Order 32 Hexamagic Series Impossibility

Theorem. There are no order 32 hexamagic series.
Proof.
S1 = 16400
S2 = 11201200       ≡ 7 (mod 9)
S3 = 8606720000     ≡ 2 (mod 9)
S4 = 7054065471760  ≡ 7 (mod 9)
S5 = 6022406005760000
S6 = 5288530297464091600

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

Let 3Aj+1, j=1..9(a+b)+7,
 be the squares of entries that are 1 or 2 (mod 3).
Let 9Bj, j=1..9c+7,
 be the squares of entries that are 0 (mod 3).

Let Tn = sum (Aj)n, j=1..9(a+b)+7,  n=1,2,3.
Let Un = sum (Bj)n, j=1..9c+7,      n=1,2,3.

  S2 = 3T1 + 9(a+b)+7 + 9U1
  S6 = 27T3 + 27T2 + 9T1 + 9(a+b)+7 + 729U3

(S2 - 7)/3 =
  T1 + 3U1 + 3(a+b) = 3733731 ≡ 0 (mod 3)
thus
  T1 ≡ 0 (mod 3).

(S6 - 7)/9 =
  3T3 + 3T2 + T1 + 81U3 + (a+b) = 587614477496010177 ≡ 0 (mod 3)
thus
  (a+b) ≡ 0 (mod 3).


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

An order 32 hexamagic series must consist of
  7 entries of 2 (mod 3),
  0 entries of 1 (mod 3), and
 25 entries of 0 (mod 3).


Starting over ...

Let 3Aj+2 be an entry of 2 (mod 3), j=1..7.
Let 3Bj be an entry of 0 (mod 3), j=1..25.

Let Tn = sum (Aj)n, j=1..7,  n=1..6.
Let Un = sum (Bj)n, j=1..25, n=1..6.

  S2 = 9T2 + 12T1 + 28 + 9U2
  S3 = 27T3 + 54T2 + 36T1 + 56 + 27U3

(S2 - 28)/3 =
  3T2 + 4T1 + 3U2 = 3733724 ≡ 2 (mod 3)
thus
  T1 ≡ 2 (mod 3).

(S3 - 56)/9 =
  3T3 + 6T2 + 4T1 + 3U3 = 956302216 ≡ 1 (mod 3)
thus
  T1 ≡ 1 (mod 3).

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