Order 25 Pentamagic Series Impossibility

Theorem. There are no order 25 pentamagic series.
Proof.
S1 = 7825
S2 = 3263025
S3 = 1530765625
S4 = 765994466145
S5 = 399274190265625

Since S4 ≡ 1 (mod 16), there must be either
1 or 17 odd entries in a pentamagic series.


Case 1 of 2: 8 even entries and 17 odd entries.

Let 2Aj+1 be an odd entry, j = 1..17
Let 2Bj   be an even entry, j=1..8
Let Tn = sum (Aj)n, j=1..17, n=1..6
Let Un = sum (Bj)n, j=1..8, n=1..6

S1 = 2T1 + 17 + 2U1
S2 = 4T2 + 4T1 + 17 + 4U2
S3 = 8T3 + 12T2 + 6T1 + 17 + 8U3
S4 = 16T4 + 32T3 + 24T2 + 8T1 + 17 + 16U4
S5 = 32T5 + 80T4 + 80T3 + 40T2 + 10T1 + 17 + 32U5

(S3 - 17)/2 =
  4T3 + 6T2 + 3T1 + 4U3 = 765382804
thus
  T1 is even, which implies T1..T6 are even.

(S1 - 17)/2 =
  T1 + U1 = 3904
thus
  U1 is even, which implies U1..U6 are even.

(S3 - 17)/4 =
  2T3 + 3T2 + 3T1/2 + 2U3 = 382691402
thus
  T1/2 is even.

(S4 - 17)/16 =
  T4 + 2T3 + 3T2/2 + T1/2 + U4 = 47874654133
thus
  T2/2 is odd.

(S3 - 17)/8 =
  T3 + 3T2/2 + 3T1/4 + U3 = 191345701
thus
  T1/4 is even.

(S5 - 17)/8 =
  4T5 + 10T4 + 10T3 + 5T2 + 5T1/4 + 4U5 = 49909273783201
thus
  T1/4 is odd.

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


Case 2 of 2: 24 even entries and 1 odd entry.

Let 2Aj be an even entry, j = 1 ... 24.
Let G = 2K+1 be the odd entry.
Let Tn = sum (Aj)n, j = 1 ... 24, n = 1 ... 4.

  S1 = 2T1 + 2K + 1 = 7825
  S2 = 4T2 + 4K2 + 4K + 1 = 3263025
  S3 = 8T3 + 8K3 + 12K2 + 6K + 1 = 1530765625
  S4 = 16T4 + 16K4 + 32K3 + 24K2 + 8K + 1 = 765994466145

(S2 - 1)/4 =
  T2 + K(K+1) = 815756

K(K+1) must be even and 815756 is even, thus T2 is even.
T1...T4 are even.

(S1 - 1)/2 =
  T1 + K = 3912.

T1 is even and 3912 is even, thus K is even.

Let K = 2L.

(S3 - 1)/4 =
  2T3 + 16L3 + 12L2 + 3L = 382691406 ≡ 2 (mod 4)

Since T3 is even, 2T3 is a multiple of 4
as well as 16L3, and 12L2.
thus
  L ≡ 2 (mod 4)

Let L = 4M+2 and thus K = 8M+4 and G = 16M+9.

Rewriting the equations in terms of M, we have

  T1 + 8M = 3908 ≡ 0 (mod 4)
  T2 + 64M2 + 72M = 815736  ≡ 0 (mod 8)
  T3 + 512M3 + 864M2 + 486M = 191345612 ≡ 0 (mod 4)
  T4 + 4096M4 + 9216M3 + 5728M2 + 2916M = 47874653724 ≡ 4 (mod 8)

thus

  T1 ≡ 0 (mod 4)
  T2 ≡ 0 (mod 8)
  T3 ≡ 2 (mod 4) if M is odd
     ≡ 0 (mod 4) if M is even
  T4 ≡ 0 (mod 8) if M is odd
     ≡ 4 (mod 8) if M is even

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Argument A

An even cube ≡ 0 (mod 4), thus
T3 ≡ sum of odd cubes (mod 4).
An odd entry ≡ that entry cubed (mod 4).

Sum of even entries ≡ (T1 - sum of odd entries) (mod 4)
                    ≡ (T1 - sum of odd cubes) (mod 4)
                    ≡ (T1 - T3) (mod 4),
therefore
sum of even entries ≡ 2 (mod 4) if M is odd,
                    ≡ 0 (mod 4) if M is even.

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Argument B

An even 4th power ≡ 0 (mod 8), thus
T4 ≡ sum of odd 4th powers (mod 8).
An odd entry squared ≡ 4th power of that entry (mod 8).

Sum of even squares ≡ (T2 - sum of odd squares) (mod 8)
                    ≡ (T2 - sum of odd 4th powers) (mod 8)
                    ≡ (T2 - T4) (mod 8),
hence
sum of even squares ≡ 0 (mod 8) if M is odd,
                    ≡ 4 (mod 8) if M is even.

If M is odd,  there must be an even number of singly even entries,
if M is even, there must be an odd  number of singly even entries,
therefore
sum of even entries ≡ 0 (mod 4) if M is odd,
                    ≡ 2 (mod 4) if M is even.

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The conclusion of Argument B is the exact opposite of Argument A;
T1 and T3 (mod 4) are inconsistent with T2 and T4 (mod 8);
therefore, no order 25 pentamagic series exists
with 1 odd and 24 even entries.