Order 39 Hexamagic Series Impossibility

Theorem. There are no order 39 hexamagic series.
Proof.
S1 = 29679
S2 = 30104399         ≡ 2 (mod 9)
S3 = 34352878599      ≡ 0 (mod 9)
S4 = 41814317809823   ≡ 5 (mod 9)
S5 = 53017060913905959
S6 = 69141806363895432479

From the Modulo 9/3 Tetramagic Series Lemma,
this hexamagic series must have
  9a+4 entries of 2 (mod 3),
  9b+4 entries of 1 (mod 3), and
  9c+4 entries of 0 (mod 3),
     where a+b+c = 3.

Let 3Aj+1, j=1..9(a+b)+8,
 be the squares of entries that are 1 or 2 (mod 3).
Let 9Bj, j=1..remaining,
 be the squares of entries that are 0 (mod 3).

Let Tn = sum (Aj)n, j=1..9(a+b)+8,  n=1,2,3.
Let Un = sum (Bj)n, j=1..remaining, n=1,2,3.

  S2 = 3T1 + 9(a+b)+8 + 9U1
  S6 = 27T3 + 27T2 + 9T1 + 9(a+b)+8 + 729U3

(S2 - 8)/3 =
  T1 + 3U1 + 3(a+b) = 10034797 ≡ 1 (mod 3)
thus
  T1 ≡ 1 (mod 3)

(S6 - 8)/9 =
  3T3 + 3T2 + T1 + 81U3 + (a+b) = 7682422929321714719 ≡ 2 (mod 3)
thus
  (a+b) ≡ 1 (mod 3).

a = 1 or b = 1.


Case of a=1,b=0
where a hexamagic series consists of
  13 entries of 2 (mod 3)
   4 entries of 1 (mod 3)
  22 entries of 0 (mod 3)

Let 3Aj+2 be an entry of 2 (mod 3), j=1..13
Let 3Bj+1 be an entry of 1 (mod 3), j=1..4
Let 3Cj   be an entry of 0 (mod 3), j=1..22
Let Tn = sum (Aj)n, j=1..13, n=1..6
Let Un = sum (Bj)n, j=1..4, n=1..6
Let Vn = sum (Cj)n, j=1..22, n=1..6


S3 = 27T3 + 54T2 + 36T1 + 8x13 +
     27U3 + 27U2 + 9U1 + 4 + 27V3

S5 = 243T5 + 5x81x2T4 + 10x27x4T3 + 10x9x8T2 + 5x3x16T1 + 32x13 +
     243U5 + 5x81U4 + 10x27U3 + 10x9U2 + 5x3U1 + 4 + 243V5

(S3 - 108)/9 =
  3T3 + 6T2 + 4T1 + 3U3 + 3U2 + U1 + 3V3 = 3816986499 ≡ 0 (mod 3)
thus
  T1 + U1 ≡ 0 (mod 3).

(S5 - 420)/3 =
  81T5 + 5x27x2T4 + 10x9x4T3 + 10x3x8T2 + 5x16T1 +
  81U5 + 5x27U4 + 10x9U3 + 10x3U2 + 5U1 + 81V5 = 17672353637968513 ≡ 1 (mod 3)
thus
  2T1 + 2U1 ≡ 1 (mod 3)
or
  T1 + U1 ≡ 2 (mod 3).

T1 + U1 can't be both 0 and 2 (mod 3).


Case of a=0,b=1
where a hexamagic series consists of
  4 entries of 2 (mod 3)
 13 entries of 1 (mod 3)
 22 entries of 0 (mod 3)

Let 3Aj+2 be an entry of 2 (mod 3), j=1..4
Let 3Bj+1 be an entry of 1 (mod 3), j=1..13
Let 3Cj   be an entry of 0 (mod 3), j=1..22
Let Tn = sum (Aj)n, j=1..4, n=1..6
Let Un = sum (Bj)n, j=1..13, n=1..6
Let Vn = sum (Cj)n, j=1..22, n=1..6


S3 = 27T3 + 54T2 + 36T1 + 32 +
     27U3 + 27U2 + 9U1 + 13 + 27V3

S5 = 243T5 + 5x81x2T4 + 10x27x4T3 + 10x9x8T2 + 5x3x16T1 + 128 +
     243U5 + 5x81U4 + 10x27U3 + 10x9U2 + 5x3U1 + 13 + 243V5

(S3 - 45)/9 =
  3T3 + 6T2 + 4T1 + 3U3 + 3U2 + U1 + 3V3 = 3816986506 ≡ 1 (mod 3)
thus
  T1 + U1 ≡ 1 (mod 3).

(S5 - 141)/3 =
  81T5 + 5x27x2T4 + 10x9x4T3 + 10x3x8T2 + 5x16T1 +
  81U5 + 5x27U4 + 10x9U3 + 10x3U2 + 5U1 + 81V5 = 17672353637968606 ≡ 1 (mod 3)
thus
  2T1 + 2U1 ≡ 1 (mod 3)
or
  T1 + U1 ≡ 2 (mod 3).

T1 + U1 can't be both 1 and 2 (mod 3).