Magic Square of Cubes
Impossibility Proofs

3x3 Normal Magic Square of Cubes Theorem
4x4 Normal Magic Square of Cubes Theorem
5x5 Normal Magic Square of Cubes Theorem
6x6 Normal Magic Square of Cubes Theorem [INCOMPLETE]
7x7 Normal Magic Square of Cubes Theorem
8x8 Normal Magic Square of Cubes Theorem [INCOMPLETE]

The magic sum of an NxN magic square of cubes is
     N3(N2+1)2/4.

The proofs express the problem using various moduli and use the
following congruences.  Note that in each case, there are only
three different residue classes.  This has the effect of reducing
the number of magic series that make a given magic sum.

03 ≡ 0 (mod 7)
13 ≡ 1 (mod 7)
23 ≡ 1 (mod 7)
33 ≡ 6 (mod 7)
43 ≡ 1 (mod 7)
53 ≡ 6 (mod 7)
63 ≡ 6 (mod 7)

03 ≡ 0 (mod 9)
13 ≡ 1 (mod 9)
23 ≡ 8 (mod 9)
33 ≡ 0 (mod 9)
43 ≡ 1 (mod 9)
53 ≡ 8 (mod 9)
63 ≡ 0 (mod 9)
73 ≡ 1 (mod 9)
83 ≡ 8 (mod 9)



3x3 Normal Magic Square of Cubes Theorem
A 3x3 magic square of cubes of distinct integers in the range 1-9 is impossible.

Proof
The magic sum of cubes is 675.  9 needs to be one of the entries.
But the cube of 9 is 729, too big to be in a magic series summing to 675.


4x4 Normal Magic Square of Cubes Theorem
A 4x4 magic square of cubes of distinct integers in the range 1-16 is impossible.

Proof
The magic sum of cubes is 4624 which is 7 (mod 9).
In the range 1-16, there are
     5 values of 0 (mod 9),
     6 values of 1 (mod 9),
     5 values of 8 (mod 9).
The magic series that sum to 7 (mod 9) are
     (0,0,8,8) (mod 9),
     (1,8,8,8) (mod 9).
Four rows of these must contain at least eight values of 8 (mod 9),
but there are only five available in the range 1-16.



5x5 Normal Magic Square of Cubes Theorem
A 5x5 magic square of cubes of distinct integers in the range 1-25 is impossible.

Proof
The magic sum of cubes is 21,125 which is 2 (mod 9).
In the range 1-25, there are
     8 values of 0 (mod 9),
     9 values of 1 (mod 9),
     8 values of 8 (mod 9).
The magic series that sum to 2 (mod 9) are
     (0,0,0,1,1) (mod 9),
     (0,1,1,1,8) (mod 9).
Five rows of these must contain at least ten values of 1 (mod 9),
but there are only nine available in the range 1-25.



6x6 Normal Magic Square of Cubes Theorem
A 6x6 magic square of cubes of distinct integers in the range 1-36 is impossible.

[INCOMPLETE PROOF ATTEMPT]

Proof Attempt - Modulo 7
The magic sum of cubes is 73,926 which is 6 (mod 7).
In the range 1-36, there are
      5 values of 0 (mod 7),
     16 values of 1 (mod 7),
     15 values of 6 (mod 7).
The magic series that sum to 6 (mod 7) are
     (0,0,0,0,0,6) (mod 7),
     (0,0,0,1,6,6) (mod 7),
     (0,1,1,6,6,6) (mod 7),
     (1,1,1,1,1,1) (mod 7).
The series combination of six rows must be
     (0,1,1,6,6,6) (mod 7)
     (0,1,1,6,6,6) (mod 7)
     (0,1,1,6,6,6) (mod 7)
     (0,1,1,6,6,6) (mod 7)
     (0,1,1,6,6,6) (mod 7)
     (1,1,1,1,1,1) (mod 7)

   [We now need a proof that there are no series in the range 1-36
    that have the form (1,1,1,1,1,1) (mod 7)].



7x7 Normal Magic Square of Cubes Theorem
A 7x7 magic square of cubes of distinct integers in the range 1-49 is impossible.

