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Statement

Here $\tau(n)=\sum_{d\mid n}1$ is the number of divisors of $n$ (p. 6).

Proposition 1.4 (Average value of $\tau(kab^2+1)$), p. 6, states:

For any $A,B>1$, and any positive integer $k\ll(AB)^{O(1)}$, one has

$$> \sum_{a\le A}\sum_{b\le B}\tau(kab^2+1)\ll AB\log(A+B)\log(1+k). >$$

Remark 1.5 (p. 6) says that the heuristic $\tau(n)\sim\log n$ on average suggests the true bound $O(AB\log(A+B))$, and that the factor $\log(1+k)$ can be reduced, for some ranges at least, with further tools such as the Pólya--Vinogradov inequality; the authors say the stated bound suffices for their applications.

Source. Elsholtz and Tao, arXiv:1107.1010v6, p. 6; read on the page image. Proved in Section 7 (pp. 25--34), the proof proper on p. 30. Published as J. Aust. Math. Soc. 94 (2013), no. 1, 50--105, DOI 10.1017/S1446788712000468; the published version was not compared.

Read depth. Claims checked: the statement and Remark 1.5 were read clause by clause; the proof was not read.

Proof pointer

The paper derives the bound from a quantitative form of a classical bound of Erdős on $\sum_{n\le N}\tau(P(n))$ for polynomials $P$ (Theorem 7.1, p. 25, the "Erdős-type bound"). For $A\ge B$ it sums over $a$ for each fixed $b$ (through Corollary 7.4); for $A\le B$ it applies Theorem 7.1 to the quadratic $b\mapsto kab^2+1$ for each fixed $a$ and bounds the local root counts with quadratic reciprocity (p. 30). Variants of the estimate follow in the same section (p. 32).

Dependencies

Theorem 7.1 of the paper and the number-theoretic facts collected in its Appendix A; not examined here.

Bears on

• Problem 242: only through Theorem 1.1, whose upper bounds for the Type I counts the paper obtains with it (Section 8, pp. 34--35); the proposition itself is a divisor-sum estimate and says nothing about the equation.