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

X_{0,0}X_{0,1}X_{0,2}X_{0,3}.. X_{0,M}X_{1,0}X_{1,1}X_{1,2}X_{1,3}.. X_{1,M}X_{2,0}X_{2,1}X_{2,2}X_{2,3}.. X_{2,M}X_{3,0}X_{3,1}X_{3,2}X_{3,3}.. X_{3,M}.. X_{L,0}X_{L,1}X_{L,2}X_{L,3}.. X_{L,M}X_{M,0}X_{M,1}X_{M,2}X_{M,3}.. X_{M,M}

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

X_{0,0}+ X_{1,0}= X_{0,2}+ X_{1,2}X_{0,0}+ X_{1,0}= X_{0,4}+ X_{1,4}.. X_{0,0}+ X_{3,0}= X_{0,2}+ X_{3,2}..

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

X_{r,c}+ X_{r+1+2i,c}= X_{r,c+2j}+ X_{r+1+2i,c+2j}(1)

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

(X_{0,0}+ X_{1,0}) + (X_{2,2}+ X_{3,2}) + .. + (X_{L,L}+ X_{M,L}) + (X_{1,1}+ X_{2,1}) + (X_{3,3}+ X_{4,3}) + .. + (X_{M,M}+ X_{0,M})

which by (1) equals:

(X_{0,0}+ X_{1,0}) + (X_{2,0}+ X_{3,0}) + .. + (X_{L,0}+ X_{M,0}) + (X_{1,1}+ X_{2,1}) + (X_{3,1}+ X_{4,1}) + .. + (X_{M,1}+ X_{0,1})

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

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

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

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

X_{0,0}X_{0,1}X_{0,2}X_{0,3}.. X_{0,G}.. X_{1,0}X_{1,1}X_{1,2}X_{1,3}.. X_{1,G}.. X_{2,0}X_{2,1}X_{2,2}X_{2,3}.. X_{2,G}.. X_{3,0}X_{3,1}X_{3,2}X_{3,3}.. X_{3,G}.. .. X_{F,0}X_{F,1}X_{F,2}X_{F,3}.. X_{F,G}.. X_{G,0}X_{G,1}X_{G,2}X_{G,3}.. X_{G,G}.. ..

From the bent diagonal sums and main diagonal sums of the full square, we have
that the main back 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 back 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 forward diagonals, we can first show that the sum of the main forward diagonal
is **Σ/2**. The sum of the main back diagonal plus the sum of the main
forward diagonal is:

(X_{0,0}+ X_{G,0}) + (X_{2,2}+ X_{E,2}) + .. + (X_{F,F}+ X_{1,F}) + (X_{1,1}+ X_{F,1}) + (X_{3,3}+ X_{D,3}) + .. + (X_{G,G}+ X_{0,G})

which by (1) equals:

(X_{0,0}+ X_{G,0}) + (X_{2,0}+ X_{E,0}) + .. + (X_{F,0}+ X_{1,0}) + (X_{1,1}+ X_{F,1}) + (X_{3,1}+ X_{D,1}) + .. + (X_{G,1}+ X_{0,1})

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

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:

X_{0,0}+ X_{0,0}= X_{0,2}+ X_{0,2}or X_{0,0}= X_{0,2}

which is impossible.