The Magic Squares

The Problem

Tui has realised that there are nine positions in a magic square. So can she make up a magic square using each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 once and only once?

Teaching sequence

1. Talk about square ‘arrays’ of numbers like the ones in A Square of Circles (Level 2). Ask the class if you can put numbers into these arrays so that the rows have the same sum; the columns have the same sum; all of the rows, columns and diagonals have the same sum.

2. Show them a magic square such as the one below.

3. Get them to check that the rows all have the same sum (of 12); that the columns all have the same sum; and that the diagonals have the same sum.
4. Tell them that these things are called magic squares and that the sum of a magic square is the common sum of the rows, columns and diagonals.
5. Tell them Tui’s problem.
6. Ask them to go away in pairs and see if they can find one with the required numbers. Note that they should not be given the hint that the sum of a magic square is three times the centre entry. If they need this piece of information (and they do), then they have to work it out for themselves.
7. Get some of the pairs to report back. Can they justify every step in the production of the magic square? How did they solve the problem? (There is more than one way.)
8. How many different magic squares can they find? How many do they think there are?
9. Ask the students to write up what they have discovered.
10. >As the first part of the Extension problem is not so different from the original problem, most of the class might be asked to try it.

Extension to the problem

What other magic squares can you make with nine consecutive numbers?

Solution

Suppose that the magic square is as in the diagram below.

Let s = the sum of three numbers in a row. Therefore:

s   = A + B + C   =    D + E + F      =    G + H  + I
     = A + D + G   =    B + E + H     =    C + F + I
     = A + E + I    =     C + E + G

We know that:
A + B + C + D + E + F + G + H + I = 1 + 2 +…+ 9 = 45
Can we use this information to establish s?

We can see that the first three equations for s are A + B + C; D + E + F; and G + H  + I.  If we sum these three equations, we get:
3s = A + B +C + D + E + F + G + H + I = 45. So 3s = 45, which means that s = 15.

There are two ways to go from here. Call this point step *. Thus the sum has to be 15.  How many ways can this sum be attained?  Let's see how many ways each number 1 to 9 can be used in a sum of three to get 15.
9 + 5 + 1
9 + 4 + 2
8 + 6 + 1
8 + 5 + 2
8 + 4 + 3
7 + 6 + 2
7 + 5 + 3
6 + 5 + 4

As can be seen, there are only eight possible equations of three numbers using the numbers 1 to 9 that give a sum of 15.  Thus, all these eight equations have to be used together in the solution.  We can see from the eight algebraic equations that each corner square is in three equations (i.e., a, c, g, i), the centre square is in four equations (i.e., e) and the remaining middle row squares are in two equations (i.e., b, d, f, h).

We can see from the eight equations using the numbers 1 to 9, that 5 is the only number which appears in four equations, thus 5 must be the centre square, E:

We can also see that 9, 7, 3, and 1 only appear twice each in these eight equations, thus they must be B, D, F, and H. We can also see that 9 and 1 have to be in an equation together, as do 7 and 3. Therefore, they are restricted to the following arrangements (and their rotations about the centre square):
 

The remaining numbers 2, 4, 6, and 8 must therefore be in the corners, i.e., A, C, G and I. Checking the eight algebraic equations, we can see that this is possible, as all of these numbers each appear three times in the eight equations using the numbers 1 to 9. As these numbers are in the corners, they all appear in one equation with each of the other corner numbers. We can also see that 2 is in equations with 9, 7, and 5; 4 is in equations with 9, 5 and 3; 6 is in equations with 7, 5 and 1; and 8 is in equations with 1, 3 and 5. Therefore, the only following arrangements can be formed (and their rotations about the centre square):

This is the same arrangement, reflected through B, E and H. Thus, 15 is the only sum, and this sum can only be attained using the above eight equations of three numbers using the numbers 1 to 9. Due to the symmetry of a 3 by 3 grid, reflecting or rotating the solution can form 8 different arrangements of this solution:

We have therefore proved that there is only one magic square that can be made from the numbers 1 to 9 inclusive (subject to rotations and reflections of the square).

But earlier, at step * we could have used more algebra. Think about all of the rows, columns and diagonals through the centre square. So, if s is the sum, then
A + E + I = s
B + E + H = s
C + E + G = s
D + E + F = s

Add all of these up and you’ll get

(A + D + G) + (B + E + H) + (C + F + I) + 3E = 4s.

Now each term in the brackets equals s as they are the column sums. So
s + s + s + 3E = 4s.

This means that 3E = s. And since s = 15, then e = 5.

Now one way to make 15 is to 1 + 5 + 9, so suppose that 1, 5, 9 are down the main diagonal.

Then B + C = 14 so {B, C} = {5, 9} or {6, 8}. But we’ve already used 5 and 9 so {B, C} = {6, 8}. This presents a problem. If C = 6 then G = 0 and if C = 8, then G = -2. This can only mean that 1, 5, 9 are not on the main diagonal.