Does there exist a sequence in $\mathbb{R}^2$ such that $\|a_n-a_m\|=\sqrt[4]{n^3-m^3}$? Does there exist a non-trivial sequence in $\mathbb{R}^2$ such that  $\|a_n-a_m\|=|\sqrt[4]{n^3-m^3}|\ \ \forall n\ge m$, where $a_i$ are terms of the sequence?
I think no. For, when we substitute the value of the distance, we get that  $(n_1-m_1)^2+(n_2-m_2)^2=|\sqrt{n^3-m^3}|$. Is this possible for any $n,m$? Any hints? Thanks beforehand.
 A: You may just consider the instances $m=n-1, m=n-2, m=n-3$. Let us define $b_n$ as $a_n-a_0$.
The instance $m=0$ gives $\|b_n\|=n^{3/4}$ while the instances $m=n-1,m=n-2,m=n-3$ give
$$ \|b_n-b_{n-1}\|\sim\|b_{n-1}-b_{n-2}\|\sim\|b_{n-2}-b_{n-3}\|\sim 3^{1/4} \sqrt{n} $$
$$ \|b_{n}-b_{n-2}\|\sim\|b_{n-1}-b_{n-3}\| \sim 6^{1/4} \sqrt{n} $$
$$ \|b_n-b_{n-3}\|\sim 9^{1/4}\sqrt{n}. $$
Now you might wonder about the existence of a quadrilateral $ABCD$ such that
$$ AB=BC=CD= 3^{1/4} $$
$$ AC=BD= 6^{1/4} $$
$$ AD = \sqrt{3} $$
and use the cosine theorem to prove it doesn't exist: the first five constraints indeed ensure $AD=3^{1/4}\sqrt{1+2\sqrt{2}}\gg\sqrt{3}$ or $AD=3^{1/4}\sqrt{3-2\sqrt{2}}\ll\sqrt{3}$:


Also an alternative approach by circle packing. Let us assume that such sequence exists and consider the $b_n$s for $n\in[N^2,(N+1)^2]$. We have $2N+2$ points in the annulus $N^{3/2}\leq \rho \leq (N+1)^{3/2}$, whose area is $\sim 3\pi N^2$. The distance between two points in the annulus is at least $3^{1/4}N$, so the circles centered at these $b_n$s with radius $\frac{3}{5}N$ are disjoint. The total area of $2N+2$ circles with radius $\frac{3}{5}N$ is $\Theta(N^3)$, while the area of the annulus $N^{3/2}-\frac{3}{5}N\leq \rho \leq (N+1)^{3/2}+\frac{3}{5}N$ is $\Theta(N^2)$, so we have reached a contradiction.
