No Perfect Magic Cube of Order Less than 5 Exists

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Theorem

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


Proof

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.

$\Box$


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:

\(\ds 4 C\) \(=\) \(\ds \paren {a + e + i} + \paren {b + e + h} + \paren {c + e + g} + \paren {d + e + f}\) These are all lines passing through the center
\(\ds \) \(=\) \(\ds \paren {a + b + c + d + e + f + g + h + i} + 3 e\) Center counted $4$ times
\(\ds \) \(=\) \(\ds 3 C + 3 e\)

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.

$\Box$


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:

\(\ds 6 C\) \(=\) \(\ds \paren {a + b + c + d} + \paren {d + h + l + p} + \paren {m + n + o + p} + \paren {a + e + i + m} + \paren {a + f + k + p} + \paren {d + g + j + m}\) These are all the diagonals and edges
\(\ds \) \(=\) \(\ds \paren {a + b + c + d + e + f + g + h + i + j + k + l + m + n + o + p} + 2 \paren {a + d + m + p}\) Corners counted thrice
\(\ds \) \(=\) \(\ds 4 C + 2 \paren {a + d + m + p}\)

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:

\(\ds a + b\) \(=\) \(\ds \frac 1 2 \paren {\paren {a + b + c + d} + \paren {a + b + e + f} - \paren {c + d + e + f} }\)
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {C + C - C}\)
\(\ds \) \(=\) \(\ds \frac C 2\)

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.

$\blacksquare$


Sources

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