Problem: x x ⟨ x ⟩ ⟨ x ⟩ x x ⟨ x ⟩ = x − ⌊ x ⌋ ⟨ x ⟩ = x − ⌊ x ⌋ ⌊ x ⌋ ⌊ x ⌋ x x a a ⟨ a − 1 ⟩ = ⟨ a 2 ⟩ ⟨ a − 1 ⟩ = ⟨ a 2 ⟩ 2 < a 2 < 3 2 < a 2 < 3 a 12 − 144 a − 1 a 1 2 − 1 4 4 a − 1
Solution:
Notice first that the given data imply that ⟨ a − 1 ⟩ = a − 1 ⟨ a − 1 ⟩ = a − 1 ⟨ a 2 ⟩ = a 2 − 2 ⟨ a 2 ⟩ = a 2 − 2 a a a − 1 = a 2 − 2 a − 1 = a 2 − 2 a 3 − 2 a − 1 = 0 a 3 − 2 a − 1 = 0
( a + 1 ) ( a 2 − a − 1 ) = 0 ( a + 1 ) ( a 2 − a − 1 ) = 0
whose only positive root is a = 1 2 ( 1 + 5 ) a = 2 1 ​ ( 1 + 5 ​ ) a 2 = a + 1 a 2 = a + 1 a 3 = 2 a + 1 a 3 = 2 a + 1
a 6 = 8 a + 5 a 12 = 144 a + 89 a 13 = 233 a + 144 a 6 a 1 2 a 1 3 ​ = 8 a + 5 = 1 4 4 a + 8 9 = 2 3 3 a + 1 4 4 ​
from which it follows that a 12 − 144 a − 1 = a 13 − 144 a = 233 a 1 2 − 1 4 4 a − 1 = a a 1 3 − 1 4 4 ​ = 2 3 3 ​
OR OR
As above, show that a = 1 2 ( 1 + 5 ) a = 2 1 ​ ( 1 + 5 ​ ) a 3 = 2 a + 1 = 2 + 5 a 3 = 2 a + 1 = 2 + 5 ​ a 6 = 9 + 4 5 a 6 = 9 + 4 5 ​ a 12 = 161 + 72 5 a 1 2 = 1 6 1 + 7 2 5 ​ 144 a − 1 = 144 ( a 2 − 2 ) = − 72 + 72 5 1 4 4 a − 1 = 1 4 4 ( a 2 − 2 ) = − 7 2 + 7 2 5 ​ a 12 − 144 a − 1 = 233 a 1 2 − 1 4 4 a − 1 = 2 3 3 ​
OR OR
As above, a = ( 1 + 5 ) / 2 a = ( 1 + 5 ​ ) / 2 b = − a − 1 b = − a − 1
F n = a n − b n 5 = a n − b n a − b = a n − 1 + a n − 2 b + ⋯ + a b n − 2 + b n − 1 F n ​ = 5 ​ a n − b n ​ = a − b a n − b n ​ = a n − 1 + a n − 2 b + ⋯ + a b n − 2 + b n − 1
Thus
F n = a n − 1 + F n − 1 b = a n − 1 − F n − 1 a − 1 F n ​ = a n − 1 + F n − 1 ​ b = a n − 1 − F n − 1 ​ a − 1
In particular,
233 = F 13 = a 12 − F 12 a − 1 = a 12 − 144 a − 1 2 3 3 ​ = F 1 3 ​ = a 1 2 − F 1 2 ​ a − 1 = a 1 2 − 1 4 4 a − 1
The problems on this page are the property of the MAA's American Mathematics Competitions