Magic squares and number arrangements

A magic square is a square grid filled with numbers so that every row, every column, and both diagonals add up to the same total , called the magic sum.

Idea

The smallest interesting magic square is order $3$, using numbers $1$ to $9$ once each:

2 7 6
9 5 1
4 3 8

Every row, every column, and each diagonal sums to $15$. (Quick check: $2+7+6 = 15$; $2+9+4=15$; $2+5+8=15$, etc.)

The magic sum equals $\frac{n(n^2+1)}{2}$ for an $n \times n$ square filled with $1$ to $n^2$. So for $n=3$: $\frac{3 \times 10}{2} = 15$. For $n=4$: $\frac{4 \times 17}{2} = 34$.

Symmetries. A magic square can be reflected and rotated , giving up to $8$ "different-looking" arrangements of the same numbers. So in some sense there is only one essential $3\times 3$ magic square using $1$–$9$.

Hardest task. Filling a magic square from scratch. For $n=3$, one trick: place $5$ in the middle (it must be, since it's the average), then balance pairs around it.

Magic squares have appeared in Indian, Chinese, Arabic and European traditions for over $2000$ years , both as puzzles and as charms. Mathematicians study them because they connect arithmetic, algebra and combinatorics.

Worked examples

Example 1. Find the missing number in this partly-filled magic square (sum $= 15$):

4 9 ?
3 5 7
? 1 6

Row 1: $4 + 9 + ? = 15 \Rightarrow ? = 2$. Row 3: $? + 1 + 6 = 15 \Rightarrow ? = 8$. Check column 1: $4+3+8=15$ ✓.

Example 2. Why must the centre of a $3\times 3$ magic square (using $1$–$9$) be $5$?

Add all four lines through the centre: $2$ diagonals + middle row + middle column = $4 \times 15 = 60$. Each of these counts the centre $4$ times and every other cell exactly once (the centre is on all four lines; other cells are on exactly one). Total of all $9$ cells once: $1+\dots+9 = 45$. So $4c + 45 - c = 60$? Let me re-do: $60 = 3 \cdot c + 45$ (because each line contains the centre and two others, and the four lines cover the centre $4$ times and the other $8$ cells exactly... actually a clean version: middle row + middle column + diag1 + diag2 = $4 \times 15 = 60$, and this counts the centre $4$ times plus the other $8$ cells once = $4c + (45 - c)$? That's wrong because every non-centre cell is on exactly one of the four lines... Actually middle row has $3$ cells (centre and $2$ others), same for middle column, and each diagonal has $3$ cells too. Sum of cells in these lines (with multiplicities) is $4 \cdot 3 = 12$ cells worth. The centre is in all $4$ lines (counted $4$ times); the $4$ corners each lie on $2$ lines (counted twice); the $4$ edge-midpoints each lie on $1$ line. So $4c + 2(\text{corner sum}) + (\text{edge-mid sum}) = 60$. With $\text{corner sum} + \text{edge-mid sum} = 45 - c$, eliminating gives $3c + (45 - c) + \text{corner sum} = 60$, i.e., corner sum $= 15 + 2c - 45 = 2c - 30$ ... too messy. Simpler reasoning: middle column + middle row sum $= 30$; this counts centre twice and the other $4$ middle-cells once. So $2c + (\text{sum of edge-mids}) = 30$. By symmetry, the edge-mids sum to $20$ (because the four corners sum to $20$ too, and corners + edge-mids $= 45 - c = 40$). So $2c + 20 = 30 \Rightarrow c = 5$. ✓

Example 3. What is the magic sum for a $4\times 4$ magic square using $1$ to $16$?

$\frac{4(4^2+1)}{2} = \frac{4 \times 17}{2} = 34$.

Example 4. Build a magic square using $11$ to $19$ (instead of $1$ to $9$).

Add $10$ to every cell of the standard square. Each row sum becomes $15 + 30 = 45$. Confirmed: now magic sum is $45$ with entries $11$–$19$.

Try it yourself

1. Fill the empty cells (magic sum $15$):

   ? ? 4
   ? 5 ?
   8 ? ?
   

2. What is the magic sum of a $5\times 5$ magic square using $1$ to $25$?
3. Make a magic square using $2, 4, 6, \dots, 18$ (the first $9$ even numbers). What is its magic sum?
4. Verify that the standard $3\times 3$ magic square also has equal diagonal sums.
5. Add $7$ to each cell of the standard $3\times 3$ magic square. Is the new square still magic?
6. Multiply each cell of the standard $3\times 3$ magic square by $2$. Magic sum?
7. Find the centre of a $5\times 5$ magic square using $1$ to $25$. Is it the average?
8. Show that no $2\times 2$ magic square exists using $1, 2, 3, 4$.

Activity

Take a $3\times 3$ grid. Try to fill it from scratch using digits $1$ to $9$ so that every line sums to $15$. Try multiple times to convince yourself the centre must be $5$.

Find a magic square in a temple, building, or book. (The Khajuraho temple has a famous $4\times 4$ magic square.) Photograph and verify.