D = {x ∈ ℕ| 15/(2x+1) ∈ ℕ}
2x+1 ∈ D₁₅ ⇔ 2x+1 ∈ {1,3,5,15} |-1
⇔ 2x ∈ {0,2,4,14} |:2 ⇔ x ∈ {0,1,2,7}
⇒ D = {0,1,2,7}.
