Smallest Magic Square is of Order 3
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Theorem
Apart from the trivial order $1$ magic square:
- $\begin{array}{|c|} \hline 1 \\ \hline \end{array}$
the smallest magic square is the order $3$ magic square:
- $\begin{array}{|c|c|c|} \hline 2 & 7 & 6 \\ \hline 9 & 5 & 1 \\ \hline 4 & 3 & 8 \\ \hline \end{array}$
Proof
Suppose there were an order $2$ magic square.
The row and column total is $\dfrac {1 + 2 + 3 + 4} 2 = 5$.
Any row or column with a $1$ in it must therefore also have a $4$ in it.
But there are:
- one row
- one column
- one diagonal
all of which have a $1$ in them.
Therefore the $4$ would need to go in all $3$ cells.
But $4$ appears in a magic square exactly once.
Hence there can be no order $2$ magic square.
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $3$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $3$