Bimagic Square Impossibility Proofs
3x3 Bimagic Impossibility
4x4 Bimagic Impossibility
5x5 Normal Bimagic Impossibility
6x6 Normal Bimagic Impossibility
7x7 Normal Bimagic Impossibility [INCOMPLETE]
Duplication Lemma
Given integers a,b,c,d, if
(1) a + b = c + d,
(2) a2 + b2 = c2 + d2
then either
a = c or a = d.
Proof
Rewrite (1) as
(3) (a - c) = (d - b).
Rewrite (2) as
a2 - c2 = d2 - b2
or
(4) (a - c)(a + c) = (d - b)(d + b).
Use (3) to substitute (a - c) for (d - b) in (4),
(5) (a - c)(a + c) = (a - c)(d + b).
Multiply (3) by (a - c).
(6) (a - c)(a - c) = (a - c)(d - b).
Add (5) and (6) and divide by 2,
(a - c)a = (a - c)d
or
(a - c)(a - d) = 0.
Thus a = c or a = d.
3x3 Bimagic Impossibility Theorem
A 3x3 bimagic square of distinct integers is impossible.
Proof
Given the following 3x3 bimagic square,
a b P
Q R c
S T d
Because the square is magic,
a + b + P = P + c + d
or
(1) a + b = c + d.
Because the square is bimagic, we must also have
(2) a2 + b2 = c2 + d2.
From (1) and (2) and the Duplication Lemma,
either a = c or a = d, thus
all 3x3 bimagic squares must have duplicate entries.
4x4 Bimagic Impossibility Theorem
A 4x4 bimagic square of distinct integers is impossible.
Proof
Given the following 4x4 bimagic square,
a M N b
P Q R S
T U V W
X c d Y
Because the square is magic, we have
(1) a + M + N + b = X + c + d + Y,
(2) a + Q + V + Y = M + Q + U + c,
(3) b + R + U + X = N + R + V + d.
Add (1), (2), and (3), cancel common terms, and divide by 2,
(4) a + b = c + d.
Because the square is bimagic, we must also have
(5) a2 + b2 = c2 + d2.
From (4) and (5) and the Duplication Lemma,
either a = c or a = d, thus
all 4x4 bimagic squares must have duplicate entries.
5x5 Normal Bimagic Impossibility Theorem
A 5x5 bimagic square using distinct integers in the range 1-25 is impossible.
Proof
The magic sum of squares is 1105 which is 1 (mod 8) and 1 (mod 4).
The magic sum is odd, thus a magic series must contain
an odd number of odd integers.
But the sum of squares of 5 odd integers is not 1 (mod 8) and
the sum of squares of 3 odd integers and 2 even integers is not 1 (mod 4).
So this leaves 1 odd integer and 4 even integers as the only combination
for a magic series. But this requires 20 even integers to make 5 rows
and there are only 12 even integers in the range 1-25.
6x6 Normal Bimagic Impossibility Theorem
A 6x6 bimagic square of distinct integers in the range 1-36 is impossible.
Proof
The magic sum is 111 which is 3 (mod 4).
The magic sum of squares is 2701 which is 1 (mod 4) and 5 (mod 8).
Since the magic sums are odd, a magic series must contain an odd
number of odd integers.
But the sum of squares of 3 odd integers and 3 even integers
is not 1 (mod 4), as required.
So a magic series must contain 1 or 5 odd integers.
Let A be the number of rows with 1 odd integer.
Let B be the number of rows with 5 odd integers.
Since there are 6 rows,
A + B = 6.
Since there are 18 odd integers in the range 1-36,
A + 5B = 18.
The only solution is A = B = 3, so there must be
three rows of 1 odd integer and 5 even integers combined with
three rows of 5 odd integers and 1 even integer.
For the three rows with 5 odd integers and 1 even integer,
the even integer must be 0 (mod 4) to make the sum of squares 5 (mod 8).
To make the magic sum 3 (mod 4), the magic series for each
of these three rows must be taken from
(0,1,1,1,1,3) (mod 4)
(0,1,1,3,3,3) (mod 4)
(0,3,3,3,3,3) (mod 4)
Note that these contain no integers of the form 2 (mod 4).
