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Statement

Setting: $k\ge2$ and $N$ as in Theorem 1; $b$ runs over positive integers.

Lemma 2 (p. 2). Let $f(x)=g(x/N)$, where $g$ is continuous and differentiable except at a finite number of points. Then

$$\sum_{n=1}^Nf(n)=N\int_0^1g(x)\,dx+O(1)$$

and

$$\sum_{b\le(N-a)^{1/k}}f(a+b^k)=N^{1/k}\,h\Bigl(\frac aN\Bigr)+O(1), \qquad h(x)=\int_0^{(1-x)^{1/k}}g(x+t^k)\,dt$$

(the definition of $h$ is the paper's (1)), and the constants in the error terms are independent of $N$.

The lemma does not say over which $a$ the second identity is asserted; the proof of Theorem 1 applies it to each $a\in A^N$ (the sum (2), p. 2), and $h(a/N)$ as defined by (1) requires $a\le N$.

Source. J. Cilleruelo, The additive completion of $k$th-powers, J. Number Theory 44 (1993), no. 3, 237--243, doi:10.1006/jnth.1993.1049, read in the author-typeset manuscript identified on the source card: the lemma on p. 2, in Section 1 (pp. 1--2).

Read depth. Claims checked: the statement and its hypotheses were read clause by clause on the page image. The paper gives no proof beyond the remark that it follows from Euler's identity; none was written out here. Nothing here is independently reviewed.

Proof pointer

P. 2: the paper says only that the proof is a straightforward application of Euler's identity. In outline, both sums are scaled Riemann sums: the first, divided by $N$, of $g$ on $[0,1]$ at spacing $1/N$, and the second, divided by $N^{1/k}$ after the substitution $t=bN^{-1/k}$, of $t\mapsto g(a/N+t^k)$ on $[0,(1-a/N)^{1/k}]$ at spacing $N^{-1/k}$.

Bears on

• Problem 33: the problem admits the square $0^2$, which Theorem 1 excludes by taking $b\ge1$. The problem's claim page for this paper uses the $O(1)$ error of this lemma to carry the proof of Theorem 1 over to $b=0$: the extra term $f(a)=g(a/N)$ in each inner sum is bounded by the maximum of $g$. That extension is the claim page's, not the paper's.