Order 44 Hexamagic Series Impossibility

Theorem. There are no order 44 hexamagic series.
Proof.
S1 = 42614
S2 = 55014674
S3 = 79901931824
S4 = 123784061778806
S5 = 199756507500568304
S6 = 331567221598425169214

S4 ≡ 6 (mod 16) thus a hexamagic series must have
6, 22, or 38 odd entries.


Case 1 of 3: 6 odd and 38 even entries.

Let 8Aj+1 be the square of an odd entry,  j=1..6.
Let 4Bj   be the square of an even entry, j=1..38.
Let Tn = sum (Aj)n, j=1..6,  n=1,2,3.
Let Un = sum (Bj)n, j=1..38, n=1,2,3.

  S2 = 8T1 + 6 + 4U1
  S4 = 64T2 + 16T1 + 6 + 16U2
  S6 = 512T3 + 192T2 + 24T1 + 6 + 64U3

(S2 - 6)/4 =
  2T1 + U1 = 13753667
thus
  U1 is odd and U2 is odd.

(S4 - 6)/16 =
  4T2 + T1 + U2 = 7736503861175
thus
  T1 is even.

(S6 - 6)/8 =
  64T3 + 24T2 + 3T1 + 8U3 = 41445902699803146151
thus
  T1 is odd.

T1 can't be both even and odd.


Case 2 of 3: 6 even and 38 odd entries.

Let 8Aj+1 be the square of an odd entry,  j=1..38.
Let 4Bj   be the square of an even entry, j=1..6.
Let Tn = sum (Aj)n, j=1..38, n=1,2,3.
Let Un = sum (Bj)n, j=1..6,  n=1,2,3.

  S2 = 8T1 + 38 + 4U1
  S4 = 64T2 + 16T1 + 38 + 16U2
  S6 = 512T3 + 192T2 + 24T1 + 38 + 64U3

(S2 - 38)/4 =
  2T1 + U1 = 13753659
thus
  U1 is odd and U2 is odd.

(S4 - 38)/16 =
  4T2 + T1 + U2 = 7736503861173
thus
  T1 is even.

(S6 - 38)/8 =
  64T3 + 24T2 + 3T1 + 8U3 = 41445902699803146147
thus
  T1 is odd.

T1 can't be both even and odd.


Case 3 of 3: 22 even and 22 odd entries.

Let 2Aj+1 be an odd entry,  j=1..22
Let 2Bj   be an even entry, j=1..22
Let Tn = sum (Aj)n, j=1..22, n=1..6
Let Un = sum (Bj)n, j=1..22, n=1..6

S1 = 2T1 + 22 + 2U1
S2 = 4T2 + 4T1 + 22 + 4U2
S3 = 8T3 + 12T2 + 6T1 + 22 + 8U3
S4 = 16T4 + 32T3 + 24T2 + 8T1 + 22 + 16U4
S5 = 32T5 + 80T4 + 80T3 + 40T2 + 10T1 + 22 + 32U5
S6 = 64T6 + 6x32T5 + 15x16T4 + 20x8T3 + 15x4T2 + 6x2T1 + 22 + 64U6

(S2 - 22)/4 =
  T2 + T1 + U2 = 13753663
thus
  U2 is odd since T2 + T1 must be even and
  U1..U6 are odd.

(S1 - 22)/2 =
  T1 + U1 = 21296
thus T1 is odd and
  T1..T6 are odd.

Let T1 = 2z+1
Let T2 = 2w+1
Let U2 = 2v+1

(S5 - 22)/2 =
  16T5 + 40T4 + 40T3 + 20(2w+1) + 5(2z+1) + 16U5 = 99878253750284141
((S5 - 22)/2 - 25)/2 =
  8T5 + 20T4 + 20T3 + 20w + 5z + 8U5 = 49939126875142058
thus
  z is even.

(S6 - 22)/4 =
  16T6 + 48T5 + 60T4 + 40T3 + 15(2w+1) + 3(2z+1) + 16U6 = 82891805399606292298
((S6 - 22)/4 - 18)/2 =
  8T6 + 24T5 + 30T4 + 20T3 + 15w + 3z + 8U6 = 41445902699803146140
thus
  w is even.

(S2 - 22)/4 =
  2w+1 + 2z+1 + 2v+1 = 13753663
((S2 - 22)/4 - 3)/2 =
  w + z + v = 6876830
thus
  v is even.

Since
  T4 = T2 (mod 4)
  U4 = U2 (mod 4)
Let T4 = 4r+1
Let U4 = 4q+1

((S5 - 22)/2 - 25)/2 =
  8T5 + 20(4r+1) + 20T3 + 20w + 5z + 8U5 = 49939126875142058
(((S5 - 22)/2 - 25)/2 - 20)/2 =
  4T5 + 40r + 10T3 + 10w + 5z/2 + 4U5 = 24969563437571019
thus
  z/2 is odd.

((S6 - 22)/4 - 18)/2 =
  8T6 + 24T5 + 30(4r+1) + 20T3 + 15w + 3z + 8U6 = 41445902699803146140
(((S6 - 22)/4 - 18)/2 - 30)/2 = 
  4T6 + 12T5 + 60r + 10T3 + 15w/2 + 3z/2 + 4U6 = 20722951349901573055
thus
  w/2 is even.

(S4 - 22)/8 =
  2(4r+1) + 4T3 + 3(2w+1) + 2z+1 + 2(4q+1) = 15473007722348
((S4 - 22)/8 - 8)/2 = 
  4r + 2T3 + 3w + z + 4q = 7736503861170
((S4 - 22)/8 - 8)/4 =
  2r + T3 + 3w/2 + z/2 + 2q = 3868251930585
thus
  w/2 is odd.

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