If ω is the non-real cube root of unity, then the number of ordered pairs of integers a , b , such that a ω + b = 1 , is equal to

Question

If ω is the non-real cube root of unity, then the number of ordered pairs of integers a , b , such that a ω + b = 1 , is equal to,

MARKS App · Accepted Answer

We have, a ω + b 2 = 1 ⇒ a ω + b a ω - + b = 1 ⇒ a 2 + a b ω + ω - + b 2 = 1 ⇒ a 2 - a b + b 2 = 1 ⇒ a - b 2 + a b = 1 … . . i A s, 1 + ω + ω 2 = 0 When a - b 2 = 0 and a b = 1 then 1 , 1 ; - 1 , - 1 When a - b 2 = 1 and a b = 0 then 0 , 1 ; 1 , 0 ; 0 , - 1 ; - 1 , 0 Hence, 0 , 1 ; 1 , 0 ; 0 , - 1 ; - 1 , 0 ; 1 , 1 ; - 1 , - 1 i.e., 6 ordered pairs.

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