Magic Square of Order 3 is Unique

Theorem
Up to rotations and reflections, the magic square of order $3$ is unique:

Proof
Let $M_3$ denote the magic square of order $3$.

Each row, column and diagonal of $M_3$ must be a different set of $3$ elements of $\N_9$, where $\N_9$ denotes the set $\set {1, 2, 3, 4, 5, 6, 7, 8, 9}$.

The sets of $3$ elements of $\N_9$ adding to $15$ can be stated:


 * $\set {1, 5, 9}, \set {1, 6, 8}$


 * $\set {2, 4, 9}, \set {2, 5, 8}, \set {2, 6, 7}$


 * $\set {3, 4, 8}, \set {3, 5, 7}$


 * $\set {4, 5, 6}$

The number of rows, columns and diagonals of $M_3$ passing through each cell of $M_3$ depends upon where in $M_3$ that cell is positioned.


 * The corner cells are each on one row, one column and one diagonal


 * The edge cells are each on one row and one column


 * The center cells is on one row, one column and both diagonals.

Thus the center cell has to contain an integer which is in at least $4$ of the $3$-element subsets of $\N^*_9$ above.

There is only one such element, which is $5$.

Thus $5$ goes in the center square of $M_3$.

By a similar analysis, it is seen that:
 * the corner cells contain $2, 4, 6, 8$
 * the edge cells contain $1, 3, 7, 9$.

, let $2$ be placed in the upper left corner of $M_3$.

There is no $3$-element subset of $\N^*_9$ adding to $15$ containing both $1$ and $2$.

It follows that $1$ cannot be placed on either the upper edge or left hand edge.

Thus $1$ must go into the right hand edge or the bottom edge.

, let $1$ be placed on the right hand edge.

Up to rotations and reflections, the placements of these three elements is unique.

Hence $M_3$ so far looks like this:
 * $\begin{array}{|c|c|c|}

\hline 2 &   &   \\ \hline  &  5 & 1 \\ \hline  &    &   \\ \hline \end{array}$

The population of the remaining cells is forced:
 * $\begin{array}{|c|c|c|}

\hline 2 &  &   \\ \hline 9 & 5 & 1 \\ \hline  &   & 8 \\ \hline \end{array}$


 * $\begin{array}{|c|c|c|}

\hline 2 &  & 6 \\ \hline 9 & 5 & 1 \\ \hline 4 &  & 8 \\ \hline \end{array}$


 * $\begin{array}{|c|c|c|}

\hline 2 & 7 & 6 \\ \hline 9 & 5 & 1 \\ \hline 4 & 3 & 8 \\ \hline \end{array}$