Office Applications and Entertainment, Magic Cubes Exhibit VIIIa About the Author

 8a   Perfect Concentric Magic Cubes (6 x 6 x 6)      Center Cubes 8a-1 Lemma Order 4 Almost Perfect Magic Cubes are not suitable for the construction of order 6 Perfect Concentric Magic Cubes. 8a-2 Proof The application of order 4 Almost Perfect Magic Cubes as center cubes for order 6 Perfect Concentric Magic Cubes would require a border with corner pairs which are non symmetrical over the space diagonals, say:    a( 1) + a(216)+ d(1) = s6    a( 6) + a(211)+ d(2) = s6    a(31) + a(186)+ d(3) = s6    a(36) + a(181)+ d(4) = s6 Based on the above the defining equations of the edge of an order 6 Perfect Concentric Magic Cube can - after deduction - be written as:
 ```a( 5) = a(212) + a(213) + a(214) - a(6) - a(1) a( 25) = a(192) + a(198) + a(204) - a(31) - a(1) a( 30) = a(187) + a(193) + a(199) - a(36) - a(6) a( 35) = s6 - a(185) - a(36) - a(31) - a(181) - a(186) a( 67) = s6 - a(103) - a(139) - a(175) - a(31) - a(211) a( 72) = s6 - a(108) - a(144) - a(180) - a(36) - a(216) a(145) = s6 - a(180) - a(36) - a(1) - a(181) - a(216) a(150) = s6 - a(175) - a(31) - a(6) - a(186) - a(211) a(182) = s6 - a(183) - a(184) - a(185) - a(181) - a(186) a(187) = s6 - a(193) - a(199) - a(205) - a(181) - a(211) a(192) = s6 - a(198) - a(204) - a(210) - a(186) - a(216) a(212) = s6 - a(213) - a(214) - a(215) - a(211) - a(216) a(36) = s6 - a(181) - d(4) a(31) = - s6/3 - a(186) + d(4) a(6) = s6 - a(211) - d(4) a(1) = - s6/3 - a(216) + d(4) d(1) = 4*s6/3 - d(4) d(2) = d(4) d(3) = 4*s6/3 - d(4) d(4) = 2*s6/3 ```
 a(2) = s6/3 - a(212) a(3) = s6/3 - a(213) a(4) = s6/3 - a(214) a(7) = s6/3 - a(192) a(12) = s6/3 - a(187) a(13) = s6/3 - a(198) a(18) = s6/3 - a(193) a(19) = s6/3 - a(204) a(24) = s6/3 - a(199) a(32) = s6/3 - a(182) a(33) = s6/3 - a(183) a(34) = s6/3 - a(184) a(37) = s6/3 - a(72) a( 42) = s6/3 - a(67) a( 73) = s6/3 - a(108) a( 78) = s6/3 - a(103) a(109) = s6/3 - a(144) a(114) = s6/3 - a(139)
 which proofs that the space diagonals d(1), d(2), d(3) and d(4) of the order 4 center cube should be identical. Consequently order 4 Almost Perfect Cubes are not suitable as center cubes for order 6 Perfect Concentric Magic Cubes (q.e.d.).