Order 33 Hexamagic Series Impossibility

Theorem. There are no order 33 hexamagic series.
Proof.
S1 = 17985
S2 = 13063105
S3 = 10674187425
S4 = 9303619147009
S5 = 8446907918837025
S6 = 7888201972342328065

S4 ≡ 1 (mod 16), thus there are
1, 17, or 33 odd entries in a hexamagic series.


Case 1 of 3:  33 odd entries, 0 even entries.

Let 8Aj+1 be the square of an odd entry, j=1..33.
Let Tn = sum (Aj)n, j=1..33, n=1,2,3.

  S2 = 8T1 + 33
  S4 = 64T2 + 16T1 + 33
  S6 = 512T3 + 192T2 + 24T1 + 33

(S2 - 33)/8 =
  T1 = 1632884
thus
  T1,T2,T3 are even

(S4 - 33)/32 =
  2T2 + T1/2 = 290738098343
thus
  T1/2 is odd.

(S6 - 33)/16 =
  32T3 + 12T2 + 3T1/2 = 493012623271395502
thus
  T1/2 is even.

T1/2 can't be both odd and even.


Case 2 of 3: 17 odd entries and 16 even entries.

Let 8Aj+1 be the square of an odd entry, j=1..17.
Let 4Bj   be the square of an even entry, j=1..16.
Let Tn = sum (Aj)n, j=1..17, n=1,2,3.
Let Un = sum (Bj)n, j=1..16, n=1,2,3.

  S2 = 8T1 + 17 + 4U1
  S4 = 64T2 + 16T1 + 17 + 16U2
  S6 = 512T3 + 192T2 + 24T1 + 17 + 64U3

(S6 - 17)/8 =
  64T3 + 24T2 + 3T1 + 8U3 = 986025246542791006
thus
  T1 is even.

(S4 - 17)/16 =
  4T2 + T1 + U2 = 581476196687
thus
  U2 is odd and U1 is odd.

(S2 - 17)/4 =
  2T1 + U1 = 3265772
thus
  U1 is even.

U1 can't be both odd and even.


Case 3 of 3: 1 odd and 32 even entries.

Let 2Aj be an even entry, j=1..32.
Let 2B+1 be the odd entry.
Let Tn = sum (Aj)n, j=1..32, n=1..6.

  S1 =  2T1 +  2B + 1
  S2 =  4T2 +  4B2 +   4B + 1
  S3 =  8T3 +  8B3 +  12B2 + 6B + 1
  S4 = 16T4 + 16B4 +  32B3 + 24B2 + 8B + 1
  S5 = 32T5 + 32B5 +  80B4 + 80B3 + 40B2 + 10B + 1
  S6 = 64T6 + 64B6 + 192B5 + 240B4 + 160B3 + 60B2 + 12B + 1

(S3 - 1)/2 =
  4T3 + 4B3 + 6B2 + 3B = 5337093712
thus
  B is even.

(S1 - 1)/2 =
  T1 + B = 8992
thus
  T1 is even and
  T1..T6 are even.

(S5 - 1)/4 =
  8T5 + 8B5 + 20B4 + 20B3 + 10B2 + 5B/2 = 2111726979709256
thus
  B/2 is even.

(S2 - 1)/8 =
  T2/2 + B2/2 + B/2 = 1632888
thus
  T2/2 is even.

(S1 - 1)/4 =
  T1/2 + B/2 = 4496
thus
  T1/2 is even.

(S5 - 1)/8 =
  4T5 + 4B5 + 10B4 + 10B3 + 5B2 + 5B/4 = 1055863489854628
thus
  B/4 is even.

(S4 - 1)/32 =
  T4/2 + B4/2 + B3 + 3B2/4 + B/4 = 290738098344
thus
  T4/2 is even.

(S2 - 1)/16 =
  T2/4 + B2/4 + B/4 = 816444
thus
  T2/4 is even.

(S1 - 1)/8 =
  T1/4 + B/4 = 2248
thus
  T1/4 is even.

(S6 - 1)/32 =
  2T6 + 2B6 + 6B5 + 15B4/2 + 5B3 + 15B2/8 + 3B/8 = 246506311635697752
thus
  B/8 is even.

(S4 - 1)/64 =
  T4/4 + B4/4 + B3/2 + 3B2/8 + B/8 = 145369049172
thus
  T4/4 is even.

(S3 - 1)/16 =
  T3/2 + B3/2 + 3B2/4 + 3B/8 = 667136714
thus
  T3/2 is even.

(S2 - 1)/32 =
  T2/8 + B2/8 + B/8 = 408222
thus
  T2/8 is even.

(S1 - 1)/16 =
  T1/8 + B/8 = 1124
thus
  T1/8 is even.

(S6 - 1)/64 =
  T6 + B6 + 3B5 + 15B4/4 + 5B3/2 + 15B2/16 + 3B/16 = 123253155817848876
thus
  B/16 is even.

(S5 - 1)/32 =
  T5 + B5 + 5B4/2 + 5B3/2 + 5B2/4 + 5B/16 = 263965872463657
thus
  T5 is odd,
contradicting an earlier deduction that T1..T6 are even.