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Claim. T. D. Browning and D. R. Heath-Brown, Equal sums of three powers, Invent. Math. 157 (2004), no. 3, 553--573. Their Theorem bounds, for every $d\ge25$, the number of positive integer solutions of

$$x_1^d+x_2^d+x_3^d=x_4^d+x_5^d+x_6^d$$

with $\max x_i\le B$ in which $(x_4,x_5,x_6)$ is not a permutation of $(x_1,x_2,x_3)$ (the paper calls the permutations the trivial solutions), and shows that these non-trivial solutions number $o(B^3)$ for every $d\ge33$. Their Corollary follows: with $r(n)$ the number of triples $(x_1,x_2,x_3)\in\mathbb N^3$ with $x_1^d+x_2^d+x_3^d=n$,

$$\sum_{n\le x}r(n)^2\sim6cx^{3/d}, \qquad c=\frac{\Gamma(1+1/d)^3}{\Gamma(1+3/d)},$$

and for $d\ge33$ asymptotically $\tfrac16cx^{3/d}$ integers $n\le x$ are sums of three $d$th powers, almost all of them with essentially one representation. Whichever convention the paper's $\mathbb N$ follows, $f_{k,3}$ in Problem 325 counts sums of three nonnegative $k$th powers and so counts at least these integers, hence $f_{k,3}(x)\ge(c/6+o(1))x^{3/k}$ for every $k\ge33$, and the answer to both forms of the question is yes for these $k$.

Covers. Both forms of the question, $f_{k,3}(x)\gg x^{3/k}$ and $f_{k,3}(x)\gg_\epsilon x^{3/k-\epsilon}$, for every $k\ge33$. The paper does not treat $3\le k\le32$; for $k=3$ the problem page records Wooley's bound.

Depends on. No page of this wiki.

Acceptance. Refereed: Inventiones Mathematicae 157 (2004), no. 3, 553--573, the DOI linked above; the page is dated by the article's online publication date in the Crossref record, 17 March 2004. Not reviewed: the paper reached the problem's thread through two comments of 2026-03-09, the first of which credits ChatGPT Deep Research with locating it and quotes the Corollary, and the second of which gives the published reference. Comments on the thread are not curator credit, the site's commentary does not mention the paper, and the site labels the problem OPEN. The proof is not checked in this corpus.