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Statement

Setting (p. 11). Fix an integer $p\ge2$. For an integer $N\ge1$, a minimal additive complement of the $p$-th powers up to $N$ is a subset $B$ of $\{0,1,\ldots,N\}$ of smallest cardinality such that every integer $n$ with $1\le n\le N$ is $b+k^p$ for some $b\in B$ and some integer $k$. Its cardinality is $b_p(N)$, and $\alpha(p)$ is the liminf, as $N\to\infty$, of $b_p(N)/N^{1-1/p}$; the print writes $b(N)$ in this definition where the context calls for $b_p(N)$.

Theorem (p. 11, unnumbered, quoted). "If, for a $\delta$ in the interval $(0,1)$ and all large integers $N$, there is a minimal additive complement of the $p$-th powers up to $N$ contained in the interval $[0,\delta N]$, then one has the following inequality." The inequality, displayed as (20):

$$\alpha(p)\ge\frac{p}{\dfrac{1-\delta^{1-\frac1p}}{(1-\delta)^{1-\frac1p}} +\Bigl(1-\dfrac1p\Bigr)\displaystyle\int_0^\delta \frac{dt}{t^{1-\frac1p}(1-t)^{\frac1p}}}. \qquad(20)$$

The paper calls this the generalization of Theorem 1 to higher powers, says the method of the note adapts easily, and records the statement for completeness; it gives no proof.

Consistency checks (observations of this page, not of the paper). For $p=2$ the right side of (20) is the right side of Theorem 1's inequality (1), since $(1-\sqrt\delta)/\sqrt{1-\delta}=\sqrt{1-\delta}/(1+\sqrt\delta)$ and $\tfrac12\int_0^\delta dt/\sqrt{t(1-t)}=\sin^{-1}\sqrt\delta$, the two identities the paper uses on p. 10. As $\delta\to0$ the right side tends to $p$; as $\delta\to1$ the first term of the denominator tends to $0$ and the integral to $\pi/\sin(\pi/p)$, so the right side tends to $p^2\sin(\pi/p)/((p-1)\pi)$, which is $4/\pi$ at $p=2$.

Source. R. Balasubramanian and D. S. Ramana, Additive complements of the squares, C. R. Math. Rep. Acad. Sci. Canada 23 (2001), no. 1, 6-11: Section 5 (Concluding Remarks), p. 11. 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 page. The paper prints no proof, so none was checked. Nothing here is independently reviewed.

Proof pointer

None in the paper. It states (p. 11) that the method used for Theorem 1 adapts easily to $p$-th powers.

Dependencies

The method of Theorem 1 of the same paper.

Bears on

• Problem 33: only through the case $p=2$, where the statement coincides with Theorem 1, whose page states the relation. The cases $p\ge3$ concern complements of higher powers, which Problem 33 does not ask about.