Order 43 Hexamagic Series Impossibility Theorem. There are no order 43 hexamagic series. Proof. S1 = 39775 S2 = 49042575 ≡ 0 (mod 9) S3 = 68028176875 ≡ 4 (mod 9) S4 = 100654480695735 ≡ 0 (mod 9) S5 = 155133699475586875 S6 = 245931201634915761975 From the Modulo 9/3 Tetramagic Series Lemma, there must be 9a+7 entries of 2 (mod 3), 9b+2 entries of 1 (mod 3), and 9c+7 entries of 0 (mod 3), where a+b+c = 3. Let 3Aj+2 be an entry of 2 (mod 3), j=1..9a+7 Let 3Bj+1 be an entry of 1 (mod 3), j=1..9b+2 Let 3Cj be an entry of 0 (mod 3), j=1..9c+7 where a+b+c = 3 Let Tn = sum (Aj)n, j=1..9a+7, n=1..6 Let Un = sum (Bj)n, j=1..9b+2, n=1..6 Let Vn = sum (Cj)n, j=1..9c+7, n=1..6 S2 = 9T2 + 12T1 + 4(9a+7) + 9U2 + 6U1 + (9b+2) + 9V2 S3 = 27T3 + 54T2 + 36T1 + 8(9a+7) + 27U3 + 27U2 + 9U1 + (9b+2) + 27V3 S5 = 243T5 + 810T4 + 1080T3 + 720T2 + 240T1 + 32(9a+7) + 243U5 + 405U4 + 270U3 + 90U2 + 15U1 + (9b+2) + 243V5 S6 = 729T6 + 2916T5 + 4860T4 + 4320T3 + 2160T2 + 576T1 + 64(9a+7) + 729U6 + 1458U5 + 1215U4 + 540U3 + 135U2 + 18U1 + (9b+2) + 729V6 (S2 - 30)/3 = 3T2 + 4T1 + 12a + 3U2 + 2U1 + 3b + 3U6 = 16347515 ≡ 2 (mod 3) thus T1 + 2U1 ≡ 2 (mod 3) (S5 - 226)/3 = 81T5 + 270T4 + 360T3 + 240T2 + 80T1 + 96a + 81U5 + 135U4 + 90U3 + 30U2 + 5U1 + 3b + 81V5 = 51711233158528883 ≡ 2 (mod 3) thus 2T1 + 2U1 ≡ 2 (mod 3) and when combined with T1 + 2U1 ≡ 2 (mod 3), T1 ≡ 0 (mod 3) U1 ≡ 1 (mod 3) (S3 - 58)/9 = 3T3 + 6T2 + 4T1 + 8a + 3U3 + 3U2 + U1 + b + 3V3 = 7558686313 ≡ 1 (mod 3) thus 2a + b ≡ 0 (mod 3) (S6 - 450)/9 = 81T6 + 324T5 + 540T4 + 480T3 + 240T2 + 64T1 + 64a + 81U6 + 162U5 + 135U4 + 60U3 + 15U2 + 2U1 + b + 81V6 = 27325689070546195725 ≡ 0 (mod 3) thus a + b ≡ 1 (mod 3) and when combined with 2a + b ≡ 0 (mod 3), a ≡ 2 (mod 3) b ≡ 2 (mod 3) which contradicts a+b <= 3