Proof
The magic sum of cubes is 214,375 which is 4 (mod 9).
In the range 1-49, there are
     16 values of 0 (mod 9),
     17 values of 1 (mod 9),
     16 values of 8 (mod 9).
The magic series that sum to 4 (mod 9) are
     (0,0,0,1,1,1,1) (mod 9)
     (0,0,8,8,8,8,8) (mod 9)
     (0,1,1,1,1,1,8) (mod 9)
     (1,8,8,8,8,8,8) (mod 9)
The series combinations of seven rows are
     (0,0,0,1,1,1,1)     (0,0,0,1,1,1,1) (mod 9)
     (0,0,0,1,1,1,1)     (0,0,0,1,1,1,1) (mod 9)
     (0,0,0,1,1,1,1)     (0,0,0,1,1,1,1) (mod 9)
     (0,0,8,8,8,8,8)     (0,0,0,1,1,1,1) (mod 9)
     (0,0,8,8,8,8,8)     (0,0,8,8,8,8,8) (mod 9)
     (0,0,8,8,8,8,8)     (0,0,8,8,8,8,8) (mod 9)
     (0,1,1,1,1,1,8)     (1,8,8,8,8,8,8) (mod 9)
The columns must use one of these series combinations.
In each row combination, there is a row containing five value of 8 (mod 9),
so there must be five columns, each containing 8 (mod 9) to match that row,
but in each combination, there can be at most four containing 8 (mod 9),
so no row/column combination exists.



8x8 Normal Magic Square of Cubes Theorem
A 8x8 magic square of cubes of distinct integers in the range 1-64 is impossible.

[INCOMPLETE PROOF ATTEMPT]

Proof Attempt - Modulo 9
The magic sum of cubes is 540,800 which is 8 (mod 9).
In the range 1-64, there are
     21 values of 0 (mod 9),
     22 values of 1 (mod 9),
     21 values of 8 (mod 9).
The magic series that sum to 8 (mod 9) are
     (0,0,0,0,0,0,0,8) (mod 9),
     (0,0,0,0,0,1,8,8) (mod 9),
     (0,0,0,1,1,8,8,8) (mod 9),
     (0,1,1,1,8,8,8,8) (mod 9),
     (1,1,1,1,1,1,1,1) (mod 9).
The last series containing all 1 (mod 9) entries must be used
otherwise (0,1,1,1,8,8,8,8) must be used at least seven times
to make the 22 1 (mod 9) entries, but this uses too many 8 (mod 9) entries.
Since there is one row containing all 1 (mod 9) entries, each column
must contain at least one 1 (mod 9) entry, so (0,0,0,0,0,0,0,8)
can't be used for a row or a column.

The combinations of eight series that must be used for both rows and columns are
     (0,0,0,1,1,8,8,8) (mod 9)   (0,0,0,0,0,1,8,8) (mod 9)
     (0,0,0,1,1,8,8,8) (mod 9)   (0,0,0,1,1,8,8,8) (mod 9)
     (0,0,0,1,1,8,8,8) (mod 9)   (0,0,0,1,1,8,8,8) (mod 9)
     (0,0,0,1,1,8,8,8) (mod 9)   (0,0,0,1,1,8,8,8) (mod 9)
     (0,0,0,1,1,8,8,8) (mod 9)   (0,0,0,1,1,8,8,8) (mod 9)
     (0,0,0,1,1,8,8,8) (mod 9)   (0,0,0,1,1,8,8,8) (mod 9)
     (0,0,0,1,1,8,8,8) (mod 9)   (0,1,1,1,8,8,8,8) (mod 9)
     (1,1,1,1,1,1,1,1) (mod 9)   (1,1,1,1,1,1,1,1) (mod 9)

     (0,0,0,0,0,1,8,8) (mod 9)   (0,0,0,0,0,1,8,8) (mod 9)
     (0,0,0,0,0,1,8,8) (mod 9)   (0,0,0,0,0,1,8,8) (mod 9)
     (0,0,0,1,1,8,8,8) (mod 9)   (0,0,0,0,0,1,8,8) (mod 9)
     (0,0,0,1,1,8,8,8) (mod 9)   (0,0,0,1,1,8,8,8) (mod 9)
     (0,0,0,1,1,8,8,8) (mod 9)   (0,1,1,1,8,8,8,8) (mod 9)
     (0,1,1,1,8,8,8,8) (mod 9)   (0,1,1,1,8,8,8,8) (mod 9)
     (0,1,1,1,8,8,8,8) (mod 9)   (0,1,1,1,8,8,8,8) (mod 9)
     (1,1,1,1,1,1,1,1) (mod 9)   (1,1,1,1,1,1,1,1) (mod 9)