Completeness theorem on order-8 Nasik magic cubes (1/3)

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Completeness theorem on order-8 Nasik magic cubes (1/3)

The proof of the theorem needs the following lemma on order-8 panmagic squares.

Lemma 1
Let (a_ij) be an order-8 panmagic square (normal or non-normal), where i and j are integers from 0 to 7,
then the sum of eight cells a₀₀, a₀₄, a₂₂, a₂₆, a₄₀, a₄₄, a₆₂, and a₆₆ equals to the magic constant.
The sum of eight cells a₀₁, a₀₅, a₂₃, a₂₇, a₄₁, a₄₅, a₆₃, and a₆₇ also does.

a₀₀	a₀₁	a₀₂	a₀₃	a₀₄	a₀₅	a₀₆	a₀₇
a₁₀	a₁₁	a₁₂	a₁₃	a₁₄	a₁₅	a₁₆	a₁₇
a₂₀	a₂₁	a₂₂	a₂₃	a₂₄	a₂₅	a₂₆	a₂₇
a₃₀	a₃₁	a₃₂	a₃₃	a₃₄	a₃₅	a₃₆	a₃₇
a₄₀	a₄₁	a₄₂	a₄₃	a₄₄	a₄₅	a₄₆	a₄₇
a₅₀	a₅₁	a₅₂	a₅₃	a₅₄	a₅₅	a₅₆	a₅₇
a₆₀	a₆₁	a₆₂	a₆₃	a₆₄	a₆₅	a₆₆	a₆₇
a₇₀	a₇₁	a₇₂	a₇₃	a₇₄	a₇₅	a₇₆	a₇₇

a₀₀	a₀₁	a₀₂	a₀₃	a₀₄	a₀₅	a₀₆	a₀₇
a₁₀	a₁₁	a₁₂	a₁₃	a₁₄	a₁₅	a₁₆	a₁₇
a₂₀	a₂₁	a₂₂	a₂₃	a₂₄	a₂₅	a₂₆	a₂₇
a₃₀	a₃₁	a₃₂	a₃₃	a₃₄	a₃₅	a₃₆	a₃₇
a₄₀	a₄₁	a₄₂	a₄₃	a₄₄	a₄₅	a₄₆	a₄₇
a₅₀	a₅₁	a₅₂	a₅₃	a₅₄	a₅₅	a₅₆	a₅₇
a₆₀	a₆₁	a₆₂	a₆₃	a₆₄	a₆₅	a₆₆	a₆₇
a₇₀	a₇₁	a₇₂	a₇₃	a₇₄	a₇₅	a₇₆	a₇₇

Proof
   It is enough to prove the first feature only, because the second feature is reduced to the first one by shifting the first column to the last (the cube that the shift is applied to is still pandiagonal).
Let s be the magic constant.
The conditions of orthogonals provide the following equations:
(a₀₀ + a₀₁ + a₀₂ + a₀₃ + a₀₄ + a₀₅ + a₀₆ + a₀₇) +
(a₂₀ + a₂₁ + a₂₂ + a₂₃ + a₂₄ + a₂₅ + a₂₆ + a₂₇) +
(a₄₀ + a₄₁ + a₄₂ + a₄₃ + a₄₄ + a₄₅ + a₄₆ + a₄₇) +
(a₆₀ + a₆₁ + a₆₂ + a₆₃ + a₆₄ + a₆₅ + a₆₆ + a₆₇) = 4s,    (1.1)

(a₀₁ + a₁₁ + a₂₁ + a₃₁ + a₄₁ + a₅₁ + a₆₁ + a₇₁) +
(a₀₃ + a₁₃ + a₂₃ + a₃₃ + a₄₃ + a₅₃ + a₆₃ + a₇₃) +
(a₀₅ + a₁₅ + a₂₅ + a₃₅ + a₄₅ + a₅₅ + a₆₅ + a₇₅) +
(a₀₇ + a₁₇ + a₂₇ + a₃₇ + a₄₇ + a₅₇ + a₆₇ + a₇₇) = 4s.    (1.2)

