The well-known magic square
has the property that the three numbers in each of the three
rows, the three columns and the two diagonals all add up to the
same number $15$. This number is called the magic sum of the
square.
It is less well known that if you multiply the three numbers in
each row together and add the three products:
$8\times 1\times 6+3\times 5\times 7+4\times 9\times 2=225$
you get the same total as when you multiply the three numbers in
each column together and add the three products:
$8\times 3\times 4+1\times 5\times 7+6\times 7\times
2=225$.
This number is called the magic product of the square.
Now take the "pairwise products'' in each row and add them all
up:
$(8\times 1+1\times 6+6\times 8)+(3\times 5+5\times 7+7\times 3)+
(4\times 9+9\times 2+2\times 6)=195$.
Do the same with the columns:
$(8\times 3+3\times 4+4\times 8)+(1\times 5+5\times 9+9\times 1)+
(6\times 7+7\times 2+2\times 6)=195$.
Again the results are the same! This number is called the magic
pairwise product of the square.
The only other $3\times 3$ magic squares using the numbers from
$1$ to $9$ are just reflections and rotations of the magic square
above, such as
These examples all have the same magic sum, magic product and
magic pairwise product as the first one. However, magic squares
can be formed using different sets of nine numbers. Three
examples are given below.
\begin{eqnarray} \textrm{Magic number:} \qquad & 18 &
\qquad & 18 & \qquad & 24 & \\ \textrm{Magic
product:} \qquad & 468 & \qquad & 414 & \qquad
& 840 &\\ \textrm{Magic pairwise product:} \qquad &
294 & \qquad & 285 &\qquad & 489 &
\end{eqnarray}
Can you find other $3\times 3$ magic squares?
Of course, you could just rotate and reflect the examples above,
and you could multiply every number in the square by the same
constant. The new squares would still be magic, and it is not
difficult to see that the magic squares you create would again
have the magic product property and the magic pairwise product
property.
You could also add the same number to each of the nine numbers in
a magic square, and the result would clearly be a new magic
square. In this case it is not at all clear that the new magic
square will still have the magic product and the magic pairwise
product properties.
Try to make up your own magic squares by assigning a few numbers
at random to some of the cells and then filling the other cells
in so that the rows, columns and diagonals add up to the same
number. Each time you succeed in making a magic square, you
should check that the magic product and magic pairwise product
properties also work.
After having worked out a few more examples, you may notice that
the magic number is always three times the middle number of the
magic square. In particular, it seems that the magic number is
always a multiple of $3$.
Now let's work out a systematic way of finding all $3\times 3$
magic squares.
Let the number in the middle be $x$ and the magic sum of each
row, diagonal and column be $R$.
Add the middle column and middle row and both diagonals giving
$4R$.
This sum includes the middle number four times and all the
other numbers once so it must add up to the total of all the
numbers ($3R$) plus three times the middle number. So
$4R=3R+3x$
$R=3x$
This also tells us that the total of all the numbers in the
square is $9x$.
Now you can make up your own magic squares. You just choose the
middle number and pick two numbers in other positions and then
you complete the whole magic square so each line adds up to three
times the middle number.
For example, with $7$ in the middle, the row total is $21$
so
must be
If you choose the `corners' to be $x+a$ and $x+b$
then the magic square is:
It is now possible, using only elementary algebra, to check
that the general $3\times 3$ magic square satisfies the magic
product property and the magic pairwise product property. We
leave that as an exercise for the energetic reader.
What happens with $5\times 5$ magic squares? Try the example
below.
What can you discover?