On the parity of k-th powers modulo p

Transcription

1 On the parity of k-th powers modulo p Jennifer Paulhus Kansas State University paulhus@math.ksu.edu paulhus

2 This is joint work with Todd Cochrane and Chris Pinner of Kansas State University and Jean Bourgain of the Institute for Advanced Study.

3 Notation p will denote an odd prime O ={1, 3, 5,..., p 2} Z/pZ E={2, 4, 6,..., p 1} Z/pZ e p ( ) = e 2πi /p the odd residues the even residues We consider N k = N k (A) = #{x E Ax k O} where k and A are any integers with p A. In previous work, we showed N k > 0 for p sufficiently large and (k, p 1) = 1, resolving a conjecture of Goresky and Klapper.

4 A Generalization Lehmer posed the problem of determining N 1 (1), the number of even residues with odd multiplicative inverse (mod p). The expectation is that N 1 (1) p/4, which was later proven by Zhang. For example when p = 13, N 1 (1) = 3. Residue Inverse Residue Inverse In the general setting we no longer always have N k p/4.

5 A general estimate for N k Φ(k) = max a Z/pZ a 0 e p (ax k ), x 0 Φ (k) = max a Z/pZ a 0 (p 1)/2 x=1 e p (ax k ) Theorem For any integer k N k p < 1 ( ) 356p 4 π Φ (k) min{log Φ, log(5p)} (k) When k is even, Φ (k) = 1 2 Φ(k) and so N k p/4 when Φ(k) = o(p).

6 Ideas of Proof Let S = {x n } be a sequence defined by x n A2 k 1 n k mod p for 1 n p 1 2 and I = { 1, 2,..., p 1 2 }. x n I precisely when A(2n) k O so N k = p 1 2 n=1 χ I (x n ). Using the Fourier expansion of χ I and bounds on exponential sums we obtain the second estimate in the theorem: N k p/4 < 1 π Φ (k) log(5p). A better bound is obtained by using a smooth approximation to the characteristic function which leads to an Erdös-Turán type inequality.

7 Ideas of Proof, continued Theorem For any sequence S = (x 1,..., x N ) Z/pZ, interval I = {a + 1, a + 2,..., a + M} Z/pZ, and positive H < p N χ I (x n ) MN p N H + 1 N p + 2 ( log H + γ + π ) Φ π 2 S n=1 where γ is Euler s constant and N Φ S = max e p y p (yx n ). n=1 The left hand side corresponds to measuring the discrepancy of the sequence of points x n /p mod 1. To obtain the constants here we use a bound of Vaaler on the discrepancy of a sequence. The bound can be improved since our interval is of length 1 2.

8 Restating Φ(k) = max a Z/pZ a 0 e p (ax k ), x 0 Φ (k) = max a Z/pZ a 0 (p 1)/2 x=1 e p (ax k ) Theorem For any integer k N k p < 1 ( ) 356p 4 π Φ (k) min{log Φ, log(5p)} (k)

9 k odd Two parameters which help dictate the bias in the parity of kth powers are d = (k, p 1) and d 1 = (k 1, p 1). Similarly we have the values s = p 1 d and t = p 1 d 1. Theorem (a) If k is odd and t is even then N k p 0.35p 89/92 log 3/2 (5p). 4 (b) If k is odd and t is odd then N k p d 1 + p 4 log p. As long as d 1 = o(p) then N k p/4.

10 Small s = p 1 d An Example If k = p 1, x k = 1 identically so N k = 0 or p 1 2 depending on whether A is even or odd. If k = p 1 2, x k = ±1. A and A have opposite parity and roughly half even residues are quadratic residues so N k p 4. If k = p 1 3 then Ax k AC 1, AC 2, or AC 3 mod p where the C i are the cube roots of unity. Then N k = 0, p 1 6, p 1 p 1 3, or 2 depending on how many AC i are odd.

11 Small s = p 1 d, continued These examples suggest the following theorem. Theorem Let k, A be any integers with p A and (Z/pZ ) k = {C 1,..., C s }. (a) If k is even then N k = p 1 s 2s i=1 χ O(AC i ). In particular if k is even and s is even then N k = p 1 4. (b) If k is odd then N k p 1 4 < s 1 2π p log(5p).

12 Ideas of Proof Again, let {C 1,..., C s } be the kth powers in Z/pZ and define c i such that c k i C i mod p. N k = x χ E (x)χ O (Ax k ) = Main + Error where, after some manipulations, Error = 1 s Main = p 1 2s ψ s =ψ 0 ψ ψ 0 s s χ O (AC i ) i=1 i=1 χ O (AC i )ψ(c i ) x 0 ψ(x)χ E (x)

13 Ideas of Proof, continued If ψ( 1) = 1 then p 1 2 x=1 ψ(x) = 1 2 x ψ(x) = 0. This will happen to all ψ satisfying ψ s = ψ 0 if k is even. Hence Error = 0. If ψ( 1) = 1 we need to use a bound of Polya, Landau, Schur, and Vinogradov for incomplete character sums. (p 1)/2 ψ(x)χ E (x) = ψ(x) 1 p log(5p). π x x=1 When k is odd exactly half the ψ satisfying ψ s = ψ 0 will also satisfy ψ( 1) = 1. Error < s 1 p log(5p). 2π

14 Restating Theorem Let k, A be any integers with p A and (Z/pZ ) k = {C 1,..., C s }. (a) If k is even then N k = p 1 s 2s i=1 χ O(AC i ). In particular if k is even and s is even then N k = p 1 4. (b) If k is odd then N k p 1 4 < s 1 2π p log(5p).

15 Small t = p 1 d 1 An Example If k = p+1 2 then t = 2 and Ax k Ax or Ax mod p depending on whether x is a quadratic residue or not. So we expect about half the even residues to become odd. If k = p+2 3 then t = 3 and Ax k AC 1 x, AC 2 x or AC 3 x mod p To compute N k in this example we need to study the distribution of points on the lattices y AC i x mod p. When none of the lattices have a small nonzero point then even and odd values are equidistributed and N k p 4.

16 Bias If one of the lattices has a small point, there may be bias. An Example If k = p+2 3 then t = 3 and Ax k AC 1 x, AC 2 x or AC 3 x mod p. Our results give that, depending on the size of the smallest point in the lattices, N k is asymptotically between p 4 p 12 and p 4 + p 12. Another Example When t and A are both small odd numbers we get bias. In particular if A < (p/t) 1/2(t 1) and t log p then ( N k 1 1 ) p At 4

17 Open Problems How large does s need to be for Φ(k) = o(p)? Work of Bourgain says there is a constant c such that Φ(k) = o(p) if s > p c log log p. Montgomery, Vaughan, and Wooley conjecture Φ(k) dp log p which would give Φ(k) = o(p) if log p s 0 as p. We know from Shparlinski that Φ(k) = o(p) requires s log p as p. When is N k = 0? If p = 3s 1 2 and k = p 1 s then the k-th powers are just <3>= {1, 3, 3 2,..., 3 s 1 } which are all odd. Here s log p, are there examples with larger s?

18 The End