$\lfloor x^k \rfloor \equiv m \pmod{n}$ with $x$ irrational

Question

Let $x>1$ be an irrational number, and $n$ a positive integer. Is it true that, for each integer $m$, there exists an integer $k$ such that $$ \lfloor x^k \rfloor \equiv m \pmod{n}? $$

Answer 1

Edit: the first version incorrectly ignored the sign of the small term. I have repaired that mistake.

In a similar spirit to the comment by Travis, it's good to look at expressions for which we know the fractional part. For example:

$$A_n=\left(1+\sqrt2\right)^n+\left(1-\sqrt2\right)^n$$

these are always integral (indeed, The $A_n$ satisfy the Fibonacci-like recurrence $A_n=2A_{n-1}+A_{n-2}$) and the second term goes to $0$ rapidly. Hence the integer part of the first term is either $A_n-1$. or $A_n$ according to the parity of $n$. We will use this to show that $\alpha=1+\sqrt2$ is a counterexample.

But work $mod(7)$, where $\sqrt2 = 3$. We see that the $A_n$ are periodic. In particular, they only take the values $2,5,6,0$ $mod(7)$. Checking cases we see that the only residue classes hit by the integer part of $\alpha^n$ are $0,1,2,5$ $mod(7)$.Thus we have a counterexample.