Order 41 Octamagic Series Impossibility

Theorem. There are no order 41 octamagic series.
Proof.
S1 = 34481
S2 = 38653201       ≡ 1 (mod 9)
S3 = 48746513801    ≡ 2 (mod 9)
S4 = 65573802634465 ≡ 1 (mod 9)
S5 = 91885277400846761
S6 = 132432914251807958881
S7 = 194850161950311443496521
S8 = 291235973392830670554641185

From the Modulo 9/3 Tetramagic Series Lemma, there are
  9a+4 entries of 2 (mod 3),
  9b+6 entries of 1 (mod 3), and
  9c+4 entries of 0 (mod 3),
    where a+b+c = 3.

Let 3Aj+1 be the square of an entry of 1 or 2 (mod 3), j=1..9(a+b)+10
Let 9Bj   be the square of an entry of 0 (mod 3), j=1..9c+4

Let Tn = sum (Aj)n, j=1..9(a+b)+10, n=1..4
Let Un = sum (Bj)n, j=1..9c+4, n=1..4

S2 = 3T1 + 9(a+b)+10 + 9U1
S4 = 9T2 + 6T1 + 9(a+b)+10 + 81U2
S6 = 27T3 + 27T2 + 9T1 + 9(a+b)+10 + 729U3
S8 = 81T4 + 108T3 + 54T2 + 12T1 + 9(a+b)+10 + 6561U4

(S2 - 10)/3 =
  T1 + 3(a+b) + 3U1 = 12884397 ≡ 0 (mod 3)
thus
  T1 ≡ 0 (mod 3) and
  T3 ≡ 0 (mod 3).

(S6 - 10)/9 =
  3T3 + 3T2 + T1 + (a+b) + 81U3 = 14714768250200884319 ≡ 2 (mod 3)
thus
  a+b ≡ 2 (mod 3) and since a+b <= 3
  a+b = 2.

(S8 - 10)/3 =
  27T4 + 36T3 + 18T2 + 4T1 + 3(a+b) + 2187U4 = 97078657797610223518213725
((S8 - 10)/3 - 6)/3 =
  9T4 + 12T3 + 6T2 + 4T1/3 + 729U4 = 32359552599203407839404573 ≡ 2 (mod 3)
thus
  T1/3 ≡ 2 (mod 3).

((S6 - 10)/9 - 2)/3 =
  T3 + T2 + T1/3 + 27U3 = 4904922750066961439 ≡ 2 (mod 3)
thus
  T2 ≡ 0 (mod 3).

(S4 - 10)/9 =
  T2 + 2T1/3 + (a+b) + 9U2 = 7285978070495 ≡ 2 (mod 3)
(S4 - 10)/9 - 2 =
  T2 + 2T1/3 + 9U2 = 7285978070493 ≡ 0 (mod 3)
thus
  T2 ≡ 1 (mod 3)

T2 can't be both 0 and 1 (mod 3)