Saudi Arabia Mathematical Competitions

We have $$a+\frac{1}{a}=[a]+\left[\frac{1}{a}\right]+\{a\}+\left\{\frac{1}{a}\right\}=[a]+\left[\frac{1}{a}\right]+1$$ is an integer and denote this integer by $k$. Let $S_n=a^n+\frac{1}{a^n}$, $n=0,1,2,\dots$ Since $a$ and $\frac{1}{a}$ are the roots of quadratic equation $x^2-kx+1=0$, it follows that $$S_{n+2}=k{S}_{n+1}-S_n,n=0,1,2,\dots$$ We have $S_0=2$, $S_1=k$ hence, by induction of step 2, we obtain that $S_n\in \mathbb{Z}$ for $n=2,3,\dots$ It follows $$\left\{a^n\right\}+\left\{\frac{1}{a^n}\right\}=a^n+\frac{1}{a^n}-\left[a^n\right]-\left[\frac{1}{a^n}\right]=S_n-\left[a^n\right]-\left[\frac{1}{a^n}\right]\in \mathbb{Z},$$ that is $\left\{a^n\right\}+\left\{\frac{1}{a^n}\right\}\in \mathbb{Z}$. We get $\left\{a^n\right\}+\left\{\frac{1}{a^n}\right\}\in \{0,1\}$. If $\left\{a^n\right\}+\left\{\frac{1}{a^n}\right\}=0$, then $a^n$ and $\frac{1}{a^n}$ are both integers, not possible. Therefore, $\left\{a^n\right\}+\left\{\frac{1}{a^n}\right\}=1$, and we are done.