Magic squares and number puzzles

A magic square is a square grid of numbers in which every row, every column, and both diagonals add to the same total. For more than two thousand years, magic squares have been admired across India, China, and the Arab world for their balance and beauty.

Concept

The smallest interesting magic square is the $3 \times 3$ magic square using the numbers $1$ to $9$ exactly once each. The unique arrangement (up to rotations and reflections) is:

Check: every row, column, and diagonal adds to $15$.

• Rows: $2+7+6 = 15$, $9+5+1 = 15$, $4+3+8 = 15$.
• Columns: $2+9+4 = 15$, $7+5+3 = 15$, $6+1+8 = 15$.
• Diagonals: $2+5+8 = 15$, $6+5+4 = 15$.

The shared total ($15$) is called the magic constant. For a $3 \times 3$ square using $1$–$9$, the magic constant must be $15$, because the sum $1+2+\dots+9 = 45$, and we are splitting it into $3$ equal rows.

For an $n \times n$ magic square using the numbers $1$ to $n^2$, the magic constant is

$M = \frac{n(n^2 + 1)}{2}$

So for $n = 4$, $M = \frac{4 \times 17}{2} = 34$. For $n = 5$, $M = 65$.

Magic squares also work with other sets of numbers. You can build a magic square using consecutive numbers like $10, 11, 12, \dots, 18$, or with any nine numbers in arithmetic progression. The arrangement is the same , only the entries shift.

There are simple rules of construction for odd-sized magic squares (called the Siamese method), but they are surprisingly hard to invent on your own. Try the $3 \times 3$ case by hand and you will see why people have stared at these for centuries.

Other number-grid puzzles:

• Kakuro , fill in digits so that segments sum to given totals.
• Sudoku , fill in a $9 \times 9$ grid with $1$–$9$ in every row, column, and $3 \times 3$ block.
• Cross sums , fill a cross-shape so all arms add to the same total.

The common idea: a grid where number rules connect cells.

Worked examples

Example 1. Verify that the magic constant for a $3 \times 3$ square using $1$–$9$ is $15$.

• Sum $1 + 2 + 3 + \dots + 9 = \frac{9 \times 10}{2} = 45$.
• Split into $3$ equal rows: $45 \div 3 = 15$. ✓

Example 2. What is the magic constant for a $5 \times 5$ magic square using $1$–$25$?

• $M = \frac{5(25 + 1)}{2} = \frac{5 \times 26}{2} = 65$.

Example 3. Complete this partial magic square (magic constant $15$):

• Middle row sums to $15$: $3 + 5 + ? = 15 \Rightarrow ? = 7$. So middle right is $7$.
• Middle column: $9 + 5 + 1 = 15$. ✓
• Top row: $? + 9 + ? = 15$. The corners must sum to $6$. From $1$–$9$, the available corners are $2, 4, 6, 8$.
• Bottom row: $? + 1 + ? = 15$, corners sum to $14$, so they must be $6$ and $8$.
• Left column: top + $3$ + bottom = $15 \Rightarrow$ top + bottom = $12$. From the corner pool $\{2, 4, 6, 8\}$ with top row pair summing to $6$ and bottom pair to $14$: top-left + top-right = $6$ and bottom-left + bottom-right = $14$.
• Try top-left = $2$, top-right = $4$, bottom-left = $6$, bottom-right = $8$: check diagonals. $2 + 5 + 8 = 15$ ✓; $4 + 5 + 6 = 15$ ✓.
• Final square:

Example 4. Add the same number $k = 10$ to every entry of the $3 \times 3$ magic square in Example 3. Is it still a magic square? What is the new magic constant?

• Every row, column, diagonal gains $3 \times 10 = 30$. New constant $= 15 + 30 = 45$. Still magic.

Try it yourself

1. Find the magic constant of a $4 \times 4$ magic square using $1$–$16$.
2. Make a magic square using $2, 3, 4, \dots, 10$ (instead of $1$–$9$). What is the magic constant?
3. Multiply every entry in a magic square by $3$. Is it still magic?
4. In a $3 \times 3$ magic square, the centre is always the average of the numbers used. Verify this for $1$–$9$.
5. Sudoku challenge: in a row of a $9 \times 9$ Sudoku, what must the sum of the digits be?
6. Tricky: design a $3 \times 3$ magic square using only odd numbers $1, 3, 5, 7, 9, 11, 13, 15, 17$. What is the constant?
7. Look up Dürer's famous magic square in his $1514$ engraving Melencolia I. What is its magic constant?

Activity

Build it. On paper, draw a $3 \times 3$ grid. Use cut paper squares numbered $1$ through $9$. By trial, find an arrangement where every row, column, and diagonal sums to $15$. You might fail many times before succeeding , that is part of the fun. Once you crack it, try the Siamese method: place $1$ in the top middle, then move up-right (wrapping around) to place $2, 3, \dots, 9$, with the rule that if you land on a filled cell, drop down one. See if you can reach the same square.