# Smallest Magic Square is of Order 3

## 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.

$\blacksquare$