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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 (aij) 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.

a00a01a02a03a04a05a06a07
a10a11a12a13a14a15a16a17
a20a21a22a23a24a25a26a27
a30a31a32a33a34a35a36a37
a40a41a42a43a44a45a46a47
a50a51a52a53a54a55a56a57
a60a61a62a63a64a65a66a67
a70a71a72a73a74a75a76a77
a00a01a02a03a04a05a06a07
a10a11a12a13a14a15a16a17
a20a21a22a23a24a25a26a27
a30a31a32a33a34a35a36a37
a40a41a42a43a44a45a46a47
a50a51a52a53a54a55a56a57
a60a61a62a63a64a65a66a67
a70a71a72a73a74a75a76a77

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)
a00a01a02a03a04a05a06a07
a10a11a12a13a14a15a16a17
a20a21a22a23a24a25a26a27
a30a31a32a33a34a35a36a37
a40a41a42a43a44a45a46a47
a50a51a52a53a54a55a56a57
a60a61a62a63a64a65a66a67
a70a71a72a73a74a75a76a77
(1.2)
a00a01a02a03a04a05a06a07
a10a11a12a13a14a15a16a17
a20a21a22a23a24a25a26a27
a30a31a32a33a34a35a36a37
a40a41a42a43a44a45a46a47
a50a51a52a53a54a55a56a57
a60a61a62a63a64a65a66a67
a70a71a72a73a74a75a76a77
(1.3)
a00a01a02a03a04a05a06a07
a10a11a12a13a14a15a16a17
a20a21a22a23a24a25a26a27
a30a31a32a33a34a35a36a37
a40a41a42a43a44a45a46a47
a50a51a52a53a54a55a56a57
a60a61a62a63a64a65a66a67
a70a71a72a73a74a75a76a77

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)
a00a01a02a03a04a05a06a07
a10a11a12a13a14a15a16a17
a20a21a22a23a24a25a26a27
a30a31a32a33a34a35a36a37
a40a41a42a43a44a45a46a47
a50a51a52a53a54a55a56a57
a60a61a62a63a64a65a66a67
a70a71a72a73a74a75a76a77

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)
a00a01a02a03a04a05a06a07
a10a11a12a13a14a15a16a17
a20a21a22a23a24a25a26a27
a30a31a32a33a34a35a36a37
a40a41a42a43a44a45a46a47
a50a51a52a53a54a55a56a57
a60a61a62a63a64a65a66a67
a70a71a72a73a74a75a76a77
(1.6)
a00a01a02a03a04a05a06a07
a10a11a12a13a14a15a16a17
a20a21a22a23a24a25a26a27
a30a31a32a33a34a35a36a37
a40a41a42a43a44a45a46a47
a50a51a52a53a54a55a56a57
a60a61a62a63a64a65a66a67
a70a71a72a73a74a75a76a77

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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