Răspuns:
a = { 0 ; 1 ; 3 ; 4 ; 10 ; 31}
Explicație pas cu pas:
a = { x ∈ N I 63/(2x+1) ∈ N } =>
2x + 1 divizibil cu D₆₃
D₆₃(naturali) = 1 ; 3 ; 7 ; 9 ; 21 ; 63
2x + 1 = 1 => 2x = 0 => x = 0 ∈ N => a = 63/1 = 63 ∈ N
2x + 1 = 3 => 2x = 2 => x = 1 ∈ N => a = 63/3 = 21 ∈ N
2x + 1 = 7 => 2x = 6 => x = 3 ∈ N => a = 63/7 = 9 ∈ N
2x + 1 = 9 => 2x = 8 => x = 4 ∈ N => a = 63/9 = 7 ∈ N
2x + 1 = 21 => 2x = 20 => x = 10 ∈ N => a = 63/21 = 3 ∈ N
2x + 1 = 63 => 2x = 62 => x = 31 ∈ N => a = 63/63 =1 ∈ N
a = { 0 ; 1 ; 3 ; 4 ; 10 ; 31}
