If equation Z - 1 n = Z n = 1 ∀ n ∈ N has solutions, then n can be

Step-by-step solution

Z - 1 n = Z n ⇒ Z - 1 = Z ⇒ x = 1 2 …(i) Z n = 1 ⇒ Z = 1 ⇒ x 2 + y 2 = 1 …(ii) From (i) & (ii), 1 4 + y 2 = 1 ⇒ y = ± 3 2 ⇒ 1 2 + 3 2 i and 1 2 - 3 2 i can be the solutions Which would exist when n is a multiple of 6 ∵ a r g 1 2 + 3 2 i = t a n - 1 3 = π 3 ⇒ l e a s t value of n is e q u a l t o 2 π π 3 = 6