For most-perfect squares, complementary cells are grouped into complementary 2x2 subsquares. Because the transforms to Franklin here always transfer adjacent rows/columns, this property is preserved in the resulting Franklin squares.

Since all order 8 Franklin magic squares can be made by transform from all order 8 most-perfect squares, all these Franklin squares have this property.

For a 2x2 subsquare X, the cells subscripted as X_{0,0} , X_{0,1} , X_{1,0} , X_{1,1} are here referred to as X0, X1, X2, X3.

The following derivations refer to the Order 8 Most-perfect figure on the Most-perfect Magic Squares page. It will be helpful to have this up in another window.

From the 2x2 subsquare sums we have:

a0 + a2 = b0 + b2 and a1 + a3 = b1 + b3 and b0 + b2 = E0 + E2 and b1 + b3 = E1 + E3 so a0 + a2 = E0 + E2 and a1 + a3 = E1 + E3 ...

In general, for 2x2 subsquares W, X in the same row:

W0 + W2 = X0 + X2 and W1 + W3 = X1 + X3(1)

Similarly, for 2x2 subsquares W, Y in the same column:

W0 + W1 = Y0 + Y1 and W2 + W3 = Y2 + Y3(2)

Also we have:

a0 + a1 = c0 + c1 so a0 - c0 = c1 - a1 and a1 + b0 = c1 + d0 so c1 - a1 = b0 - d0 and so a0 - c0 = b0 - d0 and a0 + d0 = b0 + c0 similarly b0 - d0 = E0 - G0 so b0 + G0 = d0 + E0 and so a0 - c0 = E0 - G0 and a0 + G0 = c0 + E0 and similarly c0 - e0 = d0 - f0 so c0 + f0 = d0 + e0 and so a0 - e0 = b0 - f0 and a0 + f0 = b0 + e0

In general, for 2x2 subsquares W, X in a row and Y, Z in another row and W, Y in a column and X, Z in another column, and i = 0, 1, 2, 3:

Wi - Yi = Xi - Zi and Wi + Zi = Xi + Yi(3)

The following refers to the Order 8 Most-perfect figure on the Most-perfect Magic Squares page and uses the above formulas.

For the magic square to be Franklin, we have:

half row sums, e.g., c2 + c3 + d2 + d3 = 0 half column sums, e.g., a0 + a2 + c0 + c2 = 0 half diagonal sums, e.g., a0 + a3 + d0 + d3 = 0

From the half column sum and (1):

a0 + a2 = -(c0 + c2) = -(d0 + d2) so d0 = -a0 - a2 - d2(4)

From the half diagonal sum:

a0 + a3 = - (d0 + d3) so d0 = -a0 - a3 - d3(5)

From (2) and the half row sum:

a2 + a3 = c2 + c3 = -(d2 + d3)(6)

Summing (4) and (5) and using (6):

2d0 = -2a0 - a2 - a3 - d2 - d3 so d0 = -a0

That is, for the Franklin bones the value at row 3, column 3 is
**-a0**; but, for the most-perfect bones, the value at row 5,
column 5 must be **-a0**. So the magic square cannot be Franklin
** and** most-perfect.

Of course, all of the above can be translated into actual values with
half sums equal to half the magic constant instead of 0.
Then, referring to this square representation, we have,
(Sum2 = 8^{2}+1 = 65):

for Franklin, X_{2,2}= Sum2 - X_{0,0}for most-perfect, X_{4,4}= Sum2 - X_{0,0}

From the above, we have:

d0 = -a0 a0 + a1 = -d0 - d1 = a0 - d1 so d1 = -a1 a0 + a2 = -d0 - d2 = a0 - d2 so d2 = -a2 a0 + a3 = -d0 - d3 = a0 - d3 so d3 = -a3

Thus, subsquare **d** is complementary to subsquare **a**. Similarly, all the complementary subsquares shown for
the Franklin square must be adjacent corner paired.
And, transform I gives a one-to-one correspondence between this pattern and the
most-perfect square
pattern.

The following refers to the Transform I: Order 8 Franklin figure on the Most-perfect Magic Squares page and uses the above formulas.

For the top left quarter, the sum of row 1 is:

a0 + a1 + (E0 + E1) = a0 + a1 + (A0 + A1) = 0 (2)

and similarly for the other rows.

The sum of column 1 is:

a0 + a2 + (e0 + e2) = a0 + a2 + (A0 + A2) = 0 (1)

and similarly for the other columns.

The sums of the back diagonals of the quarter are:

a0 + a3 + A0 + A3 = 0 a2 + e1 + A2 + E1 = 0 e0 + e3 + E0 + E3 = 0 e2 + a1 + E2 + A1 = 0

and similarly for the forward diagonals of the quarter.

