Honors 1: Undergraduate Math Lab^{∗}

Alex Barnett^{†}

November 5th, 2002

Courant Institute of Mathematical Sciences New York University New York, NY

Abstract

Note on conditions arising in non-trivial bound on sum of fourth powers of Kloosterman sums (Heath-Brown review).

Please refer to Sarnak lecture of 10/16/02, and Heath-Brown’s review article. There are six summations inherent in the desired sum

∑_{a=0}^{p−1} ∑_{b=0}^{p−1} |Kl(a, b, p)|^{4} ,  (1)

namely the two sums shown and one internal sum in each of the Kl’s. Using the result

∑_{a=0}^{p−1} e (\frac{an}{p}) = \{ \begin{matrix}p − 1, & n = 0 (modp) \\ 0, & otherwise \end{matrix}  (2)

twice, turns the two e(·) functions into counting conditions on the variables internal to the four Kl’s. Calling these variables w, x, y, z, we want the total number of ways that, with 1 ≤ w, x, y, z ≤ p − 1 (the four surviving sums), we can obey both

w + x − y − z = 0 (p)  (3)

∗Homepage: http://www.math.nyu.edu/Courses/V63.0393/

†E-mail: barnett@cims.nyu.edu

and

w̄ + x̄ − ȳ − z̄ = 0 (p).  (4)

Multiplying Eq. 4 by wxyz gives

(w + x)yz − (y + z)wx = 0 (p).  (5)

Substituting Eq. 3 gives

(w + x)(yz − wx) = 0 (p).  (6)

So either w + x = 0 (p) or yz − wx = 0 (p), or both.

The first set of cases has x = −w from (3) giving z = −y, so there are (p − 1)^{2} choices of w and y. For each choice x and z are fixed uniquely. Therefore these cases contribute (p − 1)^{2} ways.

For the second set, we have two equations

y + z = w + x (p)  (7)
yz = wx (p)  (8)

for two unknowns y, z, for any of the arbitrary choices of w, x. You could combine these equations into the single quadratic

y^{2} − y(w + x) + wx = 0 (p).  (9)

Two solutions for y are y = w and y = x (check by substitution). Since it is a quadratic, these are the only two solutions. Therefore the number of ways contributed is at most 2 times the (p − 1)^{2} ways of choosing w, x.

Overcounting due to the and/or is at least of order p or smaller, but also can only reduce the number of ways. Therefore the total number of ways ≤ 3(p − 1)^{2}, which is O(p^{2}). From this follows the bound on the sum given in the article and lecture.