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19.0   Special Magic Squares, Bent Diagonals

19.1   Introduction

The concept of Bent Diagonal Magic Squares as introduced in Section 8.4 (Franklin Squares) will be further elaborated in this chapter.

• Odd and double even order squares can be 1, 2, 3 and 4 way bent diagonal
• Single even order squares can be 1 and 2 way bent diagonal (due to the odd magic sum)

The Franklin Squares are an example of 4 way Bent Diagonal (Semi) Magic Squares.

Only following cases have to be considered:

• One Way   Left to Right with wrap around
• Two Way   Left to Right with wrap around and reverse
• Two Way   Left to Right with wrap around and
Top to Bottom with wrap around
• Three Way Left to Right with wrap around and reverse and
Top to Bottom with wrap around
• Four Way  Left to Right with wrap around and reverse and
Top to Bottom with wrap around and reverse

as all other possible cases can be obtained by means of rotation and (or) reflection.

19.2   Magic Squares (4 x 4)
Two Way Bent Diagonal

Although order 4 Bent Diagonal Magic Squares can be filtered from the 7040 Magic Squares as found in Section 2.2, the defining equations for the applicable case(s) will be deducted below for a better understanding.

When the equations defining the Left to Right Bent Diagonals of a fourth order Magic Square:

```a(1) + a(6) + a(10) + a(13) = s1
a(2) + a(7) + a(11) + a(14) = s1
a(3) + a(8) + a(12) + a(15) = s1
a(4) + a(5) + a( 9) + a(16) = s1
```

are added to the equations describing a Simple Magic Square of the fourth order (ref. Section 2.2), the resulting Bent Diagonal Magic Square is described by following equations:

```a(13) =   s1 -   a(14) - a(15) - a(16)
a(10) =          a(11) - a(13) + a(16)
a( 9) = 2*s1 - 2*a(11) - a(12) - a(14) - a(15) - 2*a(16)

a(8) = 0.5 * s1 - a(12)     a(4) = 0.5 * s1 - a(16)
a(7) = 0.5 * s1 - a(11)     a(3) = 0.5 * s1 - a(15)
a(6) = 0.5 * s1 - a(10)     a(2) = 0.5 * s1 - a(14)
a(5) = 0.5 * s1 - a( 9)     a(1) = 0.5 * s1 - a(13)
```

The consequential symmetry (Axial Symmetrical) is worth to be noticed and complies with Dudeney Group 6.

Further it can be deducted that:

```a(1) = s1 - a(8) - a(12) - a(13)
a(2) = s1 - a(5) - a( 9) - a(14)
a(3) = s1 - a(6) - a(10) - a(15)
a(4) = s1 - a(7) - a(11) - a(16)
```

which are the defining equations for the Right to Left Bent Diagonals. Consequently the resulting Magic Square is Two Way Bent Diagonal.

The solutions can be obtained by guessing a(11), a(12), a(14) a(15), a(16) and filling out these guesses in the equations deducted above.

An optimized guessing routine (BentDia41) produced 1216 (304 unique) Two Way Bent Diagonal Magic Squares within 3,04 seconds, of which the first 88 are shown in Attachment 19.2.1.

Note
Due to the Axial Symmetry Three Way and Four Way Bent Diagonal Magic Squares of order 4 can't exist.

19.3   Magic Squares (5 x 5)
One Way Bent Diagonal

When the equations defining the Left to Right Bent Diagonals of a fifth order Magic Square:

```a(1) + a( 7) + a(13) + a(17) + a(21) = s1
a(2) + a( 8) + a(14) + a(18) + a(22) = s1
a(3) + a( 9) + a(15) + a(19) + a(23) = s1
a(4) + a(10) + a(11) + a(20) + a(24) = s1
a(5) + a( 6) + a(12) + a(16) + a(25) = s1
```

