Order 31 Pentamagic Series Impossibility

Theorem. There are no order 31 pentamagic series.
Proof.
S1 = 14911
S2 = 9557951
S3 = 6892475551
S4 = 5301690282239
S5 = 4247977424067871

S4 ≡ 15 (mod 16) so there must be either 15 or 31 odd entries.


Case 1 of 2: 31 odd entries.

Let 8Aj+1 be the square of an odd entry, j=1..31.
Let Tn = sum (Aj)n, j=1..31, n=1,2.

S2 = 8T1 + 31
S4 = 64T2 + 16T1 + 31

(S2 - 31)/8 =
  T1 = 1194740 ≡ 0 (mod 4)

(S4 - 31)/16 =
  4T2 + T1 = 331355642638 ≡ 2 (mod 4)

T1 can't be both 0 and 2 (mod 4).


Case 2 of 2: 15 odd entries and 16 even entries.

Let 2Aj+1 be an odd entry, j=1..15
Let 2Bj be an even entry, j=1..16
Let Tn = sum (Aj)n, j = 1..15,  n=1..5
Let Un = sum (Bj)n, j = 1..16, n=1..5

S1 = 2T1 + 15 + 2U1
S2 = 4T2 + 4T1 + 15 + 4U2
S3 = 8T3 + 12T2 + 6T1 + 15 + 8U3
S4 = 16T4 + 32T3 + 24T2 + 8T1 + 15 + 16U4
S5 = 32T5 + 80T4 + 80T3 + 40T2 + 10T1 + 15 + 32U5

(S3 - 15)/2 =
  4T3 + 6T2 + 3T1 + 4U3 = 3446237768
thus
  T1 .. T5 are even.

(S1 - 15)/2 =
  T1 + U1 = 7448
thus
  U1 .. U5 are even.

(S3 - 15)/4 =
  2T3 + 3T2 + 3T1/2 + 2U3 = 1723118884
thus
  T1/2 is even.

(S4 - 15)/16 =
  T4 + 2T3 + 3T2/2 + T1/2 + U4 = 331355642639
thus
  T2/2 is odd.

Let T2/2 = 2z+1 so that T2 = 4z+2.

((S3 - 15)/4 - 6)/2 =
  T3 + 6z + 3T1/4 + U3 = 861559439
thus
  T1/4 is odd

((S5 - 15)/4 - 20)/2 =
  4T5 + 10T4 + 10T3 + 20z + 5T1/4 + 4U5 = 530997178008472
thus
  T1/4 is even

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