+------
45 31 | 3 49 | 44 26 | 6 56
34 20 | 16 62 | 39 21 | 9 59
]
The main key to the solution is the pretty law that if eight numbers sum
to 260 and their squares to 11,180, then the same will happen in the
case of the eight numbers that are complementary to 65. Thus 1 + 18 + 23
+ 26 + 31 + 48 + 56 + 57 = 260, and the sum of their squares is 11,180.
Therefore 64 + 47 + 42 + 39 + 34 + 17 + 9 + 8 (obtained by subtracting
each of the above numbers from 65) will sum to 260 and their squares to
11,180. Note that in every one of the sixteen smaller squares the two
diagonals sum to 65. There are four columns and four rows with their
complementary columns and rows. Let us pick out the numbers found in the
2nd, 1st, 4th, and 3rd rows and arrange them thus :--
[Illustration:
1 8 28 29 42 47 51 54
2 7 27 30 41 48 52 53
3 6 26 31 44 45 49 56
4 5 25 32 43 46 50 55
]
Here each column contains four consecutive numbers cyclically arranged,
four running in one direction and four in the other. The numbers in the
2nd, 5th, 3rd, and 8th columns of the square may be similarly grouped.
The great difficulty lies in discovering the conditions governing these
groups of numbers, the pairing of the complementaries in the squares of
four and the formation of the diagonals. But when a correct solution is
shown, as above, it discloses all the more important keys to the
mystery. I am inclined to think this square of two degrees the most
elegant thing that exists in magics. I believe such a magic square
cannot be constructed in the case of any order lower than 8.
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