On a Problem Related to the ABC Conjecture

Transcription

1 On a Problem Related to the ABC Conjecture Daniel M. Kane Department of Mathematics / Department of Computer Science and Engineering University of California, San Diego dakane@ucsd.edu October 18th, 2014 D. Kane (UCSD) ABC Problem October / 27

2 The ABC Conjecture For integer n 0, let the radical of n be Rad(n) := p n p. For example, Rad(12) = 2 3 = 6. The ABC-Conjecture of Masser and Oesterlé states that Conjecture For any ɛ > 0, there are only finitely many triples of relatively prime integers A, B, C so that A + B + C = 0 and max( A, B, C ) > Rad(ABC) 1+ɛ. D. Kane (UCSD) ABC Problem October / 27

3 Applications The ABC Conjecture unifies several important results and conjectures in number theory Provides new proof of Roth s Theorem (with effective bounds if ABC can be made effective) Proves