are added to the equations describing a Simple Magic Square of the fifth order (ref. Section 3.2.2), the resulting One Way Bent Diagonal Magic Square is described by following equations:

```a( 1) =        s1 - a( 7) - a(13) - a(19) - a(25)
a(17) =        s1 - a(21) - a(13) - a( 7) - a( 1)
a( 9) =  0.2 * s1 - a(17) - 0.5 * a(21) + 0.5 * a(1) + a(25)
a( 5) = -0.4 * s1 + a( 9) + a(17) + a( 7) + a(19) - a(25)
a(12) = -0.2 * s1 + a(21) + a( 1)
a(11) = -0.6 * s1 + a( 9) + a(17) + a( 7) + a(19)
a(23) =        s1 - a(19) - a(15) - a( 9) - a( 3)
a(14) =  0.8 * s1 - a(15) - a( 9) - a(19)
a(18) =           - a( 8) + a(15) + a( 9) - a(13) + a(19)
a(22) =  0.2 * s1 - a( 2) + a(13)
a( 4) =  0.4 * s1 - a( 2) - a( 3) - a( 9) + a(21) + a(13) - a(19) + a(25)
a(24) = -0.2 * s1 + a( 2) + a(15) + a( 3) + a( 9) - a(21) - a(13) + a(19) - a(25)
a(20) =  1.4 * s1 - a(10) - a(15) - a( 9) - a(17) - a( 7) - a(19)
a(16) = -0.4 * s1 + a(10) + a( 8) + a( 7) + a(13) - a(19)
a( 6) =        s1 - a(10) - a( 8) - a( 9) - a( 7)
```

The solutions can be obtained by guessing the 10 parameters:

a(i) for i = 2. 3, 7, 8, 10, 13, 15, 19, 21, 25

and filling out these guesses in the equations deducted above.

An optimized guessing routine (BentDia51) produced 18380 (9190 unique) One Way Bent Diagonal Magic Squares within 115 seconds, of which the first 654 are shown in Attachment 19.3.1.

Notes
It can be proven that One Way Bent Diagonal Pan Magic Squares of order 5 can't exist.
It can be proven that One Way Bent Diagonal Associated Magic Squares of order 5 can't exist.
It can be proven that Two Way Bent Diagonal Magic Squares of order 5 can't exist.

19.4   Magic Squares (6 x 6)

19.4.1 One Way Bent Diagonal

When the equations defining the Left to Right Bent Diagonals of a sixth order Magic Square:

```a(1) + a( 8) + a(15) + a(21) + a(26) + a(31) = s1
a(2) + a( 9) + a(16) + a(22) + a(27) + a(32) = s1
a(3) + a(10) + a(17) + a(23) + a(28) + a(33) = s1
a(4) + a(11) + a(18) + a(24) + a(29) + a(34) = s1
a(5) + a(12) + a(13) + a(19) + a(30) + a(35) = s1
a(6) + a( 7) + a(14) + a(20) + a(25) + a(36) = s1
```

are added to the equations describing a Simple Magic Square of the sixth order (ref. Section 6b.2.2), the resulting One Way Bent Diagonal Magic Square is described by following equations:

```a(31) =   s1 - a(32) - a(33) - a(34) - a(35) - a(36)
a(25) =   s1 - a(26) - a(27) - a(28) - a(29) - a(30)
a(21) =        a(22) - a(26) + a(29) - a(31) + a(36)
a(19) =   s1 - a(20) - a(21) - a(22) - a(23) - a(24)
a(14) = 2*s1 - a(15) - 2*a(17) - 2*a(18) - a(20) - a(22) - 2*a(23) - 2*a(24) + a(26) - a(29) + a(31) - a(36)
a(13) =      - a(16) + a(17) + a(18) - a(19) - a(21) + a(23) + a(24) - a(26) + a(29) - a(31) + a(36)
a(12) =   s1 + a(16) - a(17) - 2 * a(18) + a(21) - a(23) - 2 * a(24) + a(26) - a(29) - a(30) + a(31) - a(36)
a(11) =   s1 - a(17) - a(18) - a(23) - a(24) - a(29)
a(10) =   s1 - a(16) - a(17) - a(22) - a(23) - a(28)
a( 9) =   s1 - a(15) - a(16) - a(21) - a(22) - a(27)
a( 8) = - s1 + 2 * a(17) + 2*a(18) - a(21) + a(22) + 2*a(23) + 2*a(24) - 2*a(26) + a(29) - a(31) + a(36)
a( 7) = - s1 + a(15) + a(16) + a(17) + a(18) + a(21) + a(22) + a(23) + a(24) - a(25)
a( 6) =      - a(16) + a(17) + a(18) - a(21) + a(23) + a(24) - a(26) + a(29) - a(31)
a( 5) =        a(18) + a(24) - a(35)
a( 4) =        a(17) + a(23) - a(34)
a( 3) =        a(16) + a(22) - a(33)
a( 2) =        a(15) + a(21) - a(32)
a( 1) = 2*s1 - a(15) - 2 * a(17) - 2 * a(18) - a(22) - 2 * a(23) - 2 * a(24) + a(26) - a(29) - a(36)
```

The linear equations shown above are ready to be solved, for the magic constant s1 = 111.

