Magic Square

A representation of a magic square of order n, where M = n-1, L = n-2.

X0,0 X0,1 X0,2 X0,3  ..  X0,M
X1,0 X1,1 X1,2 X1,3  ..  X1,M
X2,0 X2,1 X2,2 X2,3  ..  X2,M
X3,0 X3,1 X3,2 X3,3  ..  X3,M
 ..
XL,0 XL,1 XL,2 XL,3  ..  XL,M
XM,0 XM,1 XM,2 XM,3  ..  XM,M

Compact Magic Square

Consider the above magic square as compact. From the 2x2 subsquare sums we have:

 X0,0 + X1,0 = X0,2 + X1,2
 X0,0 + X1,0 = X0,4 + X1,4
   ..
 X0,0 + X3,0 = X0,2 + X3,2 
   ..

In general, for r, c = 0 .. n-1 and i, j = 0,1,2,.. :

 Xr,c + Xr+1+2i,c = Xr,c+2j + Xr+1+2i,c+2j			(1)

A Compact Magic Square is Pandiagonal

Summing alternating numbers from the first and second \diagonals beginning in column 1, we have:

   (X0,0 + X1,0) + (X2,2 + X3,2) + .. + (XL,L + XM,L)
 + (X1,1 + X2,1) + (X3,3 + X4,3) + .. + (XM,M + X0,M)

which by (1) equals:

   (X0,0 + X1,0) + (X2,0 + X3,0) + .. + (XL,0 + XM,0)
 + (X1,1 + X2,1) + (X3,1 + X4,1) + .. + (XM,1 + X0,1)

equals the sum of column 1 plus the sum of column 2.

Therefore, if the sum of the main \diagonal is Σ, the second also sums to Σ. Next summing the second and third \diagonals, we have that the third sums to Σ. Continuing in this manner, we see that the sum of every \diagonal is Σ.

Similarly, starting from the main /diagonal we can show that all the /diagonals sum to Σ. Thus, the square is pandiagonal.

Each Quarter of a Franklin Magic Square is Pandiagonal

Consider the top left corner of the square. Let G = n/2 -1, F = n/2 -2, etc.:

X0,0 X0,1 X0,2 X0,3  ..  X0,G  ..  
X1,0 X1,1 X1,2 X1,3  ..  X1,G  ..  
X2,0 X2,1 X2,2 X2,3  ..  X2,G  ..  
X3,0 X3,1 X3,2 X3,3  ..  X3,G  ..  
 ..
XF,0 XF,1 XF,2 XF,3  ..  XF,G  ..  
XG,0 XG,1 XG,2 XG,3  ..  XG,G  ..  
 ..

From the bent diagonal sums and main diagonal sums of the full square, we have that the main \diagonal of the quarter square has sum Σ/2. From this, proceeding as we did above for the full square, we can show that the sum of every \diagonal is Σ/2. This is sufficient to prove that the quarter square cannot exist.

However, we can also show that the quarter square is fully pandiagonal. For the /diagonals, we can first show that the sum of the main /diagonal is Σ/2. The sum of the main \diagonal plus the sum of the main /diagonal is:

   (X0,0 + XG,0) + (X2,2 + XE,2) + .. + (XF,F + X1,F)
 + (X1,1 + XF,1) + (X3,3 + XD,3) + .. + (XG,G + X0,G)

which by (1) equals:

   (X0,0 + XG,0) + (X2,0 + XE,0) + .. + (XF,0 + X1,0)
 + (X1,1 + XF,1) + (X3,1 + XD,1) + .. + (XG,1 + X0,1)

equals the sum of column 1 plus the sum of column 2 of the quarter. Therefore, the sum of the main /diagonal is Σ/2 and we can proceed as before to show that the sum of every /diagonal is Σ/2. Thus, the quarter square is pandiagonal with magic sum Σ/2.

A Compact or Complete Magic Square is Doubly Even

A compact magic square is pandiagonal. See above proof. There are no singly even order pandiagonal squares.

There are no odd order n complete squares because n/2 is fractional. There are no odd order compact squares because from (1) we have:

   X0,0 + X0,0 = X0,2 + X0,2
or
   X0,0 = X0,2

which is impossible.