Problem:
Given a nonnegative real number x, let ⟨x⟩ denote the fractional part of x; that is, ⟨x⟩=x−⌊x⌋, where ⌊x⌋ denotes the greatest integer less than or equal to x. Suppose that a is positive, ⟨a−1⟩=⟨a2⟩, and 2<a2<3. Find the value of a12−144a−1.
Solution:
Notice first that the given data imply that ⟨a−1⟩=a−1 and ⟨a2⟩=a2−2. Hence a must satisfy the equation a−1=a2−2, or a3−2a−1=0. This factors as
(a+1)(a2−a−1)=0
whose only positive root is a=21(1+5). Now use the relations a2=a+1 and a3=2a+1 to calculate
a6a12a13=8a+5=144a+89=233a+144
from which it follows that a12−144a−1=aa13−144=233.
OR
As above, show that a=21(1+5). The cubic equation yields a3=2a+1=2+5. It follows that a6=9+45 and a12=161+725. The cubic equation also implies that 144a−1=144(a2−2)=−72+725, hence a12−144a−1=233.
OR
As above, a=(1+5)/2. Let b=−a−1 and use Binet's formula for Fibonacci numbers to calculate