Smallest Magic Cube is of Order 3

From ProofWiki
Jump to navigation Jump to search

Theorem

Apart from the trivial order $1$ magic cube:

$\begin{array}{|c|} \hline 1 \\ \hline \end{array}$


the smallest magic cube is the order $3$ magic cube:

$\begin{array}{|c|c|c|} \hline 2 & 13 & 27 \\ \hline 22 & 9 & 11 \\ \hline 18 & 20 & 4 \\ \hline \end{array} \qquad \begin{array}{|c|c|c|} \hline 16 & 21 & 5 \\ \hline 3 & 14 & 25 \\ \hline 23 & 7 & 12 \\ \hline \end{array} \qquad \begin{array}{|c|c|c|} \hline 24 & 8 & 16 \\ \hline 17 & 19 & 6 \\ \hline 1 & 15 & 26 \\ \hline \end{array}$


Proof

Suppose there were an order $2$ magic cube.

Take one row of this magic cube.

From Magic Constant of Magic Cube, the row and column total is $9$.

Any row or column with a $1$ in it must therefore also have an $8$ in it.

But there are:

one row
one column

both of which have a $1$ in them.

Therefore the $8$ would need to go in $2$ distinct cells.

But $8$ appears in a magic cube exactly once.

Hence there can be no order $2$ magic cube.

$\blacksquare$