Modulo 5 Tetramagic Series Lemma If J, K, L, M are the number of 1, 2, 3, 4 (mod 5) entries, respectively, in a tetramagic series, then J ≡ 4S1 + 4S2 + 4S3 + 4S4 (mod 5) K ≡ 2S1 + S2 + 3S3 + 4S4 (mod 5) L ≡ 3S1 + S2 + 2S3 + 4S4 (mod 5) M ≡ S1 + 4S2 + S3 + 4S4 (mod 5) Proof [1] J + 2K + 3L + 4M ≡ S1 (mod 5) [2] J + 4K + 4L + M ≡ S2 (mod 5) [3] J + 3K + 2L + 4M ≡ S3 (mod 5) [4] J + K + L + M ≡ S4 (mod 5) Solve [4] for J. [4a] J ≡ S4 + 4K + 4L + 4M (mod 5) [1a] S4 + K + 2L + 3M ≡ S1 (mod 5) [2a] S4 + 3K + 3L ≡ S2 (mod 5) [3a] S4 + 2K + L + 3M ≡ S3 (mod 5) Solve [3a] for L. [3b] L ≡ S3 + 4S4 + 3K + 2M (mod 5) [4b] J ≡ 4S3 + 2S4 + K + 2M (mod 5) [1b] 2S3 + 4S4 + 2K + 2M ≡ S1 (mod 5) [2b] 3S3 + 3S4 + 2K + M ≡ S2 (mod 5) Solve [2b] for M. [2c] M ≡ S2 + 2S3 + 2S4 + 3K (mod 5) [3c] L ≡ 2S2 + 3S4 + 4K (mod 5) [4c] J ≡ 2S2 + 3S3 + S4 + 2K (mod 5) [1c] 2S2 + S3 + 3S4 + 3K ≡ S1 (mod 5) Multiply [1c] by 2 and solve for K. [1d] K ≡ 2S1 + S2 + 3S3 + 4S4 (mod 5) [2d] M ≡ S1 + 4S2 + S3 + 4S4 (mod 5) [3d] L ≡ 3S1 + S2 + 2S3 + 4S4 (mod 5) [4d] J ≡ 4S1 + 4S2 + 4S3 + 4S4 (mod 5)