The sequences { a n } n = 1 ∞ and { b n } n = 1 ∞ are defined through a 1 = α , b 1 = β , a n + 1 = α a n − β b n , and b n + 1 = β a n + α b n for all n ≥ 2 . How many pairs of real numbers ( α , β ) are there such that a 2 0 1 4 = b 1 and b 2 0 1 4 = a 1 ?
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By setting z n = a n + i ⋅ b n we can see that a n = ℜ ( z n ) , b n = ℑ ( z n ) and z n = ( α + i ⋅ β ) n . In order to have z 2 0 1 4 = β + i ⋅ α we must have z 1 = 0 or z 1 = e i θ . In the second case, 2 0 1 4 θ + 2 k π = 2 π − θ must hold, so θ = 2 0 1 5 2 π m + 4 0 3 0 π with m ∈ [ 1 , 2 0 1 5 ] . In conclusion, we have 2 0 1 6 solutions.