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Statement
Let be a prime (the paper's standing hypothesis, p. 1). For pairs of subintervals , , the paper's congruence (2) asks for with (p. 1).
Theorem 1 (p. 2). Let , and for let be subintervals with and , the first intervals pairwise disjoint: for all . Then some has integers , with , provided
The print writes the condition with and names no constant; the implied constants in its proof (pp. 5--6) are absolute, so the condition reads for a suitable absolute constant .
The case (p. 2). The paper remarks that gives back the earlier criterion it describes on p. 1: two arbitrary subintervals contain an inverse pair when . The paper says (p. 1) that Heath-Brown highlights this as the best result to date, and does not prove it separately.
Source. T. D. Browning and A. Haynes, Incomplete Kloosterman sums and multiplicative inverses in short intervals, Int. J. Number Theory 9 (2013), 481–486; read in the arXiv version 1204.6374v1, Theorem 1 and the remark on p. 2, the earlier criterion on p. 1, the proof in Section 3 on pp. 5--6. The edition is identified on the source card.
Read depth. Claims checked: the statement and the remark were read clause by clause against the print. The proof (pp. 5--6) was read for its structure only, not verified.
Proof pointer
Section 3, pp. 5--6. The number of solutions in the -th pair is written by orthogonality as a main term plus an error ; summed over the main terms give (display (5)). The geometric-series estimate in reduces to incomplete Kloosterman sums over weighted by , and Cauchy's inequality with Theorem 2 bounds . Under the hypothesis on the main term dominates.
Dependencies
Theorem 2 of the same paper, the mean value bound for incomplete Kloosterman sums, which rests on Weil's bound for complete Kloosterman sums.
Bears on
- Problem 445: the case is the two-interval criterion that the problem page applies, through a short deduction stated there (reduce modulo and use the longer block of nonzero residues for both intervals), to get an inverse pair in every interval for each fixed and all large . The deduction is the problem page's, not the paper's. The logarithmic factor keeps it from reaching , and the criterion says nothing about .