Order 35 Hexamagic Series Impossibility

Theorem. There are no order 35 hexamagic series.
Proof.
S1 = 1455
S2 = 17528735       ≡ 2 (mod 9)
S3 = 16111095875    ≡ 8 (mod 9)
S4 = 15795314890103 ≡ 2 (mod 9)
S5 = 16130960856213875
S6 = 16944416526780308855

From the Modulo 9/3 Tetramagic Series Lemma,
this hexamagic series must have
  9a+6 entries of 2 (mod 3),
  9b+5 entries of 1 (mod 3), and
  9c+6 entries of 0 (mod 3),
     where a+b+c = 2.

Let 3Aj+1, j=1..9(a+b)+11,
 be the squares of entries that are 1 or 2 (mod 3).
Let 9Bj, j=1..9c+6,
 be the squares of entries that are 0 (mod 3).

Let Tn = sum (Aj)n, j=1..9(a+b)+11,  n=1,2,3.
Let Un = sum (Bj)n, j=1..9c+6,       n=1,2,3.

  S2 = 3T1 + 9(a+b)+11 + 9U1
  S6 = 27T3 + 27T2 + 9T1 + 9(a+b)+11 + 729U3

(S2 - 11)/3 =
  T1 + 3U1 + 3(a+b) = 5842908 ≡ 0 (mod 3)
thus
  T1 ≡ 0 (mod 3).

(S6 - 11)/9 =
  3T3 + 3T2 + T1 + 81U3 + (a+b) = 1882712947420034316 ≡ 0 (mod 3)
thus
  (a+b) ≡ 0 (mod 3).


Since a+b <= 2,
a = 0 and b = 0.

An order 35 hexamagic series must consist of
  6 entries of 2 (mod 3),
  5 entries of 1 (mod 3), and
 24 entries of 0 (mod 3).


Starting over ...

Let 3Aj+2 be an entry of 2 (mod 3), j=1..6.
Let 3Bj+1 be an entry of 1 (mod 3), j=1..5.
Let 3Cj   be an entry of 0 (mod 3), j=1..24.

Let Tn = sum (Aj)n, j=1..6,  n=1..6.
Let Un = sum (Bj)n, j=1..5,  n=1..6.
Let Vn = sum (Cj)n, j=1..24, n=1..6.

  S3 = (27T3 + 54T2 + 36T1 + 48) +
       (27U3 + 27U2 + 9U1 + 5) + 27V3
  S4 = (81T4 + 4x27x2T3 + 6x9x4T2 + 4x3x8T1 + 96) +
       (81U4 + 4x27U3 + 6x9U2 + 4x3U1 + 5) + 81V4
  S5 = (243T5 + 5x81x2T4 + 10x27x4T3 + 10x9x8T2 + 5x3x16T1 + 192) +
       (243U5 + 5x81U4 + 10x27U3 + 10x9U2 + 5x3U1 + 5) + 243V5

(S3 - 53)/9 =
  3T3 + 6T2 + 4T1 +
  3U3 + 3U2 + U1 + 3V3 = 1790121758 ≡ 2 (mod 3)
thus
  T1 + U1 ≡ 2 (mod 3).

(S4 - 101)/3 =
  27T4 + 4x9x2T3 + 6x3x4T2 + 4x8T1 +
  27U4 + 4x9U3 + 6x3U2 + 4U1 + 27V4 = 5265104963334 ≡ 0 (mod 3)
thus
  2T1 + U1 ≡ 0 (mod 3)
and
  T1 ≡ 1 (mod 3),
  U1 ≡ 1 (mod 3).

(S5 - 197)/3 =
  81T5 + 5x27x2T4 + 10x9x4T3 + 10x3x8T2 + 5x16T1 +
  81U5 + 5x27U4 + 10x9U3 + 10x3U2 + 5U1 + 81V5 = 5376986952071226 ≡ 0 (mod 3)
thus
  2T1 + 2U1 ≡ 0 (mod 3),
which is inconsistent with T1 ≡ 1, U1 ≡ 1 (mod 3).