The sums of the back diagonals of the square are:

a0 + a3 + A0 + A3 + d0 + d3 + D0 + D3 = 0 a2 + e1 + A2 + G1 + d2 + h1 + D2 + F1 = e1 + G1 + h1 + F1 = e1 + h1 + (G1 + F1) = e1 + h1 + (E1 + H1) = 0 (3) e0 + e3 + G0 + G3 + h0 + h3 + F0 + F3 = e0 + h0 + (G0 + F0) + e3 + h3 + (G3 + F3) = e0 + h0 + (E0 + H0) + e3 + h3 + (E3 + H3) = 0 (3) e2 + c1 + G2 + C1 + h2 + b1 + F2 + B1 = e2 + G2 + h2 + F2 = e2 + h2 + (G2 + F2) = e2 + h2 + (E2 + H2) = 0 (3) ...

and similarly for the forward diagonals of the square.

The following refers to the Transform I: Order 16 Franklin figure on the Most-perfect Magic Squares page and uses the above formulas.

For the top left quarter, the sum of row 1 is:

cells 0,1 of aQcS = aAcC = 0 (2)

and similarly for the other rows.

The sum of column 1 is:

cells 0,2 of aqiy = aAiI = 0 (1)

and similarly for the other columns.

The sums of the back diagonals of the quarter, using (3), are:

cells 0,3 of aAkK = 0 cell 1 of qYδS = qδSY = qδQΔ = 0, cell 2 of aAkK = 0 cells 0,3 of qYδS = 0 cell 1 of iIcC = 0, cell 2 of qYδS = 0 cells 0,3 of iIcC = 0 cell 1 of yQsΔ = ysQΔ = ysYS = 0, cell 2 of iIcC = 0 cells 0,3 of yQsΔ = 0 cell 1 of aAkK = 0, cell 2 of yQsΔ = 0

and similarly for the forward diagonals of the quarter.

And similarly for the other quarters.

The sums of the back diagonals of the square, using (3), are:

cells 0,3 of aAkKfFpP = 0 cell 1 of qYδWvΞφT = qδWYvφΞT = qδΔUvφΦR = qvRU = qvQV = 0, cell 2 of aAkKfFpP = 0 cells 0,3 of qYδWvΞφT = 0, (see cell 1) cell 1 of iIgGnNdD = 0, cell 2 of qYδWvΞφT = 0 ...

and similarly for the forward diagonals of the square.

The following refers to the Transform II: Order 16 Franklin figure on the Most-perfect Magic Squares page and uses the above formulas.

For the top left quarter, the sum of row 1 is:

cells 0,1 of abQR = abAB = 0 (2)

and similarly for the other rows.

The sum of column 1 is:

cells 0,2 of aequ = aeAE = 0 (1)

and similarly for the other columns.

The sums of the back diagonals of the quarter are:

cells 0,3 of afAF = 0 cell 1 of erER = 0, cell 2 of afAF = 0 ...

and similarly for the forward diagonals of the quarter.

And similarly for the other quarters.

The sums of the back diagonals of the square, using (3), are:

cells 0,3 of afAFkpKP = 0 cell 1 of erEZoθOT = rZθT = rθRΘ = 0, cell 2 of afAFkpKP = 0 cells 0,3 of erEZoθOT = 0 (see cell 1) cell 1 of qvYΞδφSX = qδSYvφΞX = qδQΔvφVΦ = 0, cell 2 of erEZoθOT = 0 ...

and similarly for the forward diagonals of the square.

The following refers to the Transform III: Order 16 Franklin figure on the Most-perfect Magic Squares page and uses the above formulas.

For the top left quarter, the sum of row 1 is:

cells 0,1 of abRQ = abBA = 0 (2)

and similarly for the other rows.

The sum of column 1 is:

cells 0,2 of aeuq = aeEA = 0 (1)

and similarly for the other columns.

The sums of the back diagonals of the quarter, using (3), are:

cells 0,3 of afFA = 0 cell 1 of evBQ = eBvQ = eBEb = 0, cell 2 of afFA = 0 ...

and similarly for the forward diagonals of the quarter.

And similarly for the other quarters.

The sums of the back diagonals of the square, using (3), are:

cells 0,3 of afFApkKP = 0 cell 1 of evBΛlδOT = eOΛvlBδT = eOEnlBcL = cnBO = cnCN = 0, cell 2 of afFApkKP = 0 cells 0,3 of evBΛlδOT = 0 (see cell 1) cell 1 of urΞYθπSX = uθXYrπΞS = uθUΘrπRΠ = 0, cell 2 of evBΛlδOT = 0 ...

and similarly for the forward diagonals of the square.