For the three rows with 5 even integers and 1 odd integer,
the odd integer must be 1 (mod 4) to make the sum of squares 1 (mod 4).
To make the magic sum 3 (mod 4), the magic series for each
of these three rows must be taken from
(0,0,0,0,1,2) (mod 4)
(0,0,1,2,2,2) (mod 4)
(1,2,2,2,2,2) (mod 4)
Note that these contain no integers of the form 3 (mod 4).
By similar reasoning, the six columns of the bimagic square must also
consist of these same magic series.
A column must have one entry in common with each row,
so a column with the magic series (0,3,3,3,3,3) must have
a 3 in common with five different rows. But there are only three rows
having a 3. So this magic series cannot be used for a column.
By similar reasoning, the magic series (0,3,3,3,3,3) and (1,2,2,2,2,2)
cannot be used in any row or column.
There are nine integers of the form 3 (mod 4) in the range 1-36
which must all be contained within three rows.
With only the magic series (0,1,1,1,1,3) and (0,1,1,3,3,3) available,
the only way to get all nine is to use (0,1,1,3,3,3) three times.
There are nine integers of the form 2 (mod 4) in the range 1-36,
so the other three rows must use (0,0,1,2,2,2) three times.
Thus, the magic series for the six rows must be the following.
(0,0,1,2,2,2) (mod 4)
(0,0,1,2,2,2) (mod 4)
(0,0,1,2,2,2) (mod 4)
(0,1,1,3,3,3) (mod 4)
(0,1,1,3,3,3) (mod 4)
(0,1,1,3,3,3) (mod 4)
The six columns must also use this same combination of magic series.
In particular, (0,0,1,2,2,2) must be used in three columns
which consist of nine 2's, six 0's, and three 1's.
The nine 2's can only be in common with entries from the first three rows,
thus the six 0's must be in common with entries from the last three rows.
But there are only a total of three 0's in the last three rows.
Therefore, a simultaneous row/column arrangement doesn't exist.
7x7 Normal Bimagic Impossibility Theorem
A 7x7 bimagic square of distinct integers in the range 1...49 is impossible.
[INCOMPLETE PROOF ATTEMPTS]
Proof Attempt 1 - Modulo 4 Approach
The magic sum is 175 which is 3 (mod 4) and 7 (mod 8).
The magic sum of squares is 5775 which is 3 (mod 4), 7 (mod 8).
Since the magic sums are odd, a magic series must contain an odd
number of odd integers: 1, 3, 5, or 7.
But the sum of squares of 1 or 5 odd integers is not 3 (mod 4), as required.
So a magic series must contain 3 or 7 odd integers.
Let A be the number of rows with 3 odd integers.
Let B be the number of rows with 7 odd integers.
Since there are 7 rows,
A + B = 7.
Since there are 25 odd integers in the range 1-49,
3A + 7B = 25.
Subtracting three times the first equation from the second,
4B = 4 or B = 1.
The only solution is A = 6, B = 1, so there must be six rows with 3 odd integers
and 4 even integers combined with one row of all odd integers.
For the six rows with 3 odd integers and 4 even integers,
an odd number of even integers must be 2 (mod 4)
in order for the sum of squares to be 7 (mod 8).
Since there are 12 integers of 0 (mod 4) and 12 integers of 2 (mod 4)
in the range 1...49, three of the six rows must contain even integers
(0,0,0,2) (mod 4) and the other three (0,2,2,2) (mod 4).
To make the magic sum 3 (mod 4),
the three odd integers must be (3,3,3) or (1,1,3).
So three rows must contain (0,0,0,2), (3,3,3) and/or (0,0,0,2), (1,1,3) (mod 4)
and three rows must contain (0,2,2,2),(3,3,3) and/or (0,2,2,2), (1,1,3) (mod 4).
1 (mod 4) 1,9,1,9 (mod 16)
3 (mod 4) 9,1,9,1 (mod 16)
For the row of all odd integers, it can't be
(1,1,1,1,1,1,1) (mod 4) or
(1,1,1,3,3,3,3) (mod 4)
[need explanation]
So this row must contain
(1,1,1,1,1,3,3) (mod 4) or
(1,3,3,3,3,3,3) (mod 4).
