A New Procedure for Magic Squares (Part II)

Loubère Block Modified Squares

Continuation using 7x7 Squares

As shown in the previous page a new Loubère method Modified Loubère Block Modified 5x5 Squares was introduced where numbers are added consecutively starting with 1 and ending with n². The final square is transformed into a modified Loubère, having numbers greater than n² by modifying a block or grid of numbers. The initial number is added either at the normal Loubère sites, i.e., (the center of the last row or last column) or on the main diagonal.

This is done by taking the non-magic squares and converting them into magic ones using a variety of means. These squares are depicted below in methods I and II. The generation of these magic squares involves an almost normal approach:
Numbers are added consecutively starting out with 1 and ending with numbers greater than n² after modification. A break involves translational moves (up, down or sideways).

Moreover, it must be stated here that the magic sum has been modified from the known equation S = ½(n³ + n) to the general equation:

which takes into account these new squares. The variable a, an odd number, is equal to 1,3,5,7 or ... and may take on + or - values. For example when a = 1 the normal magic sum S is implied. When a takes on different odd values S gives the magic sum of a modified magic square. It will be shown that the addition or subtraction of n² to some of the cells in the square gives rise to a new magic square.

Construction of 7x7 Block Modified Loubère Squares

1. Place a 1 into the lower left hand corner.
2. Add consecutive numbers up to the middle cell.
3. Move, i.e. break, one cell down.
4. Repeat the process until the square is filled as shown below in squares 1-2.
5. As shown below this square is not magic because the three columns and three rows (in grey) sum to 147 instead of 196.
6. Where the sums 147 cross corresponds to the small square in blue.
7. Since the numbers 1, 2 and 3 sit on a diagonal which sums to 196, these numbers remain unchanged.
8. Adding n² = 49 to (34, 10 and 4) or to (42, 43, 12) (color change to light green) gives the two magic squares 3 and 4 where S = ½(n³ + 7n).

1. Place a 1 into the lower left hand corner.
2. Add 4 consecutive numbers to past the center cell.
3. Move, i.e. break, one cell down.
4. Repeat the process until the square is filled as shown below in squares 1-2.
5. As shown below this square is not magic because the two columns and two rows (in grey) sum to 154 instead of 203.
6. Where the sums 154 cross corresponds to the small square in blue.
7. Since the numbers 1, 2, 3 and 4 sit on a diagonal which sums to 154, at least one of these numbers must be changed and since the left diagonal also sums to 154 the number to change is 4.
8. Adding n² = 49 to either (4, 10, 11, 34) or to (4, 12, 42, 43) gives the magic squares 3 and 4 where S = ½(n³ + 9n).

1. Place a 1 into the lower left hand corner.
2. Add 5 consecutive numbers to past the center cell.
3. Move, i.e. break, one cell down.
4. Repeat the process until the square is filled as shown below in squares 1-2.
5. As shown below this square is not magic because the two columns and two rows (in grey) sum to 70 or 168 instead of 217.
6. Where the sums 168 cross corresponds to the small square in blue.
7. Since the numbers 1, 2, 3, 4, 5 and 6 sit on a diagonal which sums to 70, at least two of these numbers must be changed and since the left diagonal also sums to 167 the number to change is 4.
8. Adding n² = 49 to either (4, 6, 11, 25,30, 43) or to (1, 4, 7, 19, 43, 46) gives the magic squares 3 and 4 where S = ½(n³ + 13n).

1. Place a 1 into the lower left hand corner.
2. Add consecutive numbers 2,3,4 up to the top cell.
3. Move, i.e. break, one cell down.
4. Repeat the process until the square is filled as shown below in squares 1-2.
5. As shown below this square is not magic because the two columns and two rows (in grey) sum to 154 instead of 203.
6. Where the sums 154 cross corresponds to the small square in blue.
7. Since the numbers 35,3 and 13 on the left diagonal which sums to 154 at least one of these numbers must be changed.
8. Two examples are obtained by adding n² = 49 to either (1, 5, 11, 35) or to (1, 2, 3, 4) giving the magic squares 3 and 4 where S = ½(n³ + 9n).

This completes this section on the new block Loubère Method (Part I). The next section deals with A New Procedure for Méziriac type magic squares (Part III). To return to Part I. To return to homepage.