Bimagic cubes of order 6 are impossible
by Walter Trump

As already proved, all the bimagic series of order 6 use:

• 5 odd numbers + 1 even number,
• or 1 odd number + 5 even numbers.

Assume that a bimagic cube of order 6 exists. Consider any x-y-plane 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 aR of rows with 1 even number number bR 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 aR+ bR = 6, and E = aR + 5bR

The number aC 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:     aC = aR

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 aR >= 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 numbersin different columns Odd numbersin the same column e e e e e e e e e e e e e e e e e e e e => aC >= 4 => aC = 5

We obtain aC = 4, 5 or 6 meaning also that aR = 4, 5 or 6.
So aR = 2 (E = 22) and aR = 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 aC = 5 or 6, and also aR = 5 or 6.
That means that aR = 4 (E = 14) is also impossible.

Combinations of 6 planes

Say plane z contains Ez even numbers, with z = 1, 2, ..., 6.
The total amount of even numbers in an order-6 cube equals 63 / 2 = 216 / 2 = 108:

• E1 + E2 + E3 + E4 + E5 + E6 = 108.

As permutations of the planes do not change the amount of even numbers in the pillars we can assume:

• E1 >= E2 >= E3 >= E4 >= E5 >= E6.

We have to solve the following system of 2 equations, k, l, m, n being positive integers:

• 30k + 26l + 10m + 6n = 108
• k + l + m + n = 6

Only 4 solutions (k, l, m, n):

• s1      (0, 3, 3, 0)    26 + 26 + 26 + 10 + 10 + 10 = 108
• s2      (1, 2, 2, 1)    30 + 26 + 26 + 10 + 10 + 6 = 108
• s3      (2, 1, 1, 2)    30 + 30 + 26 + 10 + 6 + 6 = 108
• s4      (3, 0, 0, 3)    30 + 30 + 30 + 6 + 6 + 6 = 108

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 TOTAL3 planes 21 pillarswith 3 even numbers 15 pillarswith 1 even number

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 15-i pillars with one even number.
But that's not possible because too many even numbers would occur:

• (21 + i)*5 + (15 - i)*1 = 120 + 4i >= 120 > 108

So s1 is impossible.

>>> About s2, s3, s4.

If we replace any of our 3 planes of s1 with 26 even numbers by another plane with 30 even numbers, then the total amount of even numbers would be greater than with s1, and of course greater than 108.

So s2, s3, s4 are impossible.

It is impossible to combine 6 planes, and so it is impossible to complete a bimagic order-6 cube.