However the solutions can only be obtained by guessing the 18 parameters:

a(i) for i = 15 ... 18, 20, 22, 23, 24, 26 ... 30, 32 ... 36

and filling out these guesses in the equations deducted above.

With an optimized guessing routine (BentDia61) and careful variation of the independent variables, numerous One Way Bent Diagonal Magic Squares can be produced, of which a few are shown in Attachment 19.4.1.

Solutions for more strict defined One and Two Way Bent Diagonal Magic Squares of the sixth order, will be discussed In following sections.

19.4.2 Two Way Bent Diagonal

When the equations defining the Right to Left Bent Diagonals of a sixth order Magic Square:

```a(6) + a(11) + a(16) + a(22) + a(29) + a(36) = s1
a(5) + a(10) + a(15) + a(21) + a(28) + a(35) = s1
a(4) + a( 9) + a(14) + a(20) + a(27) + a(34) = s1
a(3) + a( 8) + a(13) + a(19) + a(26) + a(33) = s1
a(2) + a( 7) + a(18) + a(24) + a(25) + a(32) = s1
a(1) + a(12) + a(17) + a(23) + a(30) + a(31) = s1
```

are added to the equations describing a One Way Bent Diagonal Magic Square of the sixth order, as deducted in previous section, the resulting Two Way Bent Diagonal Magic Square is described by following equations:

```a(31) =    s1   - a(32) - a(33) - a(34) - a(35) - a(36)
a(25) =    s1   - a(26) - a(27) - a(28) - a(29) - a(30)
a(21) =           a(22) - a(26) + a(29) - a(31) + a(36)
a(17) =  2*s1/3 - a(18) - a(23) - a(24)
a(16) =           a(18) - a(22) + a(24)
a(15) =  2*s1/3 - a(18) - a(22) - a(24) + a(26) - a(29) + a(31) - a(36)
a(14) =           a(18) - a(20) + a(24)
a(13) = -  s1/3 - a(18) + a(20) + 2 * a(22) + a(23) - a(26) + a(29) - a(31) + a(36)
a( 6) =  2*s1/3 - a(18) - a(24) - a(36)
a( 5) =           a(18) + a(24) - a(35)
a( 4) =  2*s1/3 - a(18) - a(24) - a(34)
a( 3) =           a(18) + a(24) - a(33)
a( 2) =  2*s1/3 - a(18) - a(24) - a(32)
a( 1) =           a(18) + a(24) - a(31)
```
 a(7) = s1/3 - a(25) a(8) = s1/3 - a(26) a( 9) = s1/3 - a(27) a(10) = s1/3 - a(28) a(11) = s1/3 - a(29) a(12) = s1/3 - a(30)

The consequential symmetry - Row 2 and 5 are Axial Symmetrical - is worth to be noticed.

The solutions can be obtained by guessing the 15 parameters:

a(i) for i = 18, 20, 22, 23, 24, 26 ... 30, 32 ... 36

and filling out these guesses in the equations deducted above.

With an optimized guessing routine (BentDia62) numerous Two Way Bent Diagonal Magic Squares can be produced.

Attachment 19.4.2 shows the first occurring Two Way Bent Diagonal Magic Squares for a(36) = i (i = 1 ... 36),
of which a few Axial Symmetric as discussed in Section 19.4.3 below.

Note
Due to the axial symmetrical rows (2/5), Two Way Bent Diagonal Squares with symmetrical Diagonals can't exist.

