XXI SILK ROAD MATHEMATICAL COMPETITION

Step 1. We show that there is a positive integer $l$ such that ${\alpha }^l$ is rational. Say the sequence is of length $k-1$. For any positive integer $n$ the sequence $\{{\alpha }^{nk}\}$, $\{{\alpha }^{nk+1}\}$, ..., $\{{\alpha }^{nk+k-1}\}$ contains two equal elements. Hence, there are infinitely many pairs $i,j$, $0<i-j<k$, such that $\{{\alpha }^i\}=\{{\alpha }^j\}$, i.e. ${\alpha }^j({\alpha }^{i-j}-1)$ is an integer. Since there are finitely many possible values of $i-j$, at least one of them occurs infinitely often. So we can find $m$ such that ${\alpha }^j({\alpha }^m-1)$ is an integer for infinitely many $j$. We can divide two such numbers to get ${\alpha }^l$ is rational for some positive integer $l$.

Step 2. Conclusion. Now if ${\alpha }^l$ is not an integer, say, is equal to $\frac{a}{b}$ for $b>1$, $\text{gcd}(a,b)=1$, then $\{{\alpha }^{ln}\}$ is an irreducible fraction with denominator $b^n$. This is true for any $n$ so we get infinitely many values, contradiction. So ${\alpha }^l$ is an integer. If $\alpha$ is irrational, then ${\alpha }^{nl+1}={\alpha }^{nl}\cdot \alpha$ is irrational for any natural $n$ and have distinct fractional parts (Indeed, ${\alpha }^{il+1}-{\alpha }^{jl+1}=\alpha ({\alpha }^{il}-{\alpha }^{jl})$ cannot be an integer). This contradicts finiteness as well. Thus $\alpha$ is rational, and similar to above we can conclude it is an integer.