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Source. Theorem 6, p. 458, of Kurt Mahler, On the lattice points on curves of genus 1, Proc. London Math. Soc. (2) 39 (1935), 431--466, the edition named on the source card.

Read depth. Claims checked: the statement was read clause by clause on p. 458. The paper gives no separate proof; it says the theorem follows by specializing the form and the angle in Theorem 5, whose proof was read for its structure. Nothing here is independently reviewed.

Statement

Theorem 6 (p. 458). There is an infinite set of positive integers $k_1,k_2,k_3,\ldots$ with

$$1\le k_1<k_2<k_3<\cdots$$

such that the number of representations of $k_\nu$ as a sum of two cubes of positive integers is greater than $\sqrt[4]{\log k_\nu}$.

The print does not say whether the two orders of the summands count as different representations.

Proof pointer

P. 458. The paper says only that Theorems 6 and 7 follow by specializing $F$ and $G$ in Theorem 5. One reading of that step: take $F(x,y)=x^3+y^3$, which has only simple linear factors, and an angle with $0<A<B$, so that the solutions in $G$ have $x,y$ nonzero and of one sign; since $F(-x,-y)=-F(x,y)$, changing all signs if $k<0$ gives at least $t$ representations of $|k|$ by cubes of positive integers. Taking $\gamma<1$, the bound $|k|\le e^{\gamma t^4}$ gives $t>(\log|k|)^{1/4}$, and letting $t\to\infty$ gives infinitely many such integers.

Dependencies

Theorem 5.

Bears on

• Problem 829, as a lower bound for the quantity the problem asks to bound above: with $A$ the set of cubes, every representation counted in Theorem 6 is a pair of cubes of positive integers summing to $k_\nu$, so $1_A\ast1_A(k_\nu)>(\log k_\nu)^{1/4}$ for infinitely many $\nu$, whichever way the print counts order. A bound $1_A\ast1_A(n)\ll(\log n)^{c}$ would therefore need $c\ge1/4$. The paper proves no upper bound, which is what the problem asks for.