# Perfect magic cube

In mathematics, a perfect magic cube is a magic cube in which not only the columns, rows, pillars and main space diagonals, but also the cross section diagonals sum up to the cube's magic constant.[1][2][3]

Perfect magic cubes of order one are trivial; cubes of orders two to four can be proven not to exist,[4] and cubes of orders five and six were first discovered by Walter Trump and Christian Boyer on November 13 and September 1, 2003, respectively.[5] A perfect magic cube of order seven was given by A. H. Frost in 1866, and on March 11, 1875, an article was published in the Cincinnati Commercial newspaper on the discovery of a perfect magic cube of order 8 by Gustavus Frankenstein. Perfect magic cubes of orders nine and eleven have also been constructed. The first perfect cube of order 10 has been constructed in 1988. (Li Wen, China)[6]

## An alternative definition

In recent years, an alternative definition for the perfect magic cube was proposed by John R. Hendricks. It is based on the fact that a pandiagonal magic square has traditionally been called 'perfect', because all possible lines sum correctly. This is not the case with the above definition for the cube. See Nasik magic hypercube for an unambiguous alternative term.[7]

This same reasoning may be applied to hypercubes of any dimension. Simply stated; if all possible lines of m cells (m = order) sum correctly, the hypercube is perfect. All lower dimension hypercubes contained in this hypercube will then also be perfect. This is not the case with the original definition, which does not require that the planar and diagonal squares be a pandiagonal magic cube.

The original definition is applicable only to magic cubes, not tesseracts, dimension 5 cubes, etc.

Example: A perfect magic cube of order 8 has 244 correct lines by the old definition, but 832 correct lines by this new definition.

Order 8 is the smallest possible perfect magic cube. None can exist for double odd orders.

Gabriel Arnoux constructed an order 17 perfect magic cube in 1887. F.A.P.Barnard published order 8 and order 11 perfect cubes in 1888.[6]

By the modern (Hendricks) definition, there are actually six classes of magic cube; simple magic cube, pantriagonal magic cube, diagonal magic cube, pantriagonal diagonal magic cube, pandiagonal magic cube, and perfect magic cube.[7]

Nasik; A. H. Frost (1866) referred to all but the simple magic cube as Nasik! C. Planck (1905) redefined Nasik to mean magic hypercubes of any order or dimension in which all possible lines summed correctly.

i.e. Nasik is an alternative, and unambiguous term for the perfect class of any dimension of magic hypercube.

## First known Perfect Pandiagonaal Semi-magisch Magic Cube

Thomas Krijgsman, 1982 March, 21 number 5 / link: http://www.pythagoras.nu/pyth/nummer.php?id=253

-Row 1 (4x4)- - - - - -Row 2 (4x4)- - - - - -Row 3 (4x4)- - - - - - Row 4 (4x4)

[32]-[05]-[52]-[41] = [10]-[35]-[22]-[63] = [49]-[28]-[45]-[08] = [39]-[62]-[11]-[18]

[03]-[42]-[31]-[54] = [37]-[64]-[09]-[20] = [30]-[07]-[50]-[43] = [60]-[17]-[40]-[13]

[61]-[24]-[33]-[12] = [27]-[02]-[55]-[46] = [36]-[57]-[16]-[21] = [06]-[47]-[26]-[51]

[34]-[59]-[14]-[23] = [56]-[29]-[44]-[01] = [15]-[38]-[19]-[58] = [25]-[04]-[53]-[48]

3D solution in my head, fill the numbers on graph paper, that all.

Walter Trump and Christian Boyer, 2003-11-13

This cube consists of all numbers from 1 to 125. The sum of the 5 numbers in each of the 25 rows, 25 columns, 25 pillars, 30 diagonals and 4 triagonals (space diagonals) equals the magic constant 315.

1° level 2° level
${\displaystyle {\begin{bmatrix}25&16&80&104&90\\115&98&4&1&97\\42&111&85&2&75\\66&72&27&102&48\\67&18&119&106&5\\\end{bmatrix}}}$ ${\displaystyle {\begin{bmatrix}91&77&71&6&70\\52&64&117&69&13\\30&118&21&123&23\\26&39&92&44&114\\116&17&14&73&95\\\end{bmatrix}}}$
3° level 4° level
${\displaystyle {\begin{bmatrix}47&61&45&76&86\\107&43&38&33&94\\89&68&(63)&58&37\\32&93&88&83&19\\40&50&81&65&79\\\end{bmatrix}}}$ ${\displaystyle {\begin{bmatrix}31&53&112&109&10\\12&82&34&87&100\\103&3&105&8&96\\113&57&9&62&74\\56&120&55&49&35\\\end{bmatrix}}}$
5° level
${\displaystyle {\begin{bmatrix}121&108&7&20&59\\29&28&122&125&11\\51&15&41&124&84\\78&54&99&24&60\\36&110&46&22&101\\\end{bmatrix}}}$