Order 32 Hexamagic Series Impossibility Theorem. There are no order 32 hexamagic series. Proof. S1 = 16400 S2 = 11201200 ≡ 7 (mod 9) S3 = 8606720000 ≡ 2 (mod 9) S4 = 7054065471760 ≡ 7 (mod 9) S5 = 6022406005760000 S6 = 5288530297464091600 From the Modulo 9/3 Tetramagic Series Lemma, this hexamagic series must have 9a+7 entries of 2 (mod 3), 9b entries of 1 (mod 3), and 9c+7 entries of 0 (mod 3), where a+b+c = 2. Let 3Aj+1, j=1..9(a+b)+7, be the squares of entries that are 1 or 2 (mod 3). Let 9Bj, j=1..9c+7, be the squares of entries that are 0 (mod 3). Let Tn = sum (Aj)n, j=1..9(a+b)+7, n=1,2,3. Let Un = sum (Bj)n, j=1..9c+7, n=1,2,3. S2 = 3T1 + 9(a+b)+7 + 9U1 S6 = 27T3 + 27T2 + 9T1 + 9(a+b)+7 + 729U3 (S2 - 7)/3 = T1 + 3U1 + 3(a+b) = 3733731 ≡ 0 (mod 3) thus T1 ≡ 0 (mod 3). (S6 - 7)/9 = 3T3 + 3T2 + T1 + 81U3 + (a+b) = 587614477496010177 ≡ 0 (mod 3) thus (a+b) ≡ 0 (mod 3). Since a+b <= 2, a = 0 and b = 0. An order 32 hexamagic series must consist of 7 entries of 2 (mod 3), 0 entries of 1 (mod 3), and 25 entries of 0 (mod 3).
Starting over ... Let 3Aj+2 be an entry of 2 (mod 3), j=1..7. Let 3Bj be an entry of 0 (mod 3), j=1..25. Let Tn = sum (Aj)n, j=1..7, n=1..6. Let Un = sum (Bj)n, j=1..25, n=1..6. S2 = 9T2 + 12T1 + 28 + 9U2 S3 = 27T3 + 54T2 + 36T1 + 56 + 27U3 (S2 - 28)/3 = 3T2 + 4T1 + 3U2 = 3733724 ≡ 2 (mod 3) thus T1 ≡ 2 (mod 3). (S3 - 56)/9 = 3T3 + 6T2 + 4T1 + 3U3 = 956302216 ≡ 1 (mod 3) thus T1 ≡ 1 (mod 3). T1 can't be both 1 and 2 (mod 3).