The sum of each k x k block, (including wrap-around), is equal to k^{2}/n of the
magic constant.

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 \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 \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.

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 \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:

(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 /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 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.