Order 40 Octamagic Series Impossibility Theorem. There are no order 40 octamagic series. Proof. S1 = 32020 S2 = 34165340 ≡ 8 (mod 9) S3 = 41011216000 ≡ 7 (mod 9) S4 = 52510754133332 ≡ 8 (mod 9) S5 = 70036206933328000 S6 = 96079651986268647620 S7 = 134553516987685546672000 S8 = 191424753682082110600533332 From the Modulo 9/3 Tetramagic Series Lemma, there are 5+9a entries of 2 (mod 3), 3+9b entries of 1 (mod 3), and 5+9c 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)+8 Let 9Bj be the square of an entry of 0 (mod 3), j=1..9c+5 Let Tn = sum (Aj)n, j=1..9(a+b)+8, n=1..4 Let Un = sum (Bj)n, j=1..9c+5, n=1..4 S2 = 3T1 + 9(a+b)+8 + 9U1 S4 = 9T2 + 6T1 + 9(a+b)+8 + 81U2 S6 = 27T3 + 27T2 + 9T1 + 9(a+b)+8 + 729U3 S8 = 81T4 + 108T3 + 54T2 + 12T1 + 9(a+b)+8 + 6561U4 (S2 - 8)/3 = T1 + 3(a+b) + 3U1 = 11388444 ≡ 0 (mod 3) thus T1 ≡ 0 (mod 3) and T3 ≡ 0 (mod 3). (S6 - 8)/9 = 3T3 + 3T2 + T1 + (a+b) + 81U3 = 10675516887363183068 ≡ 2 (mod 3) thus a+b ≡ 2 (mod 3) and since a+b <=2 a+b = 2. (S8 - 8)/9 = 9T4 + 12T3 + 6T2 + 4T1/3 + 729U4 = 21269417075786901177837034 ≡ 1 (mod 3) thus T1/3 ≡ 1 (mod 3). (S4 - 8)/9 = T2 + 2T1/3 + 9U2 = 5834528237034 ≡ 0 (mod 3) thus T2 ≡ 1 (mod 3). (S6 - 8)/27 = T3 + T2 + T1/3 + 27U3 = 3558505629121061022 ≡ 0 (mod 3) thus T3 ≡ 1 (mod 3). T3 can't be both 0 and 1 (mod 3).