The smallest possible bimagic cube
See also the smallest possible bimagic square


The smallest bimagic cubes currently known are of order 16 constructed in January 2003. But is it possible to construct smaller bimagic cubes?


Order 3

Proof #1: There are only 4 different magic cubes of order 3.

None of them is bimagic, so a bimagic cube of order 3 cannot exist.

Proof #2: There are only 4 bimagic series of order 3 (S1=42, S2=770).

A bimagic cube of order 3 needs 3*3 + 4 = 31 different bimagic series.

A bimagic cube of order 3 cannot exist.


Order 4

There are only 8 bimagic series of order 4 (S1=130, S2=5 590)

A bimagic cube of order 4 needs 3*4 + 4 = 52 different bimagic series.

A bimagic cube of order 4 cannot exist.


Order 5

There are 272 bimagic series of order 5 (S1=315, S2=26 355)

A bimagic cube of order 5 needs 3*5 + 4 = 79 different magic series. 272 series may be sufficient.

The squares of even numbers being of the 4k form, and the squares of odd numbers being of the 4k+1 form, a sum of 5 squares can be of the 4k+3 form if and only if the series contains 3 odd numbers.

So all these series use 3 odd numbers + 2 even numbers.

This implies that any set of 25 bimagic series will use 75 odd numbers + 50 even numbers. So it is impossible to split all the numbers (from 1 to 125) in 25 bimagic series.

A bimagic cube of order 5 cannot exist.


Order 6

There are 25,270 bimagic series of order 6 (S1=651, S2=93 961)

The squares of even numbers being of the 4k form, and the squares of odd numbers being of the 4k+1 form, a sum of 6 squares can be of the 4k+1 form if and only if the series contains 1 or 5 odd numbers.

So all these series use:

Using this first fact, Walter Trump has proved that any plane of a bimagic cube of order 6 would contain 6, 10, 26 or 30 even numbers, and has finally proved that it is impossible to complete a bimagic cube with such planes. See his proof.

A bimagic cube of order 6 cannot exist.


Order 7

There are 5,152,529 bimagic series of order 7 (S1=1 204, S2=275 716)

The squares of even numbers being of the 4k form, and the squares of odd numbers being of the 4k+1 form, a sum of 7 squares can be of the 4k form if and only if the series contains 0 or 4 odd numbers.

If the series contains 0 odd number, then all the numbers are 4k or 4k+2.

  1. The sum S1 being of the 4k form, then such a series contains an even number of 4k+2 numbers.
  2. The sum S2 being of the 16k+4 form, the squares of 4k numbers being of the 16k form, the squares of 4k+2 numbers being of the 16k+4 form, then such a series contains 1 or 5 times 4k+2 numbers.

1) & 2) = impossible: a bimagic series of order 7 cannot contain 0 odd number.

So, ALL of the bimagic series of order 7 use 4 odd numbers + 3 even numbers. This implies that any set of 49 bimagic series will use 196 odd numbers and 147 even numbers. So it is impossible to split all the numbers (from 1 to 343) in 49 bimagic series.

A bimagic cube of order 7 cannot exist.


Order 8

There is a huge number (1.6*109) of bimagic series of order 8 (S1=2 052, S2=701 100)

The squares of even numbers being of the 4k form, and the squares of odd numbers being of the 4k+1 form, a sum of 8 squares can be of the 4k form if and only if the series contains 0 or 4 or 8 odd numbers.

If the series contains 0 odd numbers, then all the numbers are 4k or 4k+2.

  1. The sum S1 being of the 4k form, then such a series contains an even number of 4k+2 numbers.
  2. The sum S2 being of the 16k+12 form, the squares of 4k numbers being of the 16k form, the squares of 4k+2 numbers being of the 16k+4 form, then such a series contains 3 or 7 times 4k+2 numbers.

1) & 2) = impossible. A bimagic series of order 8 cannot contain 0 odd number.

By the same reasoning using numbers n-1 (from 0 to 511: S1=2 044, S2=697 004) instead of standard numbers n (from 1 to 512), a bimagic series of order 8 cannot contain 0 odd numbers n-1, so cannot contain 8 odd numbers n.

So, ALL the bimagic series of order 8 use 4 odd numbers + 4 even numbers.

Perhaps that a bimagic cube of order 8 may exist, using bimagic series with (4o+4e) ranks.

And perhaps that directly a TRImagic cube of order 8 may exist! There are 17,218 trimagic series of order 8. But a tetramagic cube of order 8 or less cannot exist: there is no tetramagic series from order 3 to 8.


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