Modulo 9/3 Tetramagic Series Lemma If A, D, B, E, C, F are the number of 1, 2, 4, 5, 7, 8 (mod 9) entries, respectively, in a tetramagic series. and thus, (A+B+C) is the number of entries of 1 (mod 3) and (D+E+F) is the number of entries of 2 (mod 3) then (A+B+C) ≡ 4(S2 + S3) + S4 + S3 (mod 9) and (D+E+F) ≡ 4(S2 + S3) + S4 (mod 9). Proof [1] A + 7B + 4C + 4D + 7E + F ≡ S2 (mod 9) [2] A + B + C + 8D + 8E + 8F ≡ S3 (mod 9) [3] A + 4B + 7C + 7D + 4E + F ≡ S4 (mod 9) Subtract [2] from [1]. [4] 6B + 3C + 5D + 8E + 2F ≡ S2 - S3 (mod 9) Subtract [2] from [3]. [5] 3B + 6C + 8D + 5E + 2F ≡ S4 - S3 (mod 9) Multiply [5] by 2. [6] 6B + 3C + 7D + E + 4F ≡ 2(S4 - S3) (mod 9) Subtract [4] from [6]. [7] 2D + 2E + 2F ≡ 2S4 - S3 - S2 (mod 9) Multiply [7] by 5. [8] (D+E+F) ≡ 4(S2 + S3) + S4 (mod 9). Substitute [8] into [2]. [9] (A+B+C) ≡ 4(S2 + S3) + S4 + S3 (mod 9). Equations [8] and [9] give the desired result.