A square grid in which distinct whole numbers are written into each cell of the grid in such a way that each horizontal row and each vertical column, and each diagonal add up to the same number.

Magic Squares

2x2 Magic Squares

Assign each box of the 2x2 grid a distinct number.

Recall that the numbers in each box of the grid must be distinct and that the sum of the columns, rows, and diagonals must all be the same.
Therefore, according to the grid above,
x1+x2 = x3+x4 = x1+x3 = x2+x4 = x1+x4 = x2+x3
Then, x1+x2 = x1+x3, which implies x2 = x3.
Or, x3+x4 = x2+x3, which implies x2+x4.
Either way, the numbers are not distinct, so the 2x2 grid does not work.

1x1 Magic Squares

Although this is technically a magic square, it is not a very interesting example since there is only one number to calculate.

3x3 Magic Squares

In a normal 3x3 magic square, the grid will consist of 9 boxes
which are filled with the consecutive numbers 1 thru 9.
First, find the value of the entire square.
Since each number is used only once, this value is
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45.
Because the sum of each row is equal, the sum of the rows are
R1 + R2 + R3 = 45.
Therefore the sum of each row is 45/3 = 15.
This also means that the sum of the columns and the diagonals is also 15.

R1    R2    R3

15      15      15

Assign a value to the center box on the grid.
This box is included in one row, one column, and both diagonals.
This number is also adjacent to every other number in the square

Find the sum of all of the rows, columns, and diagonals which involve x

R2 + C2 + D1 + D2 = 15 + 15 + 15 + 15 = 60
Notice that all of the numbers, 1 thru 9, have been used
and that the number which is represented by x has been used 4 times.
Recall that the sum of all of the numbers equals 45.
Since the x-value has been used 4 times in the sum, it has been added 3 extra times.
Therefore, the sum of all of the rows, columns, and diagonals involving x is
[the sum of all the numbers] + [the three additional x-values] = 60
45 + 3x = 60
Now solve for x
3x = 60 - 45
3x = 15
x = 5
Therefore the value of the center box must be 5.
In order to find the other values in the boxes,
simply ensure that all rows, columns and diagonals sum to 15.

Variations on the 3x3 magic square can be found by rotating the entries around the center box.

Molly McKee