---

Statement

Setting (p. 292). For an integer $k\ge2$ and positive integers $M\le N$, $f_k(M,N)$ is the least size of a set $A\subset[0,M]$ such that every positive integer $n\le N$ is $a+b^k$ with $a\in A$ and $b$ a positive integer, and $M_k(N)$ is the smallest $M$ for which such a set exists (see the Proposition).

Theorem 2 (p. 293). For all integers $k\ge2$ and $M_k(N)\le M\le N$,

$$f_k(M,N)\le(B+1)^k-B^k,\qquad B=[N^{1/k}].$$

The paper introduces it as the answer to a referee's question about upper bounds for $f_k(M,N)$ (p. 293).

Source. Wenguang Zhai, The additive completion of $k$th powers, J. Number Theory 79 (1999), 292--300, doi:10.1006/jnth.1999.2441: the setting on p. 292, Theorem 2 on p. 293, the proof in Section 4 on p. 298. The edition read is identified on the source card.

Read depth. Claims checked: the statement was read clause by clause on the printed pages, and the short proof was read. Nothing here is independently reviewed.

Proof pointer

Section 4, p. 298. When $M\ge(B+1)^k-B^k-1$, the interval of integers $\{0,1,\ldots,(B+1)^k-B^k-1\}$ lies in $S_k(M,N)$ by the proof of the Proposition, and it has $(B+1)^k-B^k$ elements. For smaller $M$ the paper refers the case to the Proposition.

Dependencies

Proposition (p. 292) of the same paper.

Bears on

• Problem 33: at $k=2$ the theorem gives, for each $N$ and each $M$ with $M_2(N)\le M\le N$, a set in $[0,M]$ of at most $2B+1\le2N^{1/2}+1$ integers completing the squares $b^2$, $b\ge1$, up to $N$. The set depends on $N$; the theorem gives no infinite set as in Problem 33 and so no bound for the limsup that the problem asks for.