Order 28 Hexamagic Series Impossibility Theorem. There are no order 28 hexamagic series. Proof. S1 = 10990 S2 = 5747770 ≡ 1 (mod 9) S3 = 3381842800 ≡ 1 (mod 9) S4 = 2122443391726 ≡ 1 (mod 9) S5 = 1387546427940400 S6 = 933025104315693910 From the Modulo 9/3 Tetramagic Series Lemma, this hexamagic series must have 9a entries of 2 (mod 3), 9b+1 entries of 1 (mod 3), and 9c entries of 0 (mod 3). where a+b+c = 3. Let 3Aj+1, j=1..9(a+b)+1, be the squares of entries that are 1 or 2 (mod 3). Let 9Bj, j=1..9c, be the squares of entries that are 0 (mod 3). Let Tn = sum (Aj)n, j=1..9(a+b)+1, n=1,2,3. Let Un = sum (Bj)n, j=1..9c, n=1,2,3. S2 = 3T1 + 9(a+b)+1 + 9U1 S6 = 27T3 + 27T2 + 9T1 + 9(a+b)+1 + 729U3 (S2 - 1)/3 = T1 + 3U1 + 3(a+b) = 1915923 ≡ 0 (mod 3) thus T1 ≡ 0 (mod 3). (S6 - 1)/9 = 3T3 + 3T2 + T1 + 81U3 + (a+b) = 103669456035077101 ≡ 1 (mod 3) thus (a+b) ≡ 1 (mod 3). We divide the proof into two cases, a=0,b=1 and a=1,b=0.
Case a=0, b=1, where a hexamagic series consists of 10 entries of 1 (mod 3) and 18 entries of 0 (mod 3). Let 3Aj+1 be an entry of 1 (mod 3), j=1..10. Let 3Bj be an entry of 0 (mod 3), j=1..18. Let Tn = sum (Aj)n, j=1..10, n=1..6. Let Un = sum (Bj)n, j=1..18, n=1..6. S2 = 9T2 + 6T1 + 10 + 9U2 S3 = 27T3 + 27T2 + 9T1 + 10 + 27U3 (S2 - 10)/3 = 3T2 + 2T1 + 3U2 = 1915920 ≡ 0 (mod 3) thus T1 ≡ 0 (mod 3). (S3 - 10)/9 = 3T3 + 3T2 + T1 + 3U3 = 375760310 ≡ 2 (mod 3) thus T1 ≡ 2 (mod 3). T1 can't be both 0 and 2 (mod 3).
Case a=1, b=0, where a hexamagic series consists of 9 entries of 2 (mod 3), 1 entry of 1 (mod 3), and 18 entries of 0 (mod 3). Let 3Aj+2 be an entry of 2 (mod 3), j=1..9. Let 3Bj be an entry of 0 (mod 3), j=1..18. Let 3C+1 be the entry of 1 (mod 3). Let Tn = sum (Aj)n, j=1..9, n=1..6. Let Un = sum (Bj)n, j=1..18, n=1..6. S3 = (27T3 + 54T2 + 36T1 + 72) + 27U3 + (27C3 + 27C2 + 9C + 1) S5 = (243T5 + 810T4 + 1080T3 + 720T2 + 240T1 + 288) + 243U5 + (243C5 + 405C4 + 270C3 + 90C2 + 15C + 1) (S3 - 73)/9 = 3T3 + 6T2 + 4T1 + 3U3 + 3C3 + 3C2 + C = 375760303 ≡ 1 (mod 3) thus T1 + C ≡ 1 (mod 3). (S5 - 289)/3 = 81T5 + 270T4 + 360T3 + 240T2 + 80T1 + 81U5 + 81C5 + 135C4 + 90C3 + 30C2 + 5C = 462515475980037 ≡ 0 (mod 3) thus 2T1 + 2C ≡ 0 (mod 3) or T1 + C ≡ 0 (mod 3). T1 + C can't be both 0 and 1 (mod 3).