Question

(2n-1)il divide pe(4n+5)

Answer 1

(2n-1) | (4n+5)                            (1)
(2n-1) | (2n-1) ⇒ (2n-1) | (4n - 2)  (2)
⇒ (2n-1) | [(1) - (2)] = 7
⇒ (2n-1) ∈ D7
(2n-1) ∈ {-7,-1, 1, 1, 7}
2n ∈ {-6, 0 2, 8} ⇒  n ∈ {-3, 0, 1, 4}