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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Statement

Let pp be a prime and, for integers m,cm,c,

S(m,c;p)=∑n=1p−1e(mn+cnˉp),S(m,c;p)=\sum_{n=1}^{p-1}e\Bigl(\frac{mn+c\bar n}{p}\Bigr),

where nnˉ≡1(modp)n\bar n\equiv1\pmod p and e(t)=exp⁡(2πit)e(t)=\exp(2\pi it).

Equation (3) (printed p. 382). If p∤cp\nmid c, then for every integer mm,

∣S(m,c;p)∣<31/4p3/4.|S(m,c;p)|<3^{1/4}p^{3/4}.

The article calls this the non-trivial bound found by Kloosterman and notes that it shows SI(p;c)S_I(p;c) has some cancellation once the length of II is of order larger than p3/4log⁡pp^{3/4}\log p.

Weil's bound, display (4) (printed p. 382), cited from Weil's 1948 paper and not proved in the article: if p∤cp\nmid c, then ∣S(m,c;p)∣≤2p1/2|S(m,c;p)|\le2p^{1/2}. The article calls this essentially best possible.

Source. D. R. Heath-Brown, Arithmetic applications of Kloosterman sums, Nieuw Arch. Wiskd. (5) 1 (2000), no. 4, 380–384; the argument on printed pp. 381–382, displays (2), (3) and (4). The edition is identified on the source card.

Read depth. Claims checked: statement (3) and the hypotheses of (3) and (4) were read against the print, and the steps of the proof of (3) were read and found to give the stated inequality. Weil's bound is cited, not proved, in the article. Nothing here is independently reviewed.

Proof pointer

Printed pp. 381–382. For aa prime to pp the substitution n↦ann\mapsto an gives S(m,c;p)=S(ma,caˉ;p)S(m,c;p)=S(ma,c\bar a;p), so, since p∤cp\nmid c, the fourth moment Σ=∑r,s=0p−1∣S(r,s;p)∣4\Sigma=\sum_{r,s=0}^{p-1}|S(r,s;p)|^4 contains p−1p-1 copies of ∣S(m,c;p)∣4|S(m,c;p)|^4 (display (2)). Expanding the fourth power and summing over rr and ss by orthogonality gives p2p^2 times the number of quadruples (n1,n2,n3,n4)(n_1,n_2,n_3,n_4) of nonzero residues with n1+n2≡n3+n4n_1+n_2\equiv n_3+n_4 and nˉ1+nˉ2≡nˉ3+nˉ4\bar n_1+\bar n_2\equiv\bar n_3+\bar n_4. Such a quadruple has {n3,n4}={n1,n2}\{n_3,n_4\}=\{n_1,n_2\} or n1+n2≡n3+n4≡0n_1+n_2\equiv n_3+n_4\equiv0, so there are at most 3(p−1)23(p-1)^2 of them. Hence (p−1)∣S(m,c;p)∣4≤3p2(p−1)2<3p3(p−1)(p-1)|S(m,c;p)|^4\le3p^2(p-1)^2<3p^3(p-1), which is (3).

Dependencies

None in the article for (3). Display (4) is Weil's theorem (the article's reference [13]).

Bears on

No Erdős problem directly. The article's origin-rectangle estimate, the result the corpus cites, uses Weil's bound (4), not (3).