Similarly, the conditions of (pan)diagonals provide the following equation:
(a₀₀ + a₁₁ + a₂₂ + a₃₃ + a₄₄ + a₅₅ + a₆₆ + a₇₇) +
(a₂₀ + a₃₁ + a₄₂ + a₅₃ + a₆₄ + a₇₅ + a₀₆ + a₁₇) +
(a₄₀ + a₅₁ + a₆₂ + a₇₃ + a₀₄ + a₁₅ + a₂₆ + a₃₇) +
(a₆₀ + a₇₁ + a₀₂ + a₁₃ + a₂₄ + a₃₅ + a₄₆ + a₅₇) = 4s.    (1.3)

(1.1)
a₀₀	a₀₁	a₀₂	a₀₃	a₀₄	a₀₅	a₀₆	a₀₇
a₁₀	a₁₁	a₁₂	a₁₃	a₁₄	a₁₅	a₁₆	a₁₇
a₂₀	a₂₁	a₂₂	a₂₃	a₂₄	a₂₅	a₂₆	a₂₇
a₃₀	a₃₁	a₃₂	a₃₃	a₃₄	a₃₅	a₃₆	a₃₇
a₄₀	a₄₁	a₄₂	a₄₃	a₄₄	a₄₅	a₄₆	a₄₇
a₅₀	a₅₁	a₅₂	a₅₃	a₅₄	a₅₅	a₅₆	a₅₇
a₆₀	a₆₁	a₆₂	a₆₃	a₆₄	a₆₅	a₆₆	a₆₇
a₇₀	a₇₁	a₇₂	a₇₃	a₇₄	a₇₅	a₇₆	a₇₇

(1.2)
a₀₀	a₀₁	a₀₂	a₀₃	a₀₄	a₀₅	a₀₆	a₀₇
a₁₀	a₁₁	a₁₂	a₁₃	a₁₄	a₁₅	a₁₆	a₁₇
a₂₀	a₂₁	a₂₂	a₂₃	a₂₄	a₂₅	a₂₆	a₂₇
a₃₀	a₃₁	a₃₂	a₃₃	a₃₄	a₃₅	a₃₆	a₃₇
a₄₀	a₄₁	a₄₂	a₄₃	a₄₄	a₄₅	a₄₆	a₄₇
a₅₀	a₅₁	a₅₂	a₅₃	a₅₄	a₅₅	a₅₆	a₅₇
a₆₀	a₆₁	a₆₂	a₆₃	a₆₄	a₆₅	a₆₆	a₆₇
a₇₀	a₇₁	a₇₂	a₇₃	a₇₄	a₇₅	a₇₆	a₇₇

(1.3)
a₀₀	a₀₁	a₀₂	a₀₃	a₀₄	a₀₅	a₀₆	a₀₇
a₁₀	a₁₁	a₁₂	a₁₃	a₁₄	a₁₅	a₁₆	a₁₇
a₂₀	a₂₁	a₂₂	a₂₃	a₂₄	a₂₅	a₂₆	a₂₇
a₃₀	a₃₁	a₃₂	a₃₃	a₃₄	a₃₅	a₃₆	a₃₇
a₄₀	a₄₁	a₄₂	a₄₃	a₄₄	a₄₅	a₄₆	a₄₇
a₅₀	a₅₁	a₅₂	a₅₃	a₅₄	a₅₅	a₅₆	a₅₇
a₆₀	a₆₁	a₆₂	a₆₃	a₆₄	a₆₅	a₆₆	a₆₇
a₇₀	a₇₁	a₇₂	a₇₃	a₇₄	a₇₅	a₇₆	a₇₇

By the calculation of ((1.1)-(1.2)+(1.3))/2, we get the following equation:
(a₀₀ + a₀₂ + a₀₄ + a₀₆) + (a₂₀ + a₂₂ + a₂₄ + a₂₆) +
(a₄₀ + a₄₂ + a₄₄ + a₄₆) + (a₆₀ + a₆₂ + a₆₄ + a₆₆) = 2s. (1.4)

