What if $\alpha$ is a transcendental number? Then, none of the differences $\alpha^k-\alpha^m$ can be integer and thus, all terms in the sequence $\{\alpha^k\}$, $k=1,2\ldots$ are distinct.

Hopeooooo wrote: In an infinite sequence $\{\alpha\}, \{\alpha^2\}, \{\alpha^3\}, \cdots $ there are infinitely many distinct values$.$ Show that $\alpha$ is an integer$. (\{x\}$ denotes the fractional part of$ x.)$ (Golovanov A.S.) Wrong. If $\alpha$ is an integer, all elements of the sequence are equal to zero.

Do you mean "finitely many" instead?

Yes, I wrote.

Bump on this one

Claim: If there exist $a,b,c,d$ such that $\{\alpha^a\} = \{\alpha^b\}$ and $\{\alpha^c\} = \{\alpha^d\}$ and $a-b = c-d$ (and $a \neq c$), then we are done Proof: WLOG $a > b$ and $a > c$. We have that $n_2 = \alpha^a - \alpha^b = \alpha^{a-c}(\alpha^c - \alpha^d) = (\alpha^{a-c})n_1$ for some integers $n_1, n_2$, which gives that $\alpha^k$ is rational for some $k$. If this rational number is not an integer, then consider the subsequence $\alpha^k, \alpha^{2k}, \cdots$ which clearly cannot take finitely many values. So suppose $\alpha^{k} = n$ for some integer $n$ and let $z = \{\alpha\}$ and consider the numbers $z, nz, n^2z, n^3z$ and so on, which cannot take finitely many values if $z$ is irrational, so we have that $z = 0$ and so $a$ is an integer. $\square$ Suppose there are $k$ different values it takes, and consider the first $Nk$ terms for large $N$, there must be some value appearing at least $N$ times among these. Consider the consecutive differences between these values, if any two of them are the same, we're done by the Claim, so all of them are pairwise distinct so there must be at least $1 + 2 + \cdots + N-1 = \frac{N(N-1)}{2} > Nk$ values, a contradiction, so we're done. $\blacksquare$ Thanks @3below, I've fixed my sol I think

nice problem

edit: this solution is turbo wrong but can be fixed i think

L567 wrote: Claim: If there exist $a,b,c,d$ such that $\{\alpha^a\} = \{\alpha^b\}$ and $\{\alpha^c\} = \{\alpha^d\}$ and $a-b = c-d$ (and $a \neq c$), then we are done Proof: WLOG $a > b$ and $a > c$. We have that $n_2 = \alpha^a - \alpha^b = \alpha^{a-c}(\alpha^c - \alpha^d) = (\alpha^{a-c})n_1$ for some integers $n_1, n_2$, which forces that $\alpha$ is an rational number. But clearly in this case, it is not possible unless its an integer. $\square$ [...]$\blacksquare$ @L567 Why can't $\alpha$ be $\sqrt{x}$ such that $x$ is rational and $a-c$ is even.

L567 wrote: Claim: If there exist $a,b,c,d$ such that $\{\alpha^a\} = \{\alpha^b\}$ and $\{\alpha^c\} = \{\alpha^d\}$ and $a-b = c-d$ (and $a \neq c$), then we are done Proof: WLOG $a > b$ and $a > c$. We have that $n_2 = \alpha^a - \alpha^b = \alpha^{a-c}(\alpha^c - \alpha^d) = (\alpha^{a-c})n_1$ for some integers $n_1, n_2$, which gives that $\alpha^k$ is rational for some $k$. If this rational number is not an integer, then consider the subsequence $\alpha^k, \alpha^{2k}, \cdots$ which clearly cannot take finitely many values. So suppose $\alpha^{k} = n$ for some integer $n$ and let $z = \{\alpha\}$ and consider the numbers $z, nz, n^2z, n^3z$ and so on, which cannot take finitely many values if $z$ is irrational, so we have that $z = 0$ and so $a$ is an integer. $\square$ Suppose there are $k$ different values it takes, and consider the first $Nk$ terms for large $N$, there must be some value appearing at least $N$ times among these. Consider the consecutive differences between these values, if any two of them are the same, we're done by the Claim, so all of them are pairwise distinct so there must be at least $1 + 2 + \cdots + N-1 = \frac{N(N-1)}{2} > Nk$ values, a contradiction, so we're done. $\blacksquare$ Thanks @3below, I've fixed my sol I think A faster approach: Since there are only finitely many pairs $(\{\alpha^n\},\{\alpha^{n+1}\})$, we can conclude that there exists some $i<j$ such that $\{\alpha^i\}=\{\alpha^j\}$ and $\{\alpha^{i+1}\}=\{\alpha^{j+1}\}$.

So, we are given that $S_{\alpha}:=\{ \{\alpha\}, \, \{\alpha^2\}, \, \{\alpha^3\}, \ldots \}$ is a finite set. Then, $\{ \{\alpha\}, \, \{\alpha^3\}, \, \{\alpha^5\}, \ldots \}$ is also a finite set. That means there are integers $0\le k_1 < k_2 < \ldots $ such that $\{\alpha^{2k_1+1}\} = \{\alpha^{2k_2+1}\} =\ldots $. Now, we consider the set $\{ \{\alpha^{2k_1+2}\}, \, \{\alpha^{2k_2+2}\}, \, \{\alpha^{2k_3+2}\}, \ldots \}$, which has finitely many different elements (as being a subset of $S_{\alpha}$). Then, we have some integers $0\le k_{i_1} < k_{i_2} < \ldots $ among $k_i$' that satisfies $\{\alpha^{2k_{i_1}+2}\} = \{\alpha^{2k_{i_2}+2}\} =\ldots $. Hence, there exists a pair $m\neq n$ of positive integers such that \[ \{\alpha^{m}\} = \{\alpha^{n}\} \quad \text{and} \quad \{\alpha^{m+1}\} = \{\alpha^{n+1}\}, \]i.e. \[ \alpha^{m} - \alpha^{n} \in \mathbb{Z} \quad \text{and} \quad \alpha^{m+1} - \alpha^{n+1} \in \mathbb{Z}. \]This implies that $\alpha$ should be a rational number (taking the ratio). Now, the existing of $m\neq n$ with $\alpha^m-\alpha^n \in \mathbb{Z}$ is enough to say that $\alpha$ is an integer.