“…Assuming that v(a 1 ) ≥ 2, so a ∈ W(2, 2, 3, 3, 5), we have ∆ ≡ a 4 3 (mod 2 13 ), so n ≥ 12, with n = 12 ⇐⇒ v(a 3 ) = 3. Assuming further that v(a 3 ) ≥ 4, so a ∈ W(2, 2, 4, 3, 5), we have ∆ ≡ 2 4 (2 2 a 4 1 + 2 6 a 2 2 + a 2 6 ) ≡ 2 14 (a 4 1,2 + a 2 2,2 + a 2 6,5 ) (mod 2 15 ), so n ≥ 14, with n = 14 ⇐⇒ a 6,5 ̸ ≡ a 1,2 + a 2,2 (mod 2). Assuming that a 6,5 ≡ a 1,2 + a 2,2 (mod 2), we find that ∆ ≡ 2 15 (mod 2 16 ), so that n = 15.…”