There are 13 integers of 1 (mod 4) and 12 integers of 3 (mod 4)
in the range 1...49.
So the seven rows must consist of the following odd-number combinations.
2 rows of (3,3,3), 4 rows of (1,1,3), 1 row (1,1,1,1,1,3,3)
2 rows of (3,3,3), 6 rows of (1,1,3), 1 row (1,3,3,3,3,3,3)
There are seven possible magic series.
a (0,0,0,1,1,2,3) (mod 4)
b (0,0,0,2,3,3,3) (mod 4)
c (0,1,1,2,2,2,3) (mod 4)
d (0,2,2,2,3,3,3) (mod 4)
e (1,1,1,1,1,3,3) (mod 4)
f (1,3,3,3,3,3,3) (mod 4)
a + b + c + d + e + f = 7 [rows]
3a + 3b + c + d = 12 [integers of 0 (mod 4)]
2a + 2c + 5e + f = 13 [integers of 1 (mod 4)]
a + b + 3c + 3d = 12 [integers of 2 (mod 4)]
a + 3b + c + 3d + 2e + 6f = 12 [integers of 3 (mod 4)]
There are three solutions to this system.
a b c d e f
-----------
3 0 3 0 0 1
3 0 1 2 1 0
1 2 3 0 1 0
Here are the three magic series combinations.
A B C
0 0 0 2 1 1 3 0 0 0 2 1 1 3 0 0 0 2 1 1 3
0 0 0 2 1 1 3 0 0 0 2 1 1 3 0 0 0 2 3 3 3
0 0 0 2 1 1 3 0 0 0 2 1 1 3 0 0 0 2 3 3 3
0 2 2 2 1 1 3 0 2 2 2 1 1 3 0 2 2 2 1 1 3
0 2 2 2 1 1 3 0 2 2 2 3 3 3 0 2 2 2 1 1 3
0 2 2 2 1 1 3 0 2 2 2 3 3 3 0 2 2 2 1 1 3
1 3 3 3 3 3 3 1 1 1 1 1 3 3 1 1 1 1 1 3 3
The columns must also be composed of one of these three
magic series combinations.
If let x = 1 or 3, we can represent all three combinations as follows
for both rows and columns.
x x x 0 0 0 2
x x x 0 0 0 2
x 0 0 x x 0 2
x 2 2 x x 2 0
x 2 2 2 0 x x
x 2 2 0 2 x x
x x x x x x x
Proof Attempt 2 - Modulo 3 Approach
175 is 1 (mod 3), 1 (mod 6), 4 (mod 9), 7 (mod 12).
5775 is 0 (mod 3), 3 (mod 6), 6 (mod 9), 3 (mod 12).
Sum of squares is 0 (mod 3), so the number of integers
that are not 0 (mod 3) must be a multiple of 3,
but not 0 since 175 is 1 (mod 3).
So there must be 3 or 6 integers that are 1 or 2 (mod 3).
3 integers 1 or 2 (mod 3) that sum to 1 (mod 3).
can only be (1,1,2).
6 integers 1 or 2 (mod 3) that sum to 1 (mod 3) are
(1,1,1,1,1,2) and (1,1,2,2,2,2).
x (0,0,0,0,1,1,2) (mod 3)
y (0,1,1,2,2,2,2) (mod 3)
z (0,1,1,1,1,1,2) (mod 3)
x + y + z = 7 [rows]
4x + y + z = 16 [of 0 (mod 3)]
2x + 2y + 5z = 17 [of 1 (mod 3)]
x + 4y + z = 16 [of 2 (mod 3)]
From the first and second equations,
3x = 9 or x = 3.
From the first and third equations,
3z = 3 or z = 1.
From the first and fourth equations,
3y = 9 or y = 3.
So the only solution is
x = 3
y = 3
z = 1
(0,0,0,0,1,1,2) (mod 3)
(0,0,0,0,1,1,2) (mod 3)
(0,0,0,0,1,1,2) (mod 3)
(0,1,1,2,2,2,2) (mod 3)
(0,1,1,2,2,2,2) (mod 3)
(0,1,1,2,2,2,2) (mod 3)
(0,1,1,1,1,1,2) (mod 3)