19.4.3 Two Way Bent Diagonal
Axial Symmetric

Axial Symmetry Magic Squares are - per definition - Two Way Bent Diagonal. Consequently a Two Way Bent Diagonal Axial Symmetric Magic Square of the sixth order is described by following equations:

```a(31) = s1 - a(32) - a(33) - a(34) - a(35) - a(36)
a(25) = s1 - a(26) - a(27) - a(28) - a(29) - a(30)
a(21) =      a(22) - a(26) + a(29) - a(31) + a(36)
a(19) = s1 - a(20) - 2 * a(22) - a(23) - a(24) + a(26) - a(29) + a(31) - a(36)
```
 a(18) = s1/3 - a(24) a(17) = s1/3 - a(23) a(16) = s1/3 - a(22) a(15) = s1/3 - a(21) a(14) = s1/3 - a(20) a(13) = s1/3 - a(19) a(12) = s1/3 - a(30) a(11) = s1/3 - a(29) a(10) = s1/3 - a(28) a( 9) = s1/3 - a(27) a( 8) = s1/3 - a(26) a( 7) = s1/3 - a(25) a(6) = s1/3 - a(36) a(5) = s1/3 - a(35) a(4) = s1/3 - a(34) a(3) = s1/3 - a(33) a(2) = s1/3 - a(32) a(1) = s1/3 - a(31)

The solutions can be obtained by guessing the 14 parameters:

a(i) for i = 20, 22, 23, 24, 26 ... 30, 32 ... 36

and filling out these guesses in the equations deducted above.

With an optimized guessing routine (BentDia65) numerous Axial Symmetric Magic Squares can be produced.

Attachment 19.4.5 shows the first occurring Axial Symmetric Magic Squares for a(36) = i (i = 1 ... 36).

19.4.4 One Way Bent Diagonal
Rectangular Compact

When the equations defining the Left to Right Bent Diagonals of a sixth order Magic Square:

```a(1) + a( 8) + a(15) + a(21) + a(26) + a(31) = s1
a(2) + a( 9) + a(16) + a(22) + a(27) + a(32) = s1
a(3) + a(10) + a(17) + a(23) + a(28) + a(33) = s1
a(4) + a(11) + a(18) + a(24) + a(29) + a(34) = s1
a(5) + a(12) + a(13) + a(19) + a(30) + a(35) = s1
a(6) + a( 7) + a(14) + a(20) + a(25) + a(36) = s1
```

are added to the equations describing a Rectangular Compact Magic Square of the sixth order (ref. Section 6.10), the resulting One Way Bent Diagonal Magic Square is described by following equations:

```a(31) =   s1   - a(32) - a(33) - a(34) - a(35) - a(36)
a(28) =   s1   - a(29) - a(30) - a(34) - a(35) - a(36)
a(27) =          a(30) - a(33) + a(36)
a(26) =          a(29) - a(32) + a(35)
a(25) =        - a(29) - a(30) + a(32) + a(33) + a(34)
a(23) = - s1   - 2 * a(24) + 2 * a(32) + 2 * a(34) + a(35) + 4 * a(36)
a(22) =   s1   + a(24) - 2 * a(32) - a(34) - 3 * a(36)
a(21) =          a(24) + a(33) - a(36)
a(20) = - s1   - 2 * a(24) + 3 * a(32) + 2 * a(34) + 4 * a(36)
a(19) = 2*s1   + a(24) - 3 * a(32) - a(33) - 3 * a(34) - a(35) - 4 * a(36)
a(17) = 5*s1/3 - a(18) + a(24) - 2 * a(32) - 2 * a(34) - a(35) - 4 * a(36)
a(15) =          a(18) - a(33) + a(36)
a(14) = 5*s1/3 - a(18) + a(24) - 3 * a(32) - 2 * a(34) - 4 * a(36)
a(12) = 2*s1/3 - a(18) - a(24) - a(30)
a(10) = - s1   + a(18) + a(24) + a(29) + a(30) + a(34) + a(35) + a(36)
a( 9) = 2*s1/3 - a(18) - a(24) - a(30) + a(33) - a(36)
a( 7) =          a(18) + a(24) + a(29) + a(30) - a(32) - a(33) - a(34)
a( 5) =          a(18) + a(24) - a(35)
a( 4) = 2*s1/3 - a(18) - a(24) - a(34)
a( 2) =          a(18) + a(24) - a(32)
a( 1) = - s1/3 - a(18) - a(24) + a(32) + a(33) + a(34) + a(35) + a(36)
```
 a(6) = s1/3 - a(36) a(3) = s1/3 - a(33) a(11) = s1/3 - a(29) a( 8) = s1/3 - a(26) a(16) = s1/3 - a(22) a(13) = s1/3 - a(19)