(1.4)
a₀₀	a₀₁	a₀₂	a₀₃	a₀₄	a₀₅	a₀₆	a₀₇
a₁₀	a₁₁	a₁₂	a₁₃	a₁₄	a₁₅	a₁₆	a₁₇
a₂₀	a₂₁	a₂₂	a₂₃	a₂₄	a₂₅	a₂₆	a₂₇
a₃₀	a₃₁	a₃₂	a₃₃	a₃₄	a₃₅	a₃₆	a₃₇
a₄₀	a₄₁	a₄₂	a₄₃	a₄₄	a₄₅	a₄₆	a₄₇
a₅₀	a₅₁	a₅₂	a₅₃	a₅₄	a₅₅	a₅₆	a₅₇
a₆₀	a₆₁	a₆₂	a₆₃	a₆₄	a₆₅	a₆₆	a₆₇
a₇₀	a₇₁	a₇₂	a₇₃	a₇₄	a₇₅	a₇₆	a₇₇

On the other hand, the conditions of (pan)diagonals provide the following equations:
(a₀₀ + a₁₁ + a₂₂ + a₃₃ + a₄₄ + a₅₅ + a₆₆ + a₇₇) +
(a₄₀ + a₅₁ + a₆₂ + a₇₃ + a₀₄ + a₁₅ + a₂₆ + a₃₇) = 2s, (1.5)

(a₀₂ + a₁₁ + a₂₀ + a₃₇ + a₄₆ + a₅₅ + a₆₄ + a₇₃) +
(a₀₆ + a₁₅ + a₂₄ + a₃₃ + a₄₂ + a₅₁ + a₆₀ + a₇₇) = 2s. (1.6)

(1.5)
a₀₀	a₀₁	a₀₂	a₀₃	a₀₄	a₀₅	a₀₆	a₀₇
a₁₀	a₁₁	a₁₂	a₁₃	a₁₄	a₁₅	a₁₆	a₁₇
a₂₀	a₂₁	a₂₂	a₂₃	a₂₄	a₂₅	a₂₆	a₂₇
a₃₀	a₃₁	a₃₂	a₃₃	a₃₄	a₃₅	a₃₆	a₃₇
a₄₀	a₄₁	a₄₂	a₄₃	a₄₄	a₄₅	a₄₆	a₄₇
a₅₀	a₅₁	a₅₂	a₅₃	a₅₄	a₅₅	a₅₆	a₅₇
a₆₀	a₆₁	a₆₂	a₆₃	a₆₄	a₆₅	a₆₆	a₆₇
a₇₀	a₇₁	a₇₂	a₇₃	a₇₄	a₇₅	a₇₆	a₇₇

(1.6)
a₀₀	a₀₁	a₀₂	a₀₃	a₀₄	a₀₅	a₀₆	a₀₇
a₁₀	a₁₁	a₁₂	a₁₃	a₁₄	a₁₅	a₁₆	a₁₇
a₂₀	a₂₁	a₂₂	a₂₃	a₂₄	a₂₅	a₂₆	a₂₇
a₃₀	a₃₁	a₃₂	a₃₃	a₃₄	a₃₅	a₃₆	a₃₇
a₄₀	a₄₁	a₄₂	a₄₃	a₄₄	a₄₅	a₄₆	a₄₇
a₅₀	a₅₁	a₅₂	a₅₃	a₅₄	a₅₅	a₅₆	a₅₇
a₆₀	a₆₁	a₆₂	a₆₃	a₆₄	a₆₅	a₆₆	a₆₇
a₇₀	a₇₁	a₇₂	a₇₃	a₇₄	a₇₅	a₇₆	a₇₇

By the calculation of ((1.4)+(1.5)-(1.6)) / 2, we get the equation
a₀₀ + a₀₄ + a₂₂ + a₂₆ + a₄₀ + a₄₄ + a₆₂ + a₆₆ = s. Q.E.D.

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This page was last updated on October 1, 2007.
"Magic Cubes and Tesseracts" http://homepage2.nifty.com/googol/magcube/en/
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