Order 27 Octamagic Series Impossibility

Theorem. There are no order 27 octamagic series.
Proof.
S1 = 9855
S2 = 4792815
S3 = 2622267675
S4 = 1530354456567
S5 = 930327251647275
S6 = 581719793510911095
S7 = 371318608677643256475
S8 = 240779187283769683452567

Since S4 ≡ 7 (mod 16), there must be either
7 or 23 odd entries in a series


Case 1 of 2
20 even entries and 7 odd entries.

Let 8Aj+1 be the square of an odd entry, j=1..7.
Let 4Bj be the square of an even entry, j=1..20.
Let Tn = sum (Aj)n, j=1..7,  n=1..4.
Let Un = sum (Bj)n, j=1..20, n=1..4.

  S2 = 8T1 + 7 + 4U1
  S4 = 64T2 + 16T1 + 7 + 16U2
  S6 = 512T3 + 192T2 + 24T1 + 7 + 64U3

(S2 - 7)/4 =
  2T1 + U1 = 1198202
thus
  U1 is even
and
  U2 is even.

(S4 - 7)/16 =
  4T2 + T1 + U2 = 95647153535
thus
  T1 is odd.

(S6 - 7)/8 =
  64T3 + 24T2 + 3T1 + 8U3 = 72714974188863886
thus
  T1 is even.

T1 can't be both even and odd.


Case 2 of 2.
4 even entries and 23 odd entries

Let 8Aj+1 be the square of an odd entry, j=1..23.
Let 4Bj be the square of an even entry, j=1..4.
Let Tn = sum (Aj)n, j=1..23,  n=1..4
Let Un = sum (Bj)n, j=1..4, n=1..4

  S2 = 8T1 + 23 + 4U1
  S4 = 64T2 + 16T1 + 23 + 16U2
  S6 = 512T3 + 192T2 + 24T1 + 23 + 64U3
  S8 = 4096T4 + 2048T3 + 384T2 + 32T1 + 23 + 256U4

(S2 - 23)/4 =
  2T1 + U1 = 1198198
thus
  U1 is even and
  U1 .. U4 are even.

(S4 - 23)/16 =
  4T2 + T1 + U2 = 95647153534
thus
  T1 is even and
  T1 .. T4 are even.

(S6 - 23)/16 =
  32T3 + 12T2 + 3T1/2 + 4U3 = 36357487094431942
thus T1/2 is even

(S6 - 23)/32 =
  16T3 + 6T2 + 3T1/4 + 2U3 = 18178743547215971
thus
  T1/4 is odd

(S8 - 23)/128 =
  32T4 + 16T3 + 3T2 + T1/4 + 2U4 = 30097398410471210431568
thus T1/4 is even

T1/4 can't be both odd and even.