Order 43 Hexamagic Series Impossibility

Theorem. There are no order 43 hexamagic series.
Proof.
S1 = 39775
S2 = 49042575        ≡ 0 (mod 9)
S3 = 68028176875     ≡ 4 (mod 9)
S4 = 100654480695735 ≡ 0 (mod 9)
S5 = 155133699475586875
S6 = 245931201634915761975

From the Modulo 9/3 Tetramagic Series Lemma,
there must be
  9a+7 entries of 2 (mod 3),
  9b+2 entries of 1 (mod 3), and
  9c+7 entries of 0 (mod 3),
    where a+b+c = 3.

Let 3Aj+2 be an entry of 2 (mod 3), j=1..9a+7
Let 3Bj+1 be an entry of 1 (mod 3), j=1..9b+2
Let 3Cj   be an entry of 0 (mod 3), j=1..9c+7
  where a+b+c = 3

Let Tn = sum (Aj)n, j=1..9a+7, n=1..6
Let Un = sum (Bj)n, j=1..9b+2, n=1..6
Let Vn = sum (Cj)n, j=1..9c+7, n=1..6

S2 = 9T2 + 12T1 + 4(9a+7) + 9U2 + 6U1 + (9b+2) + 9V2
S3 = 27T3 + 54T2 + 36T1 + 8(9a+7) + 27U3 + 27U2 + 9U1 + (9b+2) + 27V3
S5 = 243T5 + 810T4 + 1080T3 + 720T2 + 240T1 + 32(9a+7) +
     243U5 + 405U4 + 270U3 + 90U2 + 15U1 + (9b+2) + 243V5
S6 = 729T6 + 2916T5 + 4860T4 + 4320T3 + 2160T2 + 576T1 + 64(9a+7) +
     729U6 + 1458U5 + 1215U4 + 540U3 + 135U2 + 18U1 + (9b+2) + 729V6

(S2 - 30)/3 =
  3T2 + 4T1 + 12a + 3U2 + 2U1 + 3b + 3U6 = 16347515 ≡ 2 (mod 3)
thus
  T1 + 2U1 ≡ 2 (mod 3)

(S5 - 226)/3 =
  81T5 + 270T4 + 360T3 + 240T2 + 80T1 + 96a +
  81U5 + 135U4 + 90U3 + 30U2 + 5U1 + 3b + 81V5 = 51711233158528883 ≡ 2 (mod 3)
thus
  2T1 + 2U1 ≡ 2 (mod 3)
and when combined with T1 + 2U1 ≡ 2 (mod 3),
  T1 ≡ 0 (mod 3)
  U1 ≡ 1 (mod 3)

(S3 - 58)/9 =
  3T3 + 6T2 + 4T1 + 8a + 3U3 + 3U2 + U1 + b + 3V3 = 7558686313 ≡ 1 (mod 3)
thus
  2a + b ≡ 0 (mod 3)

(S6 - 450)/9 =
  81T6 + 324T5 + 540T4 + 480T3 + 240T2 + 64T1 + 64a +
  81U6 + 162U5 + 135U4 + 60U3 + 15U2 + 2U1 + b + 81V6 = 27325689070546195725 ≡ 0 (mod 3)
thus
  a + b ≡ 1 (mod 3)
and when combined with 2a + b ≡ 0 (mod 3),
  a ≡ 2 (mod 3)
  b ≡ 2 (mod 3)
which contradicts
  a+b <= 3