The consequential symmetry is worth to be noticed and results in 2 additional Bent Diagonals from Right to Left, as:

a(6) + a(11) + a(16) + a(22) + a(29) + a(36) = s1
a(3) + a( 8) + a(13) + a(19) + a(26) + a(33) = s1

The solutions can be obtained by guessing the 9 parameters:

a(i) for i = 18, 24, 29, 30, 32 ... 36

and filling out these guesses in the equations deducted above.

With an optimized guessing routine (BentDia63) numerous One Way Bent Diagonal Magic Squares can be produced, including the collection of Two Way Bent Diagonal Magic Squares as discussed in Section 19.4.5 below.

Attachment 19.4.3 shows miscellaneous Rectangular Compact One Way Bent Diagonal Magic Squares, generated with the Sub Routine 'Check One Way Only' activated.

19.4.5 Two Way Bent Diagonal
Rectangular Compact

When the equations defining the Right to Left Bent Diagonals of a sixth order Magic Square:

```a(6) + a(11) + a(16) + a(22) + a(29) + a(36) = s1
a(5) + a(10) + a(15) + a(21) + a(28) + a(35) = s1
a(4) + a( 9) + a(14) + a(20) + a(27) + a(34) = s1
a(3) + a( 8) + a(13) + a(19) + a(26) + a(33) = s1
a(2) + a( 7) + a(18) + a(24) + a(25) + a(32) = s1
a(1) + a(12) + a(17) - a(23) - a(30) - a(31) = s1
```

are added to the equations describing a One Way Bent Diagonal Rectangular Compact Magic Square of the sixth order, as deducted in previous section, the resulting Two Way Bent Diagonal Magic Square is described by following equations:

```a(31) =   s1 - a(32) - a(33) - a(34) - a(35) - a(36)
a(28) =   s1 - a(29) - a(30) - a(34) - a(35) - a(36)
a(27) =        a(30) - a(33) + a(36)
a(26) =        a(29) - a(32) + a(35)
a(25) =      - a(29) - a(30) + a(32) + a(33) + a(34)
a(23) = - s1 - 2 * a(24) + 2 * a(32) + 2 * a(34) + a(35) + 4 * a(36)
a(22) =   s1 + a(24) - 2 * a(32) - a(34) - 3 * a(36)
a(21) =        a(24) + a(33) - a(36)
a(20) = - s1 - 2 * a(24) + 3 * a(32) + 2 * a(34) + 4 * a(36)
a(19) = 2*s1 + a(24) - 3 * a(32) - a(33) - 3 * a(34) - a(35) - 4 * a(36)
```
 a(18) = s1/3 - a(24) a(17) = s1/3 - a(23) a(16) = s1/3 - a(22) a(15) = s1/3 - a(21) a(14) = s1/3 - a(20) a(13) = s1/3 - a(19) a(12) = s1/3 - a(30) a(11) = s1/3 - a(29) a(10) = s1/3 - a(28) a( 9) = s1/3 - a(27) a( 8) = s1/3 - a(26) a( 7) = s1/3 - a(25) a(6) = s1/3 - a(36) a(5) = s1/3 - a(35) a(4) = s1/3 - a(34) a(3) = s1/3 - a(33) a(2) = s1/3 - a(32) a(1) = s1/3 - a(31)

The consequential symmetry (Axial Symmetrical) is worth to be noticed.

The solutions can be obtained by guessing the 8 parameters:

a(i) for i = 24, 29, 30, 32 ... 36

and filling out these guesses in the equations deducted above.

An optimized guessing routine (BentDia64) produced 17248 (4312 unique) Two Way Bent Diagonal Magic Squares within 2210 seconds, of which the first 444 are shown in Attachment 19.4.4.

19.6   Magic Squares (7 x 7)

19.6.1 Four Way Bent Diagonal

Order 7 Four Way Bent Diagonal Magic Squares - as published by Harry White in 2002 - have been discussed in detail in Section 7.9.

