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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 a_{000} and a_{444} 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 8`x` 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 (a_{ijkh}), where i, j, k, and h are integers from 0 to 7, were possible, the two 3-dimensional slices of the tesseract, (a_{0jkh}) and (a_{i0kh}), would be Nasik magic cubes.

From the Theorem 1, the pair, a_{0000} and a_{0444}, would be a complement pair, and the pair, a_{0000} and a_{4044}, would be, too. Thus a_{0444} would equal to a_{4044}, so (a_{ijkh}) 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 (a_{ijkh}), where i, j, k, and h are integers from 0 to m-1, can exist.

Define a tesseract (A_{xyzw}), where x, y, z, and w are integers from 0 to 7, as

A_{xyzw} = Σ_{i,j,k,h=0,m/8}(a_{8i+x,8j+y,8k+z,8h+w}).

(A_{xyzw}) becomes an order-8 non-normal Nasik magic tesseract whose magic constant S is given by

S = (m/8)^{4}s = (m/8)^{4}m(m^{4}+1) / 2,

where s is the magic constant of the tesseract (a_{ijkh}).^{ } Note that S is **not** divisible by 8 because m is even and **not** divisible by 16.

The 3-dimensional slices (A_{0yzw}), (A_{x4zw}), (A_{xy0w}), and (A_{xyz4}) are Nasik magic cube because the tesseract (A_{xyzw}) is Nasik, so we get the following equations

A_{0000} + A_{0444} = S/4,

A_{0444} + A_{4400} = S/4,

A_{4400} + A_{0004} = S/4,

A_{0004} + A_{4444} = S/4,

from the Theorem 1.^{ } Similarly, the 3-dimensional oblique slice (A_{xxzw}) are also pan-2,3-agonal magic cube because (A_{xyzw}) is strictly panmagic, so we have

A_{4444} + A_{0000} = S/4.

From these equations, we have

A_{0000} = 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 16`x` (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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Magic Cubes and Tesseracts