Order 29 Hexamagic Series Impossibility

Theorem. There are no order 29 hexamagic series.
Proof.
S1 = 12209
S2 = 6849249
S3 = 4322730749
S4 = 2910060970377
S5 = 2040679054718669
S6 = 1471911610303651209

S4 ≡ 9 (mod 16), thus there must be either
9 or 25 odd entries in a hexamagic series


Case 1 of 2: 4 even entries and 25 odd entries.

Let 8Aj+1 be the square of an odd entry,  j=1..25
Let 4Bj   be the square of an even entry, j=1..4
Let Tn = sum (Aj)n, j=1..25, n=1,2,3
Let Un = sum (Bj)n, j=1..4,  n=1,2,3

S2 = 8T1 + 25 + 4U1
S4 = 64T2 + 16T1 + 25 + 16U2
S6 = 512T3 + 192T2 + 24T1 + 25 + 64U3

(S2 - 25)/4 = 
  2T1 + U1 = 1712306
thus
  U1 is even
and
  U2 is even.

(S4 - 25)/16 =
  4T2 + T1 + U2 = 181878810647
thus
  T1 is odd.

(S6 - 25)/8 =
  64T3 + 24T2 + 3T1 + 8U3 = 183988951287956398
thus
  T1 is even.

T1 can't be both odd and even.


Case 2 of 2: 20 even entries and 9 odd entries.

Let 2Aj+1 be an odd entry,  j=1..9
Let 2Bj   be an even entry, j=1..20
Let Tn = sum (Aj)n, j=1..9,  n=1..6
Let Un = sum (Bj)n, j=1..20, n=1..6

S1 = 2T1 + 9 + 2U1
S2 = 4T2 + 4T1 + 9 + 4U2
S3 = 8T3 + 12T2 + 6T1 + 9 + 8U3
S4 = 16T4 + 32T3 + 24T2 + 8T1 + 9 + 16U4
S5 = 32T5 + 80T4 + 80T3 + 40T2 + 10T1 + 9 + 32U5
S6 = 64T6 + 192T5 + 240T4 + 160T3 + 60T2 + 12T1 + 9 + 64U6

(S2 - 9)/4 =
  T2 + T1 + U2 = 1712310
Since T2 + T1 is even, U2 is even and
  U1...U6 are even.

(S1 - 9)/2 =
  T1 + U1 = 6100
thus T1 is even and
  T1...T6 are even.

(S3 - 9)/4 =
  2T3 + 3T2 + 3T1/2 + 2U3 = 1080682685
thus
  T1/2 is odd.

(S4 - 9)/16 =
  T4 + 2T3 + 3T2/2 + T1/2 + U4 = 181878810648
thus
  T2/2 is odd.

(S1 - 9)/4 =
  T1/2 + U1/2 = 3050
thus
  U1/2 is odd.

(S2 - 9)/8 =
  T2/2 + T1/2 + U2/2 = 856155
thus
  U2/2 is odd.

Let
  T1/2 = 2a+1,
  T2/2 = 2b+1.
Since
  T4 = T2 = 2 (mod 4),
  U4 = U2 = 2 (mod 4),
Let
  T4 = 4e+2,
  U4 = 4f+2.

(S6 - 9)/8 =
  8T6 + 24T5 + 30(4e+2) + 20T3 + 15(2b+1) + 3(2a+1) + 8U6 = 183988951287956400
thus
((S6 - 9)/8 - 78)/2 =
  4T6 + 12T5 + 60e + 10T3 + 15b + 3a + 4U6 = 91994475643978161
thus
  a + b is odd.

(S4 - 9)/16 =
  4e+2 + 2T3 + 3(2b+1) + 2a+1 + 4f+2 = 181878810648
thus
((S4 - 9)/16 - 8)/2 =
  2e + T3 + 3b + a + 2f = 90939405320
thus
  a + b is even,

a + b can't be both even and odd.