19.6.2 Two Way Bent Diagonal
Ultra Magic

It can be proven that Four Way Bent Diagonal Magic Squares can't be Associated.

However when the equations defining the horizontal Bent Diagonals (L/R and R/L) of a seventh order Magic Square:

```a(1) + a( 9) + a(17) + a(25) + a(31) + a(37) + a(43) = s1
a(2) + a(10) + a(18) + a(26) + a(32) + a(38) + a(44) = s1
a(3) + a(11) + a(19) + a(27) + a(33) + a(39) + a(45) = s1
a(4) + a(12) + a(20) + a(28) + a(34) + a(40) + a(46) = s1
a(5) + a(13) + a(21) + a(22) + a(35) + a(41) + a(47) = s1
a(6) + a(14) + a(15) + a(23) + a(29) + a(42) + a(48) = s1
a(7) + a( 8) + a(16) + a(24) + a(30) + a(36) + a(49) = s1

a(1) + a(14) + a(20) + a(26) + a(34) + a(42) + a(43) = s1
a(2) + a( 8) + a(21) + a(27) + a(35) + a(36) + a(44) = s1
a(3) + a( 9) + a(15) + a(28) + a(29) + a(37) + a(45) = s1
a(4) + a(10) + a(16) + a(22) + a(30) + a(38) + a(46) = s1
a(5) + a(11) + a(17) + a(23) + a(31) + a(39) + a(47) = s1
a(6) + a(12) + a(18) + a(24) + a(32) + a(40) + a(48) = s1
a(7) + a(13) + a(19) + a(25) + a(33) + a(41) + a(49) = s1
```

are added to the equations describing an Ultra Magic Square of the seventh order (ref. FSection 7.3), the resulting Two Way Bent Diagonal Ultra Magic Square is described by following equations:

```a(43) =   s1     - a(44) - a(45) - a(46) - a(47) - a(48) - a(49)
a(39) = 2*s1     - 2*a(40) - 2*a(41) - 2*a(42) + 2*a(44) + a(45) - a(46) - 3*a(47) - 4*a(48) - 2*a(49)
a(38) =            a(40) - a(44) - a(45) + a(47) + a(48)
a(37) = - s1     + a(41) + a(46) + 2*a(47) + 2*a(48) + 2*a(49)
a(36) =            a(42) - a(44) + a(48)
a(35) = 6*s1/7   - a(41) - a(42) - a(47) - a(48) - a(49)
a(34) = - s1/7   - a(40) - a(42) + a(44) + a(45) + a(47) + a(49)
a(33) = - 8*s1/7 + 2*a(40) + a(41) + 2*a(42) - 2*a(44) - 2*a(45) + a(46) + 2*a(47) + 4*a(48) + a(49)
a(32) =   6*s1/7 - 2*a(40) + a(45) - a(46) - a(47) - 2*a(48)
a(31) = - 8*s1/7 + 2*a(40) + a(41) + 2*a(42) - a(44) - a(45) + a(46) + a(47) + 3*a(48) + a(49)
a(30) =   6*s1/7 - a(40) - a(42) + a(44) + a(45) - a(46) - a(47) - 2*a(48) - a(49)
a(29) =   6*s1/7 - a(41) - a(42) + a(44) - a(47) - 2*a(48) - a(49)
a(28) =   8*s1/7 - a(44) - a(45) - a(46) - a(47) - a(48) - 2*a(49)
a(27) =     s1/7 + a(44) - a(48)
a(26) =     s1/7 + a(45) - a(47)
a(25) =     s1/7
```
 a(24) = 2*s1/7 - a(26) a(23) = 2*s1/7 - a(27) a(22) = 2*s1/7 - a(28) a(21) = 2*s1/7 - a(29) a(20) = 2*s1/7 - a(30) a(19) = 2*s1/7 - a(31) a(18) = 2*s1/7 - a(32) a(17) = 2*s1/7 - a(33) a(16) = 2*s1/7 - a(34) a(15) = 2*s1/7 - a(35) a(14) = 2*s1/7 - a(36) a(13) = 2*s1/7 - a(37) a(12) = 2*s1/7 - a(38) a(11) = 2*s1/7 - a(39) a(10) = 2*s1/7 - a(40) a( 9) = 2*s1/7 - a(41) a( 8) = 2*s1/7 - a(42) a( 7) = 2*s1/7 - a(43) a(6) = 2*s1/7 - a(44) a(5) = 2*s1/7 - a(45) a(4) = 2*s1/7 - a(46) a(3) = 2*s1/7 - a(47) a(2) = 2*s1/7 - a(48) a(1) = 2*s1/7 - a(49)

The solutions can be obtained by guessing the 9 parameters:

a(i) for i = 40, 41, 42, 44 ... 49

and filling out these guesses in the equations deducted above.

