Properties of 3x3 Magic Squares
There are many properties of a 3x3 magic square (relationships among
the numbers.) I'll include a list with the solution. If you'd like
to contribute, please send them along. For example:
For each item below, if possible,
find a magic square (all rows, columns and diagonals have the same sum S)
with nine unique positive integers.
For ease in comparing solutions, minimize S, then A, then B.
If it's not possible, show why.
- A through I, arranged in increasing order,
do not form an arithmetic series. (In an arithmetic series,
the difference between successive terms is always the same value.)
- The average value of A through I is not one of the nine values.
- A through I are all odd integers.
- Exactly 2 of A through I are odd integers.
- (Are there other instructive examples to add here?)
Ignoring symmetry, if you are told the values of any three of the
nine squares in a
3x3 magic square, you can almost always determine the remaining values.
For which combinations of three squares can you not solve the rest of
the square? (Thanks to Helen Warman for showing there are 16 different
combinations of three squares.)
If asked to place 9 specific numbers, A through I, in a magic square,
a quick way to find the common sum (S) is to add them all up
and divide by 3. S=(A+...+I)/3
Solutions were received from Joseph DeVincentis, Denis Borris,
Samantha Levin, John Hewon, and Philippe Fondanaiche.
Two people pointed out that a 3x3 magic square can be represented as:
m+x m-(x+y) m+y
m-(x-y) m m+(x-y)
m-y m+(x+y) m-x
The answers to the above problems are summarized well by Philippe Fondanaiche:
John Hewson had a thorough analysis of the 18 combinations of three values:
For m=6 and S=18, we have the following magic square in which A is minimum
2 9 7
11 6 1
5 3 10
As S=3*m, the average value of the terms of the square is always equal to m.
The magic square with A through I all odd integers and S minimum, is defined
3 13 11
17 9 1
7 5 15
It is impossible to have exactly 2 odd integers. Indeed, if x and y have the
same parity, all the terms have the parity of m.
If x and y have different parities (i.e. x even and y odd), 3 terms are odd
and the 6 other ones are even (and vice versa)
Magic square with A through I all prime numbers:
7 61 43
73 37 1
31 13 67
If we are purist and consider that 1 is not a prime number,then:
17 89 71
113 59 5
47 29 101
Magic square with A through I all squares.
This square is really magic as S=21609 is too a square:
8836 9409 3364
4 5476 16129
12769 6724 2116
Denis Borris sent the smallest product Magic Multiplication Square:
2 9 12
36 6 1
3 4 18
ABC 3m=A+B+C x=A-m y=C-m
AFH F+H=2A the corresponding equations are linearly dependent.
BEH B+H=2E the corresponding equations are linearly dependent.
BGI m=I+G-b y=m-G x=m-I
ABF 2m=B+2A-F x=A-m y=A-F
ABE m=E x=A-m y=m-x-B
ABD 4m=2A+B+D x=A-m y=m-x-B
ABI 2m=A+I x=A-m y=m-x-B
ABH 2m=B+H x=A-m y=H-A
ABG m=B+A-G x=A-m y=m-x-B
ACE m=E x=A-m y=C-m
ACG 2m=C+G x=A-m y=C-m
BDE m=E 2x=2m-B-D 2y=D-B
CDE m=E y= C-m x=m+y-D
AEI A+I=2E the corresponding equations are linearly dependent.
DFH 2m=D+F 2x=F+H-2m 2y=H-F
Thus of the 16 cases only three combinations AFH, BEH, AEI do not enable
the determination of the other values in the square.
Properties of 3x3 Magic Squares
- The common sum S is one-third of the sum of all nine numbers.
- Any corner is the average of the opposite two sides. A=(F+H)/2,
C=(D+H)/2, G=(B+F)/2, I=(B+D)/2
- The center value E is:
- The average of all nine values. E=(A+B+C+D+E+F+G+H+I)/9
- The average of the four corners. E=(A+C+G+I)/4
- The average of the four sides. E=(B+D+F+H)/4
- The average of any two opposite values. E=(A+I)/2 = (B+H)/2 = (C+G)/2 =
- One-third of the common sum. E=S/3
Thanks to John Hewson, who found material in
by Maurice Kraitchik published by George Allen & Unwin Ltd in 1943/44.
He states that in order that any nine distinct numbers be capable of
forming a magic square it is necessary and sufficient that they form three
three term arithmetic progressions whose first terms are in arithmetic
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