Is $12^2$ special?

$2<a^2<3$ $\implies \{a^2\}=a^2-2$ Also, $0<\dfrac{1}{\sqrt 3}<a^{-1}<\dfrac{1}{\sqrt 2}<1$ $\implies \{a^{-1}\}=a^{-1}$ Substituting in given equation we get:- $a^2-2=a^{-1}$ $a^3-2a-1=0$ $\implies a^2-a-1=0\implies a=\phi$

$\Large {a^2=a+1}$ $\color{#EC7300}{\Large{\implies a^{n}=a^{n-1}+a^{n-2}}}$

$\Large {\begin{aligned}a^{12}&=\color{#D61F06}{1}a^{11}+\color{#3D99F6}{1}a^{10}\\&=\color{#D61F06}{2}a^{10}+\color{#3D99F6}{1}a^9\\&=\color{#D61F06}{3}a^9+\color{#3D99F6}{2}a^8\\&\cdots\\&\cdots\\&=\color{#D61F06}{144}a+\color{#3D99F6}{89}\\&=\color{#D61F06}{233}+\dfrac{\color{#3D99F6}{144}}{a}\end{aligned}}$ (Recall Fibonacci Series ( $\color{#20A900}{1,1,2,3\cdots}$ )) Hence, $\large a^{12}-144a^{-1}=\boxed{233}$

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$\Large{\text{NOTE:-}}$ Indeed we could estabilish the following relation:- $\large{\color{#69047E}{a^n=F_{n+1}+F_na^{-1}}}$ where $F_i$ denotes $ith$ Fibonnaci number and put $n=12$ to solve this question.

We know,

$2<a^2<3$

$\sqrt{2}<a<\sqrt{3}$

As,

$a>1$ then $a^{-1}<1$

So,

$\{a^{-1}\}=a^{-1}$

And,

$\{a^2\}=a^2-2$

Now equating both we get,

$a^3-2a-1=0$

Solving for $a$ we get,

$a=\frac{\sqrt{5}+1}{2}=\phi$ [we will reject the other two solutions as they don't lie between $\sqrt{2}$ and $\sqrt{3}$ ]

Now placing the values,

$\phi^{12}-144\phi^{-1}$

$161+72\sqrt{5}+72-72\sqrt{5}$

$\boxed{233}$