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Statement

Setting (p. 292). Fix an integer $k\ge2$. For positive integers $M\le N$, $S_k(M,N)$ is the family of sets $A\subset[0,M]$ such that every positive integer $n\le N$ is $a+b^k$ with $a\in A$ and $b$ a positive integer, and $f_k(M,N)=\min\{\lvert A\rvert:A\in S_k(M,N)\}$. $M_k=M_k(N)$ is the smallest integer $M$ for which $S_k(M,N)$ is non-empty.

Proposition (p. 292). With $B=[N^{1/k}]$, the integer part of $N^{1/k}$,

$$B^k-(B-1)^k-1\le M_k\le(B+1)^k-B^k-1.$$

Before the statement the paper gives, as examples, that for an integer $S\ge4k$ one has $M_k(S^k-1)=S^k-(S-1)^k-1$ and $M_k(S^k+1)=(S+1)^k-S^k-1$; in both examples $M_k$ equals the upper bound (with $B=S-1$ and $B=S$ respectively). The paper does not prove these examples.

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 and the Proposition on p. 292, the proof in Section 2 on p. 294. The edition read is identified on the source card.

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

Proof pointer

Section 2, p. 294. For the upper bound, the interval of integers $\{0,1,\ldots,(B+1)^k-B^k-1\}$ completes the $k$th powers up to $N$, since each $n\le N$ lies between consecutive $k$th powers $b^k\le n\le(b+1)^k$. For the lower bound, the integer $n=B^k-1$ can only use some $b\le B-1$, which forces an element $a\ge B^k-(B-1)^k-1$.

Bears on

• Problem 33: at $k=2$ the Proposition gives $2B-2\le M_2(N)\le2B$ with $B=[N^{1/2}]$: a finite set completing the squares $b^2$, $b\ge1$, up to $N$ has an element at least $2B-2$, and some such set lies in $[0,2B]$. It concerns finite completions only and decides neither question of the problem.