Numerous Two Way Bent Diagonal Ultra Magic Squares can be found with routine BentDia72, of which the first 33 are shown in Attachment 19.6.2.

19.7   Magic Squares (8 x 8)
Four Way Bent Diagonal

Order 8 Four Way Bent Diagonal (Semi) Magic Squares (Franklin Squares) have been discussed in detail in Section 8.4.

19.8   Magic Squares (9 x 9)
Four Way Bent Diagonal, Ultra Magic

When the equations defining the horizontal and vertical Bent Diagonals of a nineth order Magic Square, are added to the equations describing an Associated Magic Square of the nineth order, the resulting Four Way Bent Diagonal Magic Square is described by following equations:

```a(73) =     s1   - a(74) - a(75) - a(76) - a(77) - a(78) - a(79) - a(80) - a(81)
a(68) = - 4*s1/9 + a(74) + a(75) + a(77) + a(79) + a(80)
a(65) =   8*s1/9 - a(66) - a(67) - a(69) - a(70) - a(71) - a(75) - a(79)
a(64) =   5*s1/9 - a(72) - a(74) - a(77) - a(80)
a(63) =   4*s1/9 - a(71) - a(79) - a(81)
a(60) =  21*s1/9 - a(61) - a(62) - 2*a(69) - 3*a(70) - 2*a(71) - a(72) + 0.5*a(74) + 0.5*a(75) + 0.5*a(76) +
- a(77) - 2.5*a(78) - 3.5*a(79) - 3.5*a(80) - a(81)
a(59) =   6*s1/9 - a(67) - a(69) - a(75) - a(77) - a(79)
a(58) =            a(60) - a(66) - a(67) + a(69) + a(70) - a(74) - a(75) - a(76) + a(78) + a(79) + a(80)
a(57) = -17*s1/9 + a(61) + 2*a(69) + 2*a(70) + 2*a(71) + a(77) + 2*a(78) + 4*a(79) + 2*a(80) + 2*a(81)
a(56) = -13*s1/9 + a(62) + a(67) + a(69) + 2*a(70) + 2*a(71) + 2*a(72) - a(74) - a(76) + a(77) +
+ a(78) + 2*a(79) + 3*a(80)
a(55) = -13*s1/9 + a(66) + a(67) + a(69) + a(70) + a(71) + a(74) + a(75) + a(76) + a(77) + a(78) +
+ 2*a(79) + a(80) + a(81)
a(54) =   6*s1/9 - a(62) - a(70) - a(72) - a(78) - a(80)
a(53) =     s1/9 - a(61) - a(69) + a(72) + a(80)
a(52) =   5*s1/9 - a(60) - a(62) - a(70) - a(72) - a(80) + a(81)
a(51) =   4*s1/9 - a(61) - a(71) - a(81)
a(50) =  14*s1/9 - 2*a(58) - 2*a(66) - a(67) + a(69) - 3*a(74) - 2*a(75) - 2*a(76) - a(77) - a(80)
a(49) =   4*s1/9 - a(57) - a(65) - a(73)
a(48) =  (2*s1/9 + 2*a(61) + a(74) - a(75) + a(76) - a(78) - a(79) - a(80))/2
a(47) =  23*s1/9 - a(61) - a(67) - 2*a(69) - 2*a(70) - 2*a(71) - a(72) - 2*a(77) - 2*a(78) +
- 4*a(79) - 3*a(80) - 2*a(81)
a(46) =  14*s1/9 - a(62) - a(66) - a(67) - a(69) - 2*a(70) - 2*a(71) - a(72) + a(74) - a(78) - 2*a(79) - 2*a(80)
a(45) =   6*s1/9 - 2*a(72) - a(77) - 2*a(80)
a(44) =  18*s1/9 - a(66) - a(67) - a(69) - a(70) - 2*a(71) - a(75) - a(76) - a(77) - a(78) +
- 3*a(79) - 2*a(80) - 2*a(81)
a(43) =     s1/9 + a(66) - a(70) + a(74) + a(75) + a(76) - a(78) - a(79) - a(80)
a(42) =     s1/9 + a(67) - a(69) + a(75) + a(76) - a(78) - a(79)
a(41) =     s1/9
```
 a(40) = 2*s1/9- a(42) a(39) = 2*s1/9- a(43) a(38) = 2*s1/9- a(44) a(37) = 2*s1/9- a(45) a(36) = 2*s1/9- a(46) a(35) = 2*s1/9- a(47) a(34) = 2*s1/9- a(48) a(33) = 2*s1/9- a(49) a(32) = 2*s1/9- a(50) a(31) = 2*s1/9- a(51) a(30) = 2*s1/9- a(52) a(29) = 2*s1/9- a(53) a(28) = 2*s1/9- a(54) a(27) = 2*s1/9- a(55) a(26) = 2*s1/9- a(56) a(25) = 2*s1/9- a(57) a(24) = 2*s1/9- a(58) a(23) = 2*s1/9- a(59) a(22) = 2*s1/9- a(60) a(21) = 2*s1/9- a(61) a(20) = 2*s1/9- a(62) a(19) = 2*s1/9- a(63) a(18) = 2*s1/9- a(64) a(17) = 2*s1/9- a(65) a(16) = 2*s1/9- a(66) a(15) = 2*s1/9- a(67) a(14) = 2*s1/9- a(68) a(13) = 2*s1/9- a(69) a(12) = 2*s1/9- a(70) a(11) = 2*s1/9- a(71) a(10) = 2*s1/9- a(72) a( 9) = 2*s1/9- a(73) a( 8) = 2*s1/9- a(74) a( 7) = 2*s1/9- a(75) a( 6) = 2*s1/9- a(76) a( 5) = 2*s1/9- a(77) a( 4) = 2*s1/9- a(78) a( 3) = 2*s1/9- a(79) a( 2) = 2*s1/9- a(80) a( 1) = 2*s1/9- a(81)

