Order 28 Hexamagic Series Impossibility

Theorem. There are no order 28 hexamagic series.
Proof.
S1 = 10990
S2 = 5747770       ≡ 1 (mod 9)
S3 = 3381842800    ≡ 1 (mod 9)
S4 = 2122443391726 ≡ 1 (mod 9)
S5 = 1387546427940400
S6 = 933025104315693910

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

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

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

  S2 = 3T1 + 9(a+b)+1 + 9U1
  S6 = 27T3 + 27T2 + 9T1 + 9(a+b)+1 + 729U3

(S2 - 1)/3 =
  T1 + 3U1 + 3(a+b) = 1915923 ≡ 0 (mod 3)
thus
  T1 ≡ 0 (mod 3).

(S6 - 1)/9 =
  3T3 + 3T2 + T1 + 81U3 + (a+b) = 103669456035077101 ≡ 1 (mod 3)
thus
  (a+b) ≡ 1 (mod 3).

We divide the proof into two cases, a=0,b=1 and a=1,b=0.


Case a=0, b=1,
where a hexamagic series consists of
  10 entries of 1 (mod 3) and
  18 entries of 0 (mod 3).

Let 3Aj+1 be an entry of 1 (mod 3), j=1..10.
Let 3Bj be an entry of 0 (mod 3), j=1..18.

Let Tn = sum (Aj)n, j=1..10, n=1..6.
Let Un = sum (Bj)n, j=1..18, n=1..6.

  S2 = 9T2 + 6T1 + 10 + 9U2
  S3 = 27T3 + 27T2 + 9T1 + 10 + 27U3

(S2 - 10)/3 =
  3T2 + 2T1 + 3U2 = 1915920 ≡ 0 (mod 3)
thus
  T1 ≡ 0 (mod 3).

(S3 - 10)/9 =
  3T3 + 3T2 + T1 + 3U3 = 375760310 ≡ 2 (mod 3)
thus
  T1 ≡ 2 (mod 3).

T1 can't be both 0 and 2 (mod 3).


Case a=1, b=0,
where a hexamagic series consists of
   9 entries of 2 (mod 3),
   1 entry   of 1 (mod 3), and
  18 entries of 0 (mod 3).

Let 3Aj+2 be an entry of 2 (mod 3), j=1..9.
Let 3Bj be an entry of 0 (mod 3), j=1..18.
Let 3C+1 be the entry of 1 (mod 3).

Let Tn = sum (Aj)n, j=1..9,  n=1..6.
Let Un = sum (Bj)n, j=1..18, n=1..6.

  S3 = (27T3 + 54T2 + 36T1 + 72) + 27U3 + (27C3 + 27C2 + 9C + 1)
  S5 = (243T5 + 810T4 + 1080T3 + 720T2 + 240T1 + 288) + 243U5 + 
         (243C5 + 405C4 + 270C3 + 90C2 + 15C + 1)

(S3 - 73)/9 =
  3T3 + 6T2 + 4T1 + 3U3 + 3C3 + 3C2 + C = 375760303 ≡ 1 (mod 3)
thus
  T1 + C ≡ 1 (mod 3).

(S5 - 289)/3 =
  81T5 + 270T4 + 360T3 + 240T2 + 80T1 + 81U5 +
      81C5 + 135C4 + 90C3 + 30C2 + 5C = 462515475980037 ≡ 0 (mod 3)
thus
  2T1 + 2C ≡ 0 (mod 3)
or
  T1 + C ≡ 0 (mod 3).

T1 + C can't be both 0 and 1 (mod 3).