“…We have that if p satisfies p − 5 √ p − 12 > 0, which yields p > 45, then there exists an element x ∈ Z p such that x ∈ C 2 1 , x + 1 ∈ C 2 1 and x − 1 ∈ C 2 0 . When 5 ≤ p ≤ 43, it is readily checked that we may take x as (p, x) = (5, 2), (7,5), (11,6), (13,5), (17,5), (19,2), (23,10), (29,2), (31,11), (37,5), (41, 6), (43,2). ✷ Lemma 6.9 There exists a 3-SCHGDD of type (6, 2 p ) for any prime p ≥ 3.…”