Let $\{x\}$ be the fractional part of a real number $x$. If $a$ is a positive real number with $\{a^{-1}\}=\{a^2\}$ and $2<a^2<3$, find $a^{12}-144a^{-1}$.
This is probably an AIME problem. But I don't know from which year it is. Here is my solution of the problem:
Because $2<a^2<3\implies\frac1{\sqrt3}<a^{-1}<\frac1{\sqrt2}$, we have $\lfloor a^{-1}\rfloor=0$ and $\lfloor a^2\rfloor=2$. So the equation is
$\begin{align}a^{-1}=a^2-2 &\implies a^3-2a-1=0\\ &\implies (a+1)(a^2-a-1)=0\\ &\implies a=\frac{1+\sqrt5}2.\end{align}$
Now with some boring calculations, we can find $a^{12}-144a^{-1}$. But this is very much dependant on computations. Is there some other way to find $a^{12}-144a^{-1}$ that is not much dependant on computational skills? Also as we obtained $a=\phi$, can we use Fibonacci numbers here?
Best Answer
As Martin R says, use can use $\phi^{-1}=\phi-1$. You can also use the fact that $\phi^n=F_n\phi+F_{n-1}$ where $F_n$ is the $n$th Fibonacci number. (This follows by induction: $\phi^1=1\phi+0=F_1\phi+F_0$ and $\phi^{n+1}=\phi(F_n\phi+F_{n-1})=F_n\phi^2+F_{n-1}\phi=(F_n+F_{n-1})\phi+F_n=F_{n+1}\phi+F_n$, again using an identity suggested by Martin R.) This reduces $\phi^{12}-144\phi^{-1}$ to $F_{12}\phi+F_{11}-144(\phi-1)$ and $F_{11}=89$ and $F_{12}=144$ are much less cumbersome to compute.