Order 36 Hexamagic Series Impossibility

Theorem. There are no order 36 hexamagic series.
Proof.
S1 = 23346
S2 = 20178726
S3 = 19621285776
S4 = 20351193571122
S5 = 21987763270443216
S6 = 24434679943257705546

S4 ≡ 2 (mod 16) thus a hexamagic series must have
2, 18, or 34 odd entries.


Case 1 of 3: 2 odd and 34 even entries.

Let 8Aj+1 be the square of an odd entry,  j=1..2.
Let 4Bj   be the square of an even entry, j=1..34.
Let Tn = sum (Aj)n, j=1..2,  n=1,2,3.
Let Un = sum (Bj)n, j=1..34, n=1,2,3.

  S2 = 8T1 + 2 + 4U1
  S4 = 64T2 + 16T1 + 2 + 16U2
  S6 = 512T3 + 192T2 + 24T1 + 2 + 64U3

(S2 - 2)/4 =
  2T1 + U1 = 5044681
thus
  U1 is odd and U2 is odd.

(S4 - 2)/16 =
  4T2 + T1 + U2 = 1271949598195
thus
  T1 is even.

(S6 - 2)/8 =
  64T3 + 24T2 + 3T1 + 8U3 = 3054334992907213193
thus
  T1 is odd.

T1 can't be both even and odd.


Case 2 of 3: 2 even and 34 odd entries.

Let 8Aj+1 be the square of an odd entry,  j=1..34.
Let 4Bj   be the square of an even entry, j=1..2.
Let Tn = sum (Aj)n, j=1..34, n=1,2,3.
Let Un = sum (Bj)n, j=1..2,  n=1,2,3.

  S2 = 8T1 + 34 + 4U1
  S4 = 64T2 + 16T1 + 34 + 16U2
  S6 = 512T3 + 192T2 + 24T1 + 34 + 64U3

(S2 - 34)/4 =
  2T1 + U1 = 5044673
thus
  U1 is odd and U2 is odd.

(S4 - 34)/16 =
  4T2 + T1 + U2 = 1271949598193
thus
  T1 is even.

(S6 - 34)/8 =
  64T3 + 24T2 + 3T1 + 8U3 = 3054334992907213189
thus
  T1 is odd.

T1 can't be both even and odd.


Case 3 of 3: 18 even and 18 odd entries.

Let 2Aj+1 be an odd entry,  j=1..18
Let 2Bj   be an even entry, j=1..18
Let Tn = sum (Aj)n, j=1..18, n=1..6
Let Un = sum (Bj)n, j=1..18, n=1..6

S1 = 2T1 + 18 + 2U1
S2 = 4T2 + 4T1 + 18 + 4U2
S3 = 8T3 + 12T2 + 6T1 + 18 + 8U3
S4 = 16T4 + 32T3 + 24T2 + 8T1 + 18 + 16U4
S5 = 32T5 + 80T4 + 80T3 + 40T2 + 10T1 + 18 + 32U5
S6 = 64T6 + 6x32T5 + 15x16T4 + 20x8T3 + 15x4T2 + 6x2T1 + 18 + 64U6

(S2 - 18)/4 =
  T2 + T1 + U2 = 5044677
thus
  U2 is odd since T2 + T1 must be even and
  U1..U6 are odd.

(S1 - 18)/2 =
  T1 + U1 = 11664
thus T1 is odd and
  T1..T6 are odd.

Let T1 = 2z+1
Let U1 = 2y+1
Let T2 = 2w+1
Let U2 = 2v+1

(S5 - 18)/2 =
  16T5 + 40T4 + 40T3 + 20(2w+1) + 5(2z+1) + 16U5 = 10993881635221599
((S5 - 18)/2 - 25)/2 =
  8T5 + 20T4 + 20T3 + 20w + 5z + 8U5 = 5496940817610787
thus
  z is odd.

(S1 - 18)/2 =
  2z+1 + 2y+1 = 11664
((S1 - 18)/2 - 2)/2 =
  z + y = 5831
thus
  y is even.

(S6 - 18)/4 =
  16T6 + 48T5 + 60T4 + 40T3 + 15(2w+1) + 3(2z+1) + 16U6 = 6108669985814426382
((S6 - 18)/4 - 18)/2 =
  8T6 + 24T5 + 30T4 + 20T3 + 15w + 3z + 8U6 = 3054334992907213182
thus
  w is odd.

(S2 - 18)/4 =
  2w+1 + 2z+1 + 2v+1 = 5044677
((S2 - 18)/4 - 3)/2 =
  w + z + v = 2522337
thus
  v is odd.

Let z = 2u+1 so that T1 = 4u+3
Let w = 2t+1 so that T2 = 4t+3
Let v = 2s+1 so that U2 = 4s+3

Since
  T4 = T2 (mod 4)
  U4 = U2 (mod 4)
Let T4 = 4r+3
Let U4 = 4q+3

((S5 - 18)/2 - 25)/2 =
  8T5 + 20(4r+3) + 20T3 + 20(2t+1) + 5(2u+1) + 8U5 = 5496940817610787
(((S5 - 18)/2 - 25)/2 - 85)/2 =
  4T5 + 40r + 10T3 + 20t + 5u + 4U5 = 2748470408805351
thus
  u is odd.

((S6 - 18)/4 - 18)/2 =
  8T6 + 24T5 + 30(4r+3) + 20T3 + 15(2t+1) + 3(2u+1) + 8U6 = 3054334992907213182
(((S6 - 18)/4 - 18)/2 - 108)/2 = 
  4T6 + 12T5 + 60r + 10T3 + 15t + 3u + 4U6 = 1527167496453606537
thus
  t is even.

(S4 - 18)/8 =
  2T4 + 4T3 + 3(2w+1) + 2z+1 + 2U4 = 2543899196388
(S4 - 18)/8 - 4)/2 = 
  4r+3 + 2T3 + 3(2t+1) + 2u+1 + 4q+3 = 1271949598192
(S4 - 18)/8 - 4)/2 - 10)/2 =
  2r + T3 + 3t + u + 2q = 635974799091
thus
  t is odd.

t can't be both even and odd.