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)