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.