Existence of a sequence in $\mathbb R^2$ with specified distances Does there exist a sequence $\{a_n\}$ in $\mathbb R^2$ such that
$$|a_n-a_m|=\sqrt[4] {n-m}$$
for all $n\ge m$?
I tried the case for $\mathbb R$ which is trivial. For the $\mathbb R^2$ case I tried but could not obtain a solution, my intuition is that the existence is not possible, since the $(n+1)$-th term of the sequence would have to satisfy $n$ distance relations.
 A: There is no such sequence. In general, let $r_n > 0$ and consider the following question:

Q. Does there exist a a sequence $(a_n)_{n\geq 0}$ in $\mathbb{R}^2$ satisfying the following condition:
$$ \left\| a_n - a_m \right\| = \sqrt{r_{n-m}} \quad\text{for} \quad n > m \geq 0. \tag{*}$$

Note that OP's case corresponds to $r_n = \sqrt{n}$. We show that the answer is negative, if we impose the following conditions:
$$ \lim_{n\to\infty} r_n = \infty \qquad\text{and}\qquad \lim_{n\to\infty} (r_n - r_{n-1}) = 0. $$
Indeed, suppose otherwise. Since the condition is invariant under isometry, without losing the generality, we may assume that
$$ a_0 = (0, -\sqrt{r_1}/2) \qquad \text{and} \qquad a_1 = (0, \sqrt{r_1}/2). $$
Then for each $n\geq2$, by using the condition $\text{(*)}$ with $m = 0, 1$, we find that $a_n$ is one of $(\pm x_n, y_n)$, where $\delta_n = r_n - r_{n-1}$ and
$$ x_n = \sqrt{r_n -\frac{r_1}{4} - \frac{\delta_n}{2} - \frac{\delta_n^2}{4 r_1}} \qquad \text{and} \qquad y_n = \frac{\delta_n}{2\sqrt{r_1}}. $$
Now by noting that
$$ \| (\pm x_{n+1}, y_{n+1}) - (\mp x_n, y_n) \| \geq x_{n+1} + x_n \to \infty $$
we find that the signs of the $x$-coordinates of $a_{n+1}$ and $a_n$ are the same for all sufficiently large $n$, for otherwise the above computation contradicts $\| a_{n+1} - a_n \| = \sqrt{r_1}$. However, in such case,
$$ \sqrt{r_1} = \| a_{n+1} - a_n \| \leq |x_{n+1} - x_n| + |y_{n+1} - y_n| $$
and it is not hard to check that both $x_{n+1} - x_n$ and $y_{n+1} - y_n$ converges to $0$ by the assumption, a contradiction. Therefore there does not exist any such sequence.

Here is an alternative argument. By rescaling if necessary, one may assume that $r_1 = 1$. This forces that $\| a_{n+2} - a_n \| \leq 2$.

*

*If the equality holds, then $a_n$, $a_{n+1}$, $a_{n+2}$ must be colinear, which then forces that $a_n$ must be of the form $a_0 + n \mathbf{u}$ for some unit vector $\mathbf{u}$. Of course, this forces $r_n = n^2$.


*Otherwise, there exists a unique $\theta \in (0, \pi)$ such that $ \| a_{n+2} - a_n \| = 2\cos(\theta/2) $. This $\theta$ satisfies $ \langle a_{n+1} - a_n, a_{n+2} - a_{n+1} \rangle = \cos\theta$, i.e., $\theta$ describes the change in direction. Now by identifying $\mathbb{R}^2$ with $\mathbb{C}$ and applying a suitable isometry, one may assume
$$ a_0 = 0, \qquad a_1 = 1, \qquad a_2 = 1 + e^{i\theta}. $$
Then we must have either
$$ a_3 = 1 + e^{i\theta} + e^{2i\theta} \quad\text{or}\quad 2 + e^{i\theta}. $$
Notice that each of the choices leads to
$$ \sqrt{r_3} = \| a_3 - a_0 \| = \left|1 + 2\cos\theta \right| \quad\text{or}\quad \sqrt{5+4\cos \theta}. $$
It is easy to check that these two possible candidates of $\sqrt{r_3}$ always differ for $\theta \in (0, \pi)$:

So, there are only two possible choices of $r_3$. Moreover, each choice completely determines the configuration of $(a_0, a_1, a_2, a_3)$ up to isometry, which then can be used to show that all of $(a_n)_{n\geq 0}$ is determined as either
$$ a_n = \sum_{k=0}^{n-1} e^{ik\theta} \quad \text{for all} \quad n \geq 0 $$
when $\sqrt{r_3} = \left|1 + 2\cos\theta \right|$ or
$$ a_n = \lceil n/2 \rceil + \lfloor n/2 \rfloor e^{i\theta} \quad \text{for all} \quad n \geq 0 $$
when $\sqrt{r_3} = \sqrt{5+4\cos \theta}$. This of course completely determines the sequence $r_n$:

