Let the middle be 5, and divide the remainder into two groups: 1, 3, 7, 92, 4, 6, 8, and fill in 4 sides or 4 corners respectively.

The number on each side is 3 times, the number on the corner is 2 times, and the number in the middle is 4 times ***6 squares, so the sum of each square is 20.

Set an odd number on the side, check the upper left square, and find that the odd number required on the corner is opposite to the even number.

Let an even number be placed on the side and an odd number be placed on the corner, which holds.

So 2, 4, 6, 8 can be arbitrarily filled on the sides, and 20 can fill the corners according to the sum of squares.

Can you have 9 2 7?

4 5 6

3 8 1 equal method arrangement