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Completeness theorem on order-8 Nasik magic cubes (3/3)

Theorem 1 (completeness theorem on order-8 Nasik magic cubes)
   Every order-8 (normal or non-normal) Nasik (namely, pan-2,3-agonal) magic cube is complete. In particular, an order-8 normal Nasik magic cube cannot be associated.

Proof
   Let us consider any pair of cells spaced 4 apart on a common pantriagonal of an order-8 (normal or non-normal) Nasik magic cube.
   We can move the pair to the positions of a000 and a444 in the Lemma 2 by the shift of some planes from one side of the cube to the other, and the cube that the shift is applied to is still Nasik. Thus the pair is a complement pair from the Lemma 2. So it is proved that every order-8 Nasik magic cube is complete.
   Obviously, a normal magic cube cannot be both complete and associated, so an order-8 normal Nasik magic cube cannot be associated.    Q.E.D.

Note 1
   This theorem does not hold for orders 8x greater than 8. Here are counter-examples.

Impossibility theorems on magic tesseracts

   The Theorem 1 can prove the following impossibility theorems on magic tesseracts.

Theorem 2
   There cannot exist an order-8 normal pan-2,3-agonal magic tesseract. In particular, there cannot exist an order-8 normal Nasik (namely, pan-2,3,4-agonal) magic tesseract.

Proof
   If such a tesseract (aijkh), where i, j, k, and h are integers from 0 to 7, were possible, the two 3-dimensional slices of the tesseract, (a0jkh) and (ai0kh), would be Nasik magic cubes.
   From the Theorem 1, the pair, a0000 and a0444, would be a complement pair, and the pair, a0000 and a4044, would be, too. Thus a0444 would equal to a4044, so (aijkh) cannot be a normal magic tesseract.    Q.E.D.

Theorem 3
   There cannot exist an order-m normal Nasik magic tesseract if m is divisible by 8 but not by 16.

Proof
   Assume that such a tesseract (aijkh), where i, j, k, and h are integers from 0 to m-1, can exist.
   Define a tesseract (Axyzw), where x, y, z, and w are integers from 0 to 7, as
Axyzw = Σi,j,k,h=0,m/8(a8i+x,8j+y,8k+z,8h+w).

   (Axyzw) becomes an order-8 non-normal Nasik magic tesseract whose magic constant S is given by
S = (m/8)4s = (m/8)4m(m4+1) / 2,

where s is the magic constant of the tesseract (aijkh). Note that S is not divisible by 8 because m is even and not divisible by 16.

   The 3-dimensional slices (A0yzw), (Ax4zw), (Axy0w), and (Axyz4) are Nasik magic cube because the tesseract (Axyzw) is Nasik, so we get the following equations
A0000 + A0444 = S/4,
A0444 + A4400 = S/4,
A4400 + A0004 = S/4,
A0004 + A4444 = S/4,

from the Theorem 1. Similarly, the 3-dimensional oblique slice (Axxzw) are also pan-2,3-agonal magic cube because (Axyzw) is strictly panmagic, so we have
A4444 + A0000 = S/4.

From these equations, we have
A0000 = S/8,

but S/8 is not an integer. That is contradiction.    Q.E.D.

Note 2
   An order-m (normal) Nasik magic tesseract can exist if m is divisible by 16x (see the page to construct Nasik magic tesseracts). The first one is the order-16 Nasik magic tesseract of 1998 by John R. Hendricks. He also proved that (normal) Nasik magic tesseracts cannot exist for any order lower than 16.

Note 3   [Added on October 1, 2007]
   The following theorem also holds which is stronger than Theorems 2 and 3:
If m is even and not divisible by 16, an order-m normal magic tesseract can be neither pan-2,3-agonal nor pan-2,4-agonal.
   It is unknown whether a normal pan-2-agonal magic tesseract can exist or not for order divisible by 8 but not by 16.

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