=== Order $2$ === Consider a layer of the order $2$ perfect magic cube: :$\begin{array}{|c|c|} \hline  a & b \\ \hline  c & d \\ \hline \end{array}$ Then we must have $a + b = a + c$. So $b = c$, so they are not distinct, so this array cannot be a layer of a perfect magic cube. {{qed|lemma}} === Order $3$ === Let $C$ be the magic constant of an order $3$ perfect magic cube. Consider a layer of the perfect magic cube: :$\begin{array}{|c|c|c|} \hline  a & b & c \\ \hline  d & e & f \\ \hline  g & h & i \\ \hline \end{array}$ The sum of all elements in this layer is $3 C$. We also have: {{begin-eqn}} {{eqn | l = 4 C       | r = \paren {a + e + i} + \paren {b + e + h} + \paren {c + e + g} + \paren {d + e + f}       | c = These are all lines passing through the center }} {{eqn | r = \paren {a + b + c + d + e + f + g + h + i} + 3 e       | c = Center counted $4$ times }} {{eqn | r = 3 C + 3 e }} {{end-eqn}} This gives $e = \dfrac C 3$. However these equations applies for every layer of the perfect magic cube. So the center of every layer is equal to $\dfrac C 3$. They are not distinct, so there cannot exist a order $3$ perfect magic cube. {{qed|lemma}} === Order $4$ === Let $C$ be the magic constant of an order $4$ perfect magic cube. Consider a layer of the perfect magic cube: :$\begin{array}{|c|c|c|c|} \hline  a & b & c & d \\ \hline  e & f & g & h \\ \hline  i & j & k & l \\ \hline  m & n & o & p \\ \hline \end{array}$ The sum of all elements in this layer is $4 C$. We also have: {{begin-eqn}} {{eqn | l = 6 C       | r = \paren {a + b + c + d} + \paren {d + h + l + p} + \paren {m + n + o + p} + \paren {a + e + i + m} + \paren {a + f + k + p} + \paren {d + g + j + m}       | c = These are all the diagonals and edges }} {{eqn | r = \paren {a + b + c + d + e + f + g + h + i + j + k + l + m + n + o + p} + 2 \paren {a + d + m + p}       | c = Corners counted thrice }} {{eqn | r = 4 C + 2 \paren {a + d + m + p} }} {{end-eqn}} So the sum of the corners is equal to $C$. These equations apply to all layers, including the diagonal layers. Now consider the full perfect magic cube, but only the corners: :$\begin{array}{|c|c|c|c|} \hline  a & \phantom b & \phantom b & c \\ \hline   &  &  &  \\ \hline   &  &  &  \\ \hline  b &  &  & d \\ \hline \end{array} \begin{array}{|c|c|c|c|} \hline  \phantom b & \phantom b & \phantom b & \phantom b \\ \hline   &  &  &  \\ \hline   &  &  &  \\ \hline   &  &  &  \\ \hline \end{array} \begin{array}{|c|c|c|c|} \hline  \phantom b & \phantom b & \phantom b & \phantom b \\ \hline   &  &  &  \\ \hline   &  &  &  \\ \hline   &  &  &  \\ \hline \end{array} \begin{array}{|c|c|c|c|} \hline  e & \phantom b & \phantom b & * \\ \hline   &  &  &  \\ \hline   &  &  &  \\ \hline  f &  &  & * \\ \hline \end{array}$ From our discussion above, we have: :$\paren {a + b + c + d} = \paren {a + b + e + f} = \paren {c + d + e + f} = C$ Hence: {{begin-eqn}} {{eqn | l = a + b       | r = \frac 1 2 \paren {\paren {a + b + c + d} + \paren {a + b + e + f} - \paren {c + d + e + f} } }} {{eqn | r = \frac 1 2 \paren {C + C - C} }} {{eqn | r = \frac C 2 }} {{end-eqn}} Therefore any two adjacent corners sum up to $\dfrac C 2$. However this implies $a + c = \dfrac C 2 = a + b$. So $b = c$, so they are not distinct, so there cannot exist a order $4$ perfect magic cube. {{qed}}