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A Magic Square of order $n$ is an arrangement of $n^2$ numbers, usually distinct integers, in a square, such that the $n$ numbers in all rows, all columns, and both diagonals sum to the same constant.

a 3x3 magic square

How to prove that a normal $3\times 3$ magic square must have $5$ in its middle cell?

I have tried taking $a,b,c,d,e,f,g,h,i$ and solving equations to calculate $e$ but there are so many equations that I could not manage to solve them.

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11  
as this question has hit the hot list, a definition of a magic square would be helpful. – hildred Jan 13 '14 at 13:46
1  
It's also worth pointing out that this question only applies to normal magic squares (i.e. 1...n). The general magic square might not have a 5 in it at all, let alone in the center. – Bobson Jan 13 '14 at 14:46
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@ArchismanPanigrahi I don´t think so. This comment will self destruct. – hildred Jan 13 '14 at 15:43
up vote 40 down vote accepted

The row, column, diagonal sum must be $15$, e.g. because three disjoint rows must add up to $1+\ldots +9=45$. The sum of all four lines through the middle is therefore $60$ and is also $1+\ldots +9=45$ plus three times the middle number.

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How so for the diagonal sums? – alexis Jan 13 '14 at 10:31
    
@alexis: The diagonal sums have to be equal to the row sums, which have to be equal to 15. – user2357112 Jan 13 '14 at 11:03
    
Oh, by the problem statement you mean... got it. – alexis Jan 13 '14 at 11:06

The common sum must be $15$, because the sum of numbers from $1$ to $9$ is $45$. How can we write $15$ as sum of three distinct numbers between $1$ and $9$ (included)?

\begin{gather} 1+5+9 \\ 1+6+8 \\ 2+4+9 \\ 2+5+8 \\ 2+6+7 \\ 3+5+7 \\ 3+4+8 \\ 4+5+6 \end{gather}

We have just eight ways and we can try accommodating them in the square. The central place belongs to one row, one column and the two diagonals, so the number we put in it must appear four times in the above sums: the only one is $5$.

Similarly, in the four corners we have to place numbers that appear three times, that is: $2$, $4$, $6$ and $8$.

This also shows that basically only one $3\times3$ magic square is possible, up to symmetries of the square.


How to compose the magic square? First note how many rows, columns and diagonals each cell belongs to:

enter image description here

Now we know that $5$ must be in the center; choose arbitrarily an even number, say $2$ and place it in a corner. It could go in any corner, let's choose the upper left one; in the opposite corner we have to write $8$:

enter image description here

Now $4$ must go in one of the other corners and $6$ in the opposite one:

enter image description here

At this point, the other cells can be filled in a unique way:

enter image description here

We have four choices for placing $2$ and, for any choice we can choose two places for $4$. In total we have eight magic squares, but just one if we consider two of them identical after applying a symmetry of the square (there are eight of them).

If we subtract $5$ to each cell, we can better see the symmetry:

enter image description here

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very useful, thank you – Rustyn Aug 4 '15 at 9:56

A normal magic square of order $3$ can be represented by a $3 \times 3$ matrix

$$\begin{bmatrix} x_{11} & x_{12} & x_{13}\\ x_{21} & x_{22} & x_{23}\\ x_{31} & x_{32} & x_{33}\end{bmatrix}$$

where $x_{ij} \in \{1, 2,\dots, 9\}$. Since $1 + 2 + \dots + 9 = 45$, the sum of the elements in each column, row and diagonal must be $15$, as mentioned in the other answers. Vectorizing the matrix, these $8$ equality constraints can be written in matrix form as follows

$$\begin{bmatrix} 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 0 & 1& 1 & 1\\ 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0\\ 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 0 & 1 & 0 & 0 & 1\\ 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 0\end{bmatrix} \begin{bmatrix} x_{11}\\ x_{21}\\ x_{31}\\ x_{12}\\ x_{22}\\ x_{32}\\ x_{13}\\ x_{23}\\ x_{33}\end{bmatrix} = \begin{bmatrix} 15\\ 15\\ 15\\ 15\\ 15\\ 15\\ 15\\ 15\end{bmatrix}$$

Note that each row of the matrix above has $3$ ones and $6$ zeros. We can guess a particular solution by visual inspection of the matrix. This particular solution is $(5,5,\dots,5)$, the $9$-dimensional vector whose nine entries are all equal to $5$. To find a homogeneous solution, we compute the RREF

$$\begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & -1 & -1\\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & -1 & -2\\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 2\\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\end{bmatrix}$$

Thus, the general solution is of the form

$$\begin{bmatrix} 5\\ 5\\ 5\\ 5\\ 5\\ 5\\ 5\\ 5\\ 5\end{bmatrix} + \begin{bmatrix} 0 & -1\\ -1 & 0\\ 1 & 1\\ 1 & 2\\ 0 & 0\\ -1 & -2\\ -1 & -1\\ 1 & 0\\ 0 & 1\end{bmatrix} \eta$$

where $\eta \in \mathbb{R}^2$. Un-vectorizing, we obtain the solution set

$$\left\{ \begin{bmatrix} 5 & 5 & 5\\ 5 & 5 & 5\\ 5 & 5 & 5\end{bmatrix} + \eta_1 \begin{bmatrix} 0 & 1 & -1\\ -1 & 0 & 1\\ 1 & -1 & 0\end{bmatrix} + \eta_2 \begin{bmatrix} -1 & 2 & -1\\ 0 & 0 & 0\\ 1 & -2 & 1\end{bmatrix} : (\eta_1, \eta_2) \in \mathbb{R}^2 \right\}$$

which is a $2$-dimensional affine matrix space in $\mathbb{R}^{3 \times 3}$. For example, the magic square illustrated in the question can be decomposed as follows

$$\begin{bmatrix} 2 & 9 & 4\\ 7 & 5 & 3\\ 6 & 1 & 8\end{bmatrix} = \begin{bmatrix} 5 & 5 & 5\\ 5 & 5 & 5\\ 5 & 5 & 5\end{bmatrix} - 2 \begin{bmatrix} 0 & 1 & -1\\ -1 & 0 & 1\\ 1 & -1 & 0\end{bmatrix} + 3 \begin{bmatrix} -1 & 2 & -1\\ 0 & 0 & 0\\ 1 & -2 & 1\end{bmatrix}$$

Note that the sum of the elements in each column, row and diagonal of each of the two basis matrices is equal to $0$. Note also that the $(2,2)$-th entry of each of the two basis matrices is $0$. Hence, no matter what parameters $\eta_1, \eta_2$ we choose, the $(2,2)$-th entry of the normal magic square of order $3$ will be equal to $5$.

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|A|B|C|

|D|E|F|

|G|H|I|

^Imagine this is a 3x3 square

A+E+I=15 (1)

B+E+H=15 (2)

C+E+G=15 (3)

D+E+F=15 (4)

Sum equations (1), (2), (3), and (4)

A+B+C+D+4E+F+G+H+I=60 (5)

A+B+C=15 (6)

D+E+F=15 (7)

G+H+I=15 (8)

Substitute (6), (7) and (8) equations into equation (5)

45+3E=60

E=5

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Why the repeat, 2+ years later? – Did May 17 at 18:59
    
idk i worked it out for my math homework and was bored so i typed it out – TDMB May 18 at 0:51

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