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Statement

Setting (printed pp. 445--446): $p$ is a prime, $\mathbf e_p(z)=\exp(2\pi iz/p)$, and $\mathscr P(X,Y)$ is the set of primes in the interval $[X,Y]$. Display (2) defines $\mathscr W(X)=\{w=rl:r,l\in\mathscr P(X,2X)\}$, the set of products of two primes from $[X,2X]$, and for an integer $a$

$$S_a(X)=\sum_{w\in\mathscr W(X)}\mathbf e_p\left(aw^{-1}\right),$$

where $w^{-1}$ is the inverse of $w$ modulo $p$.

Lemma 2 (printed p. 446). Let $m\ge1$ be an integer and let $X$ be defined by $m(2X)^{2m-1}=p-1$. Then

$$\max_{1\le a\le p-1}|S_a(X)|\le2m^2X^{2-1/2m^2}.$$

Here $X^{2-1/2m^2}$ means $X^{2-1/(2m^2)}$, the reading under which the proof's last display gives the bound (an authored reading of the notation).

Source. I. E. Shparlinski, On a question of Erdős and Graham, Arch. Math. (Basel) 78 (2002), no. 6, 445--448, DOI 10.1007/s00013-002-8269-2; Lemma 2 and its proof on printed p. 446. The edition read is identified in the source digest.

Read depth. Claims checked: the statement and the definitions it uses were read clause by clause on the page images. The proof was read; it rests on Theorem 2 of Friedlander and Iwaniec, which is not held, so no step was checked against its input. Nothing here is independently reviewed.

Proof pointer

Printed p. 446. The paper compares $S_a(X)$ with half the sum $\sigma_a(X)$ of $\mathbf e_p(a(rl)^{-1})$ over ordered pairs $r,l\in\mathscr P(X,2X)$; the two differ by at most $(X+1)/2$, the contribution of the diagonal $r=l$. It takes the bound $\max_{1\le a\le p-1}|\sigma_a(X)|\le m^2X^{2-1/m}p^{1/m^2}$ from Theorem 2 of Friedlander and Iwaniec (the paper's [3], based on Karatsuba's technique [4,5]), substitutes $p=m(2X)^{2m-1}+1$, and finishes with $2^{(2m-1)/2m^2}(m+1)^{1/2m^2}\le2$. Not checked here.

A filing observation, not a review verdict: in the proof's display the first line prints the factor $p^{1/m^2}$, while the next line, written as equal to it, raises $m(2X)^{2m-1}+1=p$ to the power $1/2m^2$; the lemma's exponent $2-1/2m^2$ follows from the second form, since $X^{2-1/m}X^{(2m-1)/2m^2}=X^{2-1/2m^2}$. Which exponent Theorem 2 of Friedlander and Iwaniec gives was not checked, as that paper is not held.

Dependencies

Outside the paper: Theorem 2 of J. Friedlander and H. Iwaniec, The Brun--Titchmarsh theorem, Analytic Number Theory, Lond. Math. Soc. Lecture Note Ser. 247 (1997), 363--372, which the paper describes as based on the technique of Karatsuba's two 1995 papers in Izv. Ross. Akad. Nauk Ser. Mat. 55, nos. 4 and 5 (its [4] and [5]). None of these is held.

Bears on

• Problem 1180: no direct bearing; the lemma is the exponential-sum input to Theorem 3, the paper's answer to the question, and the problem page names it as the step that rests on the Friedlander--Iwaniec theorem, which is not held.