The resulting Four Way Bent Diagonal Associated Magic Square appears to be Pan Diagonal as well (Ultra Magic).

The solutions can be obtained by guessing the 16 parameters:

a(i) for i = 61, 62, 66, 67, 69 ... 72, 74 ... 81

and filling out these guesses in the equations deducted above.

Attachment 19.8.1 shows a few Four Way Bent Diagonal Ultra Magic Squares which could be found with routine BentDia94.

Note
It can be proven that Four Way Bent Diagonal (Pan) Magic Squares of order 9 can't be 3 x 3 Compact.

19.9   Magic Squares (12 x 12)
Four Way Bent Diagonal

Order 12 Four Way Bent Diagonal Magic Squares (Franklin Like) have been discussed in detail in Section 12.2.1.

Order 12 Four Way Bent Diagonal Pan Magic Squares (Most Perfect) have been discussed in detail in Section 12.2.3.

19.10  Magic Squares (16 x 16)
Four Way Bent Diagonal

Order 16 Four Way Bent Diagonal Magic Squares (Franklin Squares) have been discussed in detail in Section 12.3.1.

Order 16 Four Way Bent Diagonal Pan Magic Squares (Most Perfect) have been discussed in detail in Section 12.2.3.

19.11  Summary

The obtained results regarding the miscellaneous types of Bent Diagonal Magic Squares as deducted and discussed in previous sections are summarized in following table:

 Order Characteristics Subroutine Results 4 Two Way Bent Diagonal 5 One Way Bent Diagonal 6 One Way Bent Diagonal Two Way Bent Diagonal Two Way Bent Diagonal, Axial Symmetric One Way Bent Diagonal, Rect Compact Two Way Bent Diagonal, Rect Compact 7 Two Way Bent Diagonal, Ultra Magic 9 Four Way Bent Diagonal, Ultra Magic
 Comparable routines as listed above, can be used to generate Lozenge Squares, which will be described in following sections.