Source: Original.
Henry Bottomley's solutions are quite comprehensive:
The solutions are: 1.1 5; 1.2 5; 1.3 5 2.1 4; 2.2 7; 2.3 7; 2.4 7 Bonus: 1.1 9; 1.2 8 2.1 impossible; 2.2 9; 2.3 8 Examples: 1. For 3x3, 4x4 or the plane as a whole, it is obvious that at least 5 colours are needed as each square must touch four other distinct colours. Since it is possible to colour the plane using five colours, this is the solution to all three questions. The following colouring involves each row being offset a couple of places from the preceedeing row: ............ .ABCDEABCDE. .CDEABCDEAB. .EABCDEABCD. .BCDEABCDEA. .DEABCDEABC. .ABCDEABCDE. .CDEABCDEAB. ............ etc. so A always has B,C,E and D around it, all in the same relative positions and similarly for the other colours. So possible solutions for 3x3 and 4x4 are ABC ABCD CDE CDEA EAB EABC BCDE while 2x2 requires exactly 4 colours for the 4 squares, e.g. AB CD 2. For 3x3x3, 4x4x4 or 3-space as a whole, it is obvious that at least 7 colours are needed as each cube must touch six other distinct colours. Since it is possible to colour the Since it is possible to colour 3-space using seven colours, this is the solution to the last three questions. The following colouring involves each row being offset two places from the preceeding row, and each plane being offset three places: ............ ............ ............ ............ .ABCDEFGABC. .DEFGABCDEF. .GABCDEFGAB. .CDEFGABCDE. .CDEFGABCDE. .FGABCDEFGA. .BCDEFGABCD. .EFGABCDEFG. .EFGABCDEFG. .ABCDEFGABC. .DEFGABCDEF. .GABCDEFGAB. .GABCDEFGAB. .CDEFGABCDE. .FGABCDEFGA. .BCDEFGABCD. .BCDEFGABCD. .EFGABCDEFG. .ABCDEFGABC. .DEFGABCDEF. .DEFGABCDEF. .GABCDEFGAB. .CDEFGABCDE. .FGABCDEFGA. .FGABCDEFGA. .BCDEFGABCD. .EFGABCDEFG. .ABCDEFGABC. .ABCDEFGABC. .DEFGABCDEF. .GABCDEFGAB. .CDEFGABCDE. ............ ............ ............ ............ etc. so A always has B,C,G and F around it and has E and D above and below, all in the same relative positions and similarly for the other colours. this gives a 3x3 solution such as ABC DEF GAB CDE FGA BCD EFG ABC DEF and a 4x4 solution such as ABCD DEFG GABC CDEF CDEF FGAB BCDE EFGA EFGA ABCD DEFG GABC GABC CDEF FGAB BCDE For 2x2x2 at least four colours are needed for the top plane (as in 2x2) and this is indeed a solution, e.g. AB DC CD BA Bonus. With the wrapping suggested, 2x2 and 2x2x2 are impossible since each square or cube touches another on more than one side (if this doesn't count then 4 are needed for 2x2 and 2x2x2 exactly as above, since every square or cube which touches another in the wrapped version - always twice - also touches it in the unwrapped version - once). For wrapped 3x3 nine colours are needed as every square is either adjacent or next to adjacent to every other, e.g. ABC DEF GHI So for wrapped 3x3x3 nine colours are needed for the top plane, but nine is also sufficent for all three planes, e.g. ABC IGH EFD DEF CAB HIG GHI FDE BCA For 4x4 at least eight colours are needed, as each coloured square prevents its four neighbours being adjacent to another square of that colour, and after two squares have been coloured the same there are no available squares for that colour. Eight is sufficient, e.g. ABCD EFGH BADC FEHG For 4x4x4 eight colours are required for the top plane as in 4x4, and eight is sufficient for all four planes, e.g. ABCD GHEF BADC HGFE EFGH CDAB FEGH DCBA BADC HGFE ABCD GHEF FEHG DCBA EFGH CDAB Lots of symmetry and knight's moves in these solutions Regards Henry Bottomley