$$\mathbf S_x := \big\{\left\lfloor x^n\right\rfloor : n\in \mathbb N\big\}$$
are Sidon for $x\ge 2$. In a recent preprint, I investigated the tails of these sequences in the range $x\in (1,2)$. In particular, one of the results I proved is that $\mathbf S_x$ has Sidon tail for almost all $x\in (1,2)$. It was only after I uploaded the preprint, and it was only after I sent the paper to a journal, that I realised that an even stronger statement is true. This is what I will prove here.
To keep this post self-contained, we will call an $\mathbf S_x$ tail Sidon if there exists $N_0$ such that the set
$$\big\{\left\lfloor x^n\right\rfloor : n\ge N_0\big\}$$
is a Sidon set.
Theorem: If $\mathbf S_x$ is not tail Sidon, then $x$ is algebraic. In particular, one can produce integers $u> w\ge v\ge 1$ such that
$$1+y^u-y^v-y^w=0 \qquad\text{or}\qquad 1-y^v-y^w=0$$
for $y:=\frac 1x \in (0,1)$.
Proof: Assume an $x$ for which $\mathbf S_x$ is not tail Sidon. Note that $\mathbf S_x$ is eventually increasing - so, from now on, we will only work in this increasing tail. Let $\mathbf x_n:=\left\lfloor x^n\right\rfloor$. It is easy to show that if there are indices $a$, $b$, $c$ and $d$ such that $\mathbf x_d+ \mathbf x_a = \mathbf x_b+ \mathbf x_c$, then $a<b\le c<d$.
Now, write $\theta_n := x^n-\lfloor x^n\rfloor \in [0,1)$. If $\mathbf x_d+\mathbf x_a=\mathbf x_b+\mathbf x_c$, then
$$x^d+x^a-x^b-x^c = (\theta_d+\theta_a)-(\theta_b+\theta_c)$$
implying $\bigl|x^d+x^a-x^b-x^c\bigr|<2$.
Now, define
$$y:=\frac1x\in(0,1),\qquad u:=d-a,\quad w:=d-b,\quad v:=d-c$$
so that
$$\bigl|1+y^u-y^w-y^v\bigr|<2y^d$$
for infinitely many quadruples $(u,v,w,d)$ with $d\to\infty$.
Choose $D_0$ for which $4y^{D_0}<1$. Consider any collision $-2y^d < 1+y^u-y^w-y^v < 2y^d$ for $d\ge D_0$. This gives
$$y^v+y^w > 1+y^u-2y^d \ge 1-2y^d > \frac12$$
since $y^u\ge 0$ and $2y^d<\tfrac12$. Since $v\le w$ and $0<y<1$, we have $y^v\ge y^w$, hence $y^v+y^w \le 2y^v$. Therefore $2y^v>\tfrac12$, i.e. $y^v>\tfrac14$. Because $v\mapsto y^v$ is strictly decreasing, there is a bound $V=V(y)$ such that for every collision with $d\ge D_0$, one has $v\le V$.
Using $y^v+y^w > 1-2y^d$ in a similar way as in the last argument, we also get a bound $W=W(y)$ such that for every collision with $d\ge D_1$, one has $w\le W$. In other words, all sufficiently large collisions lie in the finite set $\{(v,w)\in\mathbb N^2:\ 1\le v\le w\le W\}$. Since there are infinitely many collisions with $d\to\infty$, by the pigeonhole principle there exist fixed integers $v_0$, $w_0$ and an infinite subsequence of collisions along which $(v,w)=(v_0,w_0)$ and $d\to\infty$.
Let $C:=C(y):=y^{v_0}+y^{w_0}$ so that
$$\bigl|1+y^{u}-C\bigr|<2y^{d}\xrightarrow[]{d\to\infty}0$$
so that $y^{u}\to C-1$.
If $C=1$, then $y^{v_0}+y^{w_0}=1$, i.e., $1-y^{v_0}-y^{w_0}=0$.
If $C>1$, then $y^{u}\to C-1>0$. Since $y^n\to 0$, the exponents $u$ cannot tend to $\infty$; more formally, choose $M$ such that $2y^M<C-1$. For all sufficiently large terms of the subsequence, $2y^{u}>C-1$ implying $u<M$. Thus, $u$ takes values in a finite set $\{1,2,\dots,M-1\}$, so there exists $u_0$ and a further infinite subsequence with $u=u_0$ constant.On that further subsequence, we have
$$\bigl|1+y^{u_0}-C\bigr|<2y^{d}\xrightarrow[]{d\to\infty}0$$
$$\bigl|1+y^{u_0}-C\bigr|<2y^{d}\xrightarrow[]{d\to\infty}0$$
implying $1+y^{u_0}-C = 0$, and hence, $1+y^{u_0}-y^{v_0}-y^{w_0}=0$.
This completes the proof in both cases.
This completes the proof in both cases.
