Constructing 4x4x4 magic cube requires solving 76 linear equations for 64 unknowns. On top of that, we are looking for an all integer solution. The number of equations is given by 6 equations per 4x4 layer, times 4 layers per dimension, times 3 dimensions, for a total of 6x4x3 = 72. Add 4 equations from the body diagonals for a total of 76. The 76 equations may be considered as 76 vectors in a 64-dimensional space. Maximum of 64 can be linarly independent. The corresponding homogenous system (right sides of all equations equal to zero) obviously has the trivial solution (x1 = x2 = ... = x64 = 0) and the non-homogenous system (all right sides equal to the magic number M) also has the trivial solution (x1 = x2 = ... = x64 = M/4). Therefore, no linearly dependent equations have incommpatible right sides. If we find 64 linearly independent vectors (equations), the trivial solution is the only solution and it obviously does not satisfy the magic cube conditions (all numbers must be mutually different). If we cannot find 64 linearly independent vectors (equations), other (non-trivial) solutions are possible. Number the small 1x1x1 cubes on any 4x4x4 cube face (for example, the top horizontal face) as follows: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Write down the 8 vectors a1, a2, ..., a8 corresponding to the 8 possible equations for the left-right and forward-backward rows, including the right sides (the magic number M): 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 M -> a1 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 M -> a2 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 M -> a3 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 M -> a4 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 M -> a5 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 M -> a6 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 M -> a7 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 M -> a8 The remaining 48 coordinates of these vectors are all zeros. It is obvious that these 8 vectors are linearly dependent, because: a1 + a2 + a3 + a4 - (a5 + a6 + a7 + a8) = 0 So lets kick out the vector a5. 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 M -> a1 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 M -> a2 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 M -> a3 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 M -> a4 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 M -> a6 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 M -> a7 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 M -> a8 The remaining 7 vectors are linearly independent because of the triangular shape of the matrix (we may exchange the 6th - 9th and 7th - 13th columns to see the triangular shape). Add the 2 possible diagonals: 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 M -> a1 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 M -> a2 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 M -> a3 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 M -> a4 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 M -> a6 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 M -> a7 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 M -> a8 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 M -> a9 0 0 0 1 0 0 1 0 0 1 0 0 1 0 0 0 M -> a10 Transform the matrix to the triangular shape: 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 M -> a1 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 M -> a2 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 M -> a3 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 M -> a4 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 M -> a6 0 0 0 0 0 2 1 2 -1 0 1 0 -1 0 0 -1 0 -> a6-(a1-a9) 0 0 0 0 0 0 1 -1 0 1 0 -1 1 0 0 -1 M -> a10-a4 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 M -> a7 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 M -> a8 Again, we may exchange the 6th - 9th and 7th - 13th columns to see the triangular shape. We have 9 linearly independent vectors from the first (top) 4x4 layer that span a 9-dimensional subspace of the 64 dimensional space. We can get 9 linearly independent vectors from each of the three parallel horizontal 4x4 layers, for a total of 36 linearly independent vectors, because the four 9-dimensional subspaces are mutually orthogonal. Let's see if we can find 28 more vectors without causing any linear dependence. We have 16 columns, 16 vertical 4x4 layer diagonals, and 4 body diagonals of the 4x4x4 cube to work with. All these 36 remaining 64-dimensional vectors take one (and only one) non-zero coordinate from each of the four 9-dimensional subspaces. The remaining linearly independent vectors either have the 8th, 10th , 11th , 12th , 14th , 15th and/or 16th coordinates in at least one 9-dimensional subspace non-zero or we can reduce them to such vectors by separate linear combinations within the subspaces. Counting the allowed columns (7), face diagonals (12), and body diagonals (4) adds up to 23. Some of them might be linearly dependent, but it requires triangularization of a 21x23 matrix to find out how many (none, I did the triangularization). Therefore the non-trivial solutions do exist for the 4x4x4 magic cube.