Order 17 Tetramagic Series Impossibility

Theorem. There are no order 17 tetramagic series.
Proof.
S1 = 2465
S2 = 475745
S3 = 103295825
S4 = 23923217921

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


Case 1 of 2: 17 odd entries (all entries are odd)

Let 8Aj+1 be the square of an odd entry, j = 1 ... 17.
Let Tn = sum (Aj)n, j = 1 ... 17, n = 1,2.

  S2 = 8T1 + 17
thus
  (S2 - 17)/8 = T1 = 59466
and
  T1 is even.

  S4 = 64T2 + 16T1 + 17
thus
  (S4 - 17)/16 = 4T2 + T1 = (23923217921 - 17)/16 = 1495201119
and
  T1 is odd.

T1 can't be both odd and even.


Case 2 of 2: 16 even entries and 1 odd entry.

Let 2Aj be an even entry, j = 1 ... 16.
Let G = 2K+1 be the odd entry.
Let Tn = sum (Aj)n, j = 1 ... 16, n = 1,2,3,4.

  S1 = 2T1 + 2K + 1 = 2465
  S2 = 4T2 + 4K2 + 4K + 1 = 475745
  S3 = 8T3 + 8K3 + 12K2 + 6K + 1 = 103295825
  S4 = 16T4 + 16K4 + 32K3 + 24K2 + 8K + 1 = 23923217921

(S2 - 1)/4 =
  T2 + K(K+1) = 118936

K(K+1) must be even and 118936 is even, thus T2 is even.
T1..T4 are even.

(S1 - 1)/2 =
  T1 + K = 1232.

T1 is even and 1232 is even, thus K is even.

Let K = 2L.

(S3 - 1)/4 =
  2T3 + 16L3 + 12L2 + 3L = 25823956

Since T3 is even, 2T3 is a multiple of 4
as well as 16L3, 12L2, and 25823956,
thus 3L must be a multiple of 4.
Since 3 and 4 are coprime, L must be a multiple of 4.

Let L = 4M and thus K = 8M and G = 16M+1.

Rewriting the equations in terms of M, we have

  T1 + 8M = 1232 ≡ 0 (mod 4)
  T2 + 64M2 + 8M = 118936 ≡ 0 (mod 8)
  T3 + 512M3 + 96M2 + 6M = 12911978 ≡ 2 (mod 4)
  T4 + 256M4 + 1024M3 + 96M2 + 4M = 1495201120 ≡ 0 (mod 8)

thus

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

If M is odd,  there must be an odd  number of singly even entries,
if M is even, there must be an even number of singly even entries,
therefore
sum of even entries ≡ 2 (mod 4) if M is odd,
                    ≡ 0 (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 17 tetramagic series exists.