**The smallest possible bimagic cube**

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

- 3rd-order bimagic cube?
- 4th-order bimagic cube?
- 5th-order bimagic cube?
- 6th-order bimagic cube?
- 7th-order bimagic cube?
- 8th-order bimagic cube?

**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).

- 3 19 20
- 4 15 23
- 5 13 24
- 8 9 25

A bimagic cube of order 3 needs 3*3² + 4 = 31 different bimagic series.

A bimagic cube of order 3 cannot exist.

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

- 1 38 44 47
- 1 39 42 48
- 3 36 37 54
- 6 28 39 57
- 8 26 37 59
- 11 28 29 62
- 17 23 26 64
- 18 21 27 64

A bimagic cube of order 4 needs 3*4² + 4 = 52 different bimagic series.

A bimagic cube of order 4 cannot exist.

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

A bimagic cube of order 5 needs 3*5² + 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.

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:

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

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.

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.

- The sum S1 being of the 4k form, then such a series contains an even number of 4k+2 numbers.
- 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.

There is a huge number (1.6*10^{9}) 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.

- The sum S1 being of the 4k form, then such a series contains an even number of 4k+2 numbers.
- 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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