7th grader wanting to find solution to magic square: place the integers from -5 to +10 in the magic square so that the total of each row, column, and diagonal is 10. The magic square is 4 squares x 4 squares. Also 5 x 5 magic square using the integers from -9 to +15. The total of each row, column, and diagonal should be 15. Thanks Hi, I'll show you a method to construct a 3x3 magic square that should work for 4x4 and 5x5 squares. I will place the numbers from 1 to 9 in the square and the sum of each row, column and diagonal will be 15. The procedure has three steps. First the rows. Put the numbers from 1 to 9 into three rows so that the sum of each row is 15. I got 9 1 5 8 4 3 7 6 2 Next the columns. The row sums are all 15 but the column sums are not. I can, however, move the numbers in any row, as long as they stay in the same row, and the row sums will remain at 15. Move the 4 in the second row to the first column and the 2 in the third row to the first column. 9 1 5 4 8 3 2 7 6 Now the first column sums to 15. Interchange the 6 and 7 in the last row. 9 1 5 4 8 3 2 6 7 Now the row sums and column sums are all 15. Finally the diagonals. At this point I can interchange any two rows or interchange any two columns and the row sums and column sums with remain unchanged. Interchange the first two rows. 4 8 3 9 1 5 2 6 7 Interchange the second and third columns. 4 3 8 9 5 1 2 7 6 If you want a 3x3 magic square constructed from the numbers from -2 to 6 with the row, column and diagonal sums of 6 just subtract 3 from each of the numbers in the magic square above. 1  0  5  6  2 -2 -1  4  3 Cheers, Penny Go to Math Central