Let $\{a_{k}\}$ be an increasing sequence of positive real numbers. Let us define $A_{n}:=\{a_1,\ldots,a_{n}\}$ and consider the set

$$ S_{n}^{(2)}=\Big \{(x_1,x_2,x_3,x_4)\in A_{n}^4:x_1-x_2+x_3-x_4=0 \Big\}. $$

In other words, the size of the set $S_n^{(2)}$ is equal to the number of solutions to the *sum-set equation*

$$
2A_{n}-2A_{n}=0.
$$

For instance, if $a_k=k$ then it is not difficult to see that $$ |S_{n}^{(2)}|=\frac{2}{3}n^3+o(n^3), $$ and that if $a_{k}=2^k$ then $$ |S_{n}^{(2)}|=2n^2-n. $$

My question is: let $\alpha>0$ and $\alpha\neq 1$ is it true that if we define $a_{k}=k^{\alpha}$ then $$ > |S_{n}^{(2)}|=o(n^3)? $$

**Update 1:** Simulations suggest that the previous statement is true at least for $\alpha=\frac{1}{2},2,3,4$.

However, for the sequence $a_{k}=\text{the k-th prime number}$, simulations again suggest that $$ \frac{|S_n^{(2)}|}{n^3}\to \beta>0. $$ Can this be right?

*Answer: No. As Ben and Johan pointed out below this is not true and the limit is actually $\beta=0$.*

Update 2:Define for $r\geq 2$ the set $S_{n}^{(r)}$ as the solutions to the sumset equation $$ > rA_{n}-rA_{n}=0. $$Is it true that for the sets above ($a_{k}=k^{\alpha}$) $$ > |S_{n}^{(r)}|=o(n^{2r-1})? $$

Thanks!