What's known about magic cubes of order 4? An earlier question asked for a demonstration that there is no magic cube of order 4. The question was closed and deleted. I think it's worth having some information on magic cubes on m.se, so I'm reposting (a version) of the question, and posting an answer. 
A magic cube of order 4 is a $4\times4\times4$ array of the integers $1,2,\dots,64$ such that each row of cubes parallel to an edge, and also each body diagonal, has the same sum. Are there any? Are there any with interesting additional properties?
 A: Yes, there are a great many magic cubes of order 4. Here is one I have copied from Harvey Heinz' site: $$\matrix{13&36&20&61\cr51&30&46&3\cr50&31&47&2\cr16&33&17&64\cr}\qquad\matrix{56&25&41&8\cr10&39&23&58\cr11&38&22&59\cr53&28&44&5\cr}\qquad\matrix{60&21&37&12\cr6&43&27&54\cr7&42&26&55\cr57&24&40&9\cr}\qquad\matrix{1&48&32&49\cr63&18&34&15\cr62&19&35&14\cr4&45&29&52\cr}$$ Each $4\times4$ square is one layer of the cube, and the layers are given in the order in which they are to be stacked. This cube was published by Hermann Schubert in 1898; Heinz knows of no earlier published order 4 magic cube. It is an "associated" cube, meaning numbers opposite each other (with respect to the center of the cube) add up to 65. 
Heinz has several more cubes, some with additional properties (e.g., there's one where every $2\times2$ square sums to 130), and links to the literature and to pages with more order 4 cubes. 
The original question asked for a demonstration that there are no order 4 magic cubes. Perhaps what was wanted was a demonstration that there are no order 4 magic cubes with the additional property that all the planes parallel to a face are magic squares. Some authors call these "perfect magic cubes" (but not Heinz, who reserves that term for cubes with even more properties). Schroeppel proved there is no perfect magic cube of order 4 (although he didn't use the word, "perfect"; he just called it a magic cube). I copy his proof: 
Proof: Let $K = 130$ be the sum of a row.
Lemma 1: In a magic square of order four, the sum of the corners is $K$.
Proof: Add together each edge of the square and the two diagonals. This covers the square entirely, and each corner twice again. This adds to $6 K$, so twice the corner sum is $2 K$.
Lemma 2: In a magic cube of order 4, the sum of any two corners connected by an edge of the cube is $K/2$.
Proof: Call the corners $a$ and $b$. Let $c$, $d$ and $e$, $f$ be the corners of any two edges of the cube parallel to $ab$. Then $abcd$, $abef$, and $cdef$ are all the corners of magic squares. So $(a+b+c+d) + (a+b+e+f) + (c+d+e+f) = 3K$; $a+b+c+d+e+f = 3K/2$; $a+b = K/2$.
Proof of magic cube impossibility: Consider a corner $x$. There are three corners connected by an edge to $x$.
Each must have value $K/2 - x$. QED
Trump and Boyer give an example of a perfect magic cube of order 5. 
