Bimagic cubes of order 6 are impossible
by
Walter Trump
As already proved, all the bimagic series of order 6 use:
Assume that a bimagic cube of order 6 exists. Consider any xyplane of the cube. Such a plane consists of 6 rows, each with 1 or 5 even numbers. The following combinations of rows may occur in a plane:
number a_{R} of rows with 1 even number 
number b_{R} of rows with 5 even numbers 
number E of even numbers in the plane 
0 
6 
30 
1 
5 
26 
2 
4 
22 
3 
3 
18 
4 
2 
14 
5 
1 
10 
6 
0 
6 
with a_{R}+ b_{R} = 6, and E = a_{R} + 5b_{R}
The number a_{C} of columns with one even number must equal aR. Otherwise
we would obtain a different value for E by adding the even numbers of all 6
columns. Of course E is the same if we count the numbers by columns or by rows.
So: a_{C}
= a_{R}
E = 6, 10, 26, 30 are possible.
The following samples show possible arrangements of the even numbers in a plane.
E = 6 

E = 10 

E = 26 

E = 30 

e 






e 
e 
e 
e 
e 
e 






e 
e 
e 
e 
e 


e 




e 






e 
e 
e 
e 
e 
e 

e 
e 
e 
e 



e 



e 






e 
e 
e 
e 
e 
e 
e 

e 
e 
e 




e 


e 






e 
e 
e 
e 
e 
e 
e 
e 

e 
e 





e 

e 






e 
e 
e 
e 
e 
e 
e 
e 
e 

e 






e 
e 






e 
e 
e 
e 
e 
e 
e 
e 
e 
e 

E = 14, 18, 22 are impossible.
Consider a plane with a_{R} >= 2. Two rows with 5 even numbers cause at least 4 columns with more than one even number (and more than one even number means automatically 5 even numbers in these columns):
Odd numbers 

Odd numbers 

e 
e 
e 
e 
e 

e 
e 
e 
e 
e 


e 
e 
e 
e 

e 
e 
e 
e 
e 
e 






















































=> a_{C} >= 4 
=> a_{C} = 5 
We obtain a_{C} = 4, 5 or 6 meaning also that a_{R} = 4,
5 or 6.
So a_{R} = 2 (E = 22) and a_{R} = 3 (E = 18) are
impossible.
If we add a third row with 5 even numbers to the left figure above then we
get at least 5 columns with more than one even number (another 'e' must occur
in column 5 or 6). Hence a_{C} = 5 or 6, and also a_{R} = 5
or 6.
That means that a_{R} = 4 (E = 14) is also impossible.
Combinations of 6 planes
Say plane z contains E_{z} even numbers, with z = 1, 2, ..., 6.
The total amount of even numbers in an order6 cube equals 6^{3} / 2 = 216 /
2 = 108:
As permutations of the planes do not change the amount of even numbers in the pillars we can assume:
We have to solve the following system of 2 equations, k, l, m, n being positive integers:
Only 4 solutions (k, l, m, n):
>>> About s_{1}.
3 planes with 26 even numbers each. We can obtain a maximum of 15 pillars with one (or none) even number:
plane 1 
26 even 
10 odd 

plane 2 
21 even 
5 odd 
5 even 
5 odd 
plane 3 
21 even 
10 odd 
5 even 

TOTAL 
21 pillars 
15 pillars 
It means that our 3 planes have 3 even numbers in at least 21 pillars.
So,
with 0 <= i <= 15, the whole bimagic cube with 6 planes must
contain 21+i pillars with 5 even numbers, and additionally 15i pillars with
one even number.
But that's not possible because too many even numbers would
occur:
So s_{1} is impossible.
>>> About s_{2}, s_{3}, s_{4}.
If we replace any of our 3 planes of s_{1} with 26 even numbers by another plane with 30 even numbers, then the total amount of even numbers would be greater than with s_{1}, and of course greater than 108.
So s_{2}, s_{3}, s_{4} are impossible.
It is impossible to combine 6 planes, and so it is impossible to complete a bimagic order6 cube.
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