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Source. Unnumbered argument on p. 2 of Sándor Z. Kiss and Csaba Sándor, Generalized Sidon sets of perfect powers, The Ramanujan Journal 59 (2022), no. 2, 351--363, doi:10.1007/s11139-022-00622-z. Pages are those of arXiv:2006.02783v1 (4 June 2020), the edition named on the source card. The paper gives these bounds no label.

Read depth. Claims checked: the statements and the short derivations were read on the printed page. Nothing here is independently reviewed.

Statement

Setting as in Theorem 2: a $B_h[g]$ set $A$ of positive integers has $R^*_{A,h}(n)\le g$ for every $n$, counting solutions of $a_1+\cdots+a_h=n$ with $a_1\le\cdots\le a_h$.

General bound (p. 2). If $A$ is a $B_h[g]$ set, then $A(x)\le\sqrt[h]{hgx\cdot h!}+h-1$. If moreover $A\subseteq(\mathbb Z^+)^k$, then $A(x)\le x^{1/k}$ as well, so

$$A(x)\ll x^{\min\{\frac1k,\frac1h\}}.$$

Squares, $k=h=2$ (p. 2). If $A$ is a $B_2[g]$ set of squares, then

$$A(x)\ll\frac{\sqrt x}{\sqrt[4]{\log x}}.$$

The second bound comes from Landau's theorem that the integers up to $x$ which are sums of two squares number asymptotically $cx/\sqrt{\log x}$: every sum of two members of $A$ up to $2x$ is such an integer. The displayed chain on p. 2 writes $\binom{A(x)}2\le\sum_{n\le2x}R^*_{A,2}(n)\le(c+o(1))\,2x/\sqrt{\log 2x}$ without the factor $g$ that the bound $R^*_{A,2}(n)\le g$ contributes to the middle step; with that factor the conclusion holds, its implied constant depending on $g$.

These bounds motivate the paper's Conjecture 1 (p. 2): for every $k\ge1$, $h\ge2$ and $\varepsilon>0$ there is a $B_h[g]$ set $A\subseteq(\mathbb Z^+)^k$ with $A(x)\gg x^{\min\{1/k,1/h\}-\varepsilon}$.

Proof pointer

Page 2: count the $h$-element subsets of $A\cap[1,x]$, whose sums lie up to $hx$ and each value of which is hit at most $g$ times.

Dependencies

Landau's theorem on sums of two squares (1908), cited by the paper.

Bears on

• Problem 158: the bound for squares with $g=2$ shows that a $B_2[2]$ set of squares, with representations counted with $a\le b$ as the problem counts them, has $\lvert A\cap\{1,\ldots,N\}\rvert/N^{1/2}\to0$; no such set is a counterexample. This settles the problem's question only for sets of squares, and the paper does not mention the problem.