---

Source. Theorem 1.1, p. 1, of Trevor D. Wooley, Sums of three cubes, II, Acta Arith. 170 (2015), 73--100, read in the arXiv version arXiv:1502.01944v1 named on the source card; labels and pages here are that version's.

Read depth. Claims checked: the statement and the definition of $N(X)$ were read clause by clause on p. 1, and the deduction on p. 24 for its structure. Nothing here is independently reviewed.

Statement

Let $N(X)$ be the number of integers not exceeding $X$ that are the sum of three cubes of natural numbers (p. 1).

Theorem 1.1 (p. 1). "One has $N(X)\gg X^\beta$, where $\beta=0.91709477$."

For comparison the paper records (pp. 1--2) the author's earlier lower bound $N(X)\gg X^{1-\xi/3-\varepsilon}$ (its reference [26]), with $\xi=(\sqrt{2833}-43)/41=0.24941301\ldots$ and $1-\xi/3=0.91686232\ldots$, and the conditional estimate $N(X)\gg X^{1-\varepsilon}$ of Hooley and Heath-Brown, which assumes an unproved Riemann Hypothesis for a certain Hasse--Weil $L$-function.

Proof pointer

Section 7, p. 24. The paper calls the theorem a standard consequence of Theorem 1.2 after Cauchy's inequality: the argument of the author's earlier paper (its reference [26], Theorem 1.1 and §2) gives $N(X)\gg X^{1-\delta_6/3-\varepsilon}$ whenever $\delta_6$ is an associated sixth-moment exponent, and the exponent $\delta_6=0.24871567$ of Theorem 1.2 gives $1-\delta_6/3=0.917094776\ldots$, so $\beta=0.91709477$ absorbs the $\varepsilon$.

Dependencies

Theorem 1.2 of the same paper; T. D. Wooley, Sums of three cubes, Mathematika 47 (2000), 53--61 (the paper's reference [26]), for the deduction.

Bears on

• Problem 325: since a sum of three cubes of natural numbers is a sum of three nonnegative cubes, $f_{3,3}(x)\ge N(x)\gg x^{0.91709477}$. This is a lower bound for the case $k=3$, below the exponent $3/k=1$ the problem asks for, and it does not give $f_{3,3}(x)\gg x^{1-\epsilon}$.