Show that the central cell of a normal magic square of the third order must be occupied by 5.

Show that in a normal magic square of the third order 1 can never occur in a corner cell.

To show that the central cell of a normal magic square of the third order must be occupied by 5, we need to understand the properties of a magic square and the specific rules that apply to magic squares of the third order.

A magic square is a square grid of numbers, where the sum of each row, column, and diagonal is the same. In a normal magic square, the numbers used are consecutive positive integers starting from 1.

A magic square of the third order is a 3x3 grid, and it follows certain rules:

1. The numbers used must be consecutive positive integers starting from 1 to 9 in this case.
2. Each number can only occur once in the magic square.
3. The sum of each row, column, and diagonal must be the same.

Let's start by considering the sum of the rows, columns, and diagonals:

1st row sum: a + b + c = x
2nd row sum: d + e + f = x
3rd row sum: g + h + i = x

1st column sum: a + d + g = x
2nd column sum: b + e + h = x
3rd column sum: c + f + i = x

Main diagonal sum: a + e + i = x
Secondary diagonal sum: c + e + g = x

Now, let's analyze the possibilities for the central cell:

The central cell must be an odd number since it equally divides the sum of the rows, columns, and diagonals. Therefore, it can be 1, 3, 5, 7, or 9.

If the central cell were 1, the possible values for the other cells would be:

1 2 x
x x x
x x x

In this case, it is not possible to form a magic square because the row and column sums would equal 1 + 2 + x = x + 2, which can never be equal to the other sums in the magic square.

Similar reasoning can be used for 7 and 9, as they will also lead to contradictions and cannot form a magic square.

If the central cell were 3, the possible values for the other cells would be:

x x x
x 3 x
x x x

In this case, it is still not possible to create a magic square because the row and column sums would equal x + 3, which can never be equal to the other sums in the magic square.

Finally, if the central cell were 5, the possible values for the other cells would be:

x x x
x 5 x
x x x

In this case, we can create a valid magic square. The sum of each row, column, and diagonal will be 15.

Therefore, by examining the possibilities for the central cell, we have shown that in a normal magic square of the third order, the central cell must be occupied by 5.

To prove that 1 can never occur in a corner cell of a normal magic square of the third order, we can use a similar line of reasoning.

Let's consider the possibilities for the corner cells:

Top-left corner:

1 x x
x x x
x x x

In this case, there is no number that can be placed in the remaining cells to create a valid magic square. The row and column sums involving the top-left corner will always be less than 1, which cannot be equal to the other sums in the magic square.

Similar reasoning can be applied to other corners (e.g., top-right, bottom-left, and bottom-right).

Therefore, by examining the possibilities for the corner cells, we can conclude that in a normal magic square of the third order, 1 can never occur in a corner cell.