No Perfect Magic Cube of Order Less than 5 Exists

Theorem
Apart from the trivial order $1$ case, no perfect magic cube exists whose order is $4$ or less.

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.

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:

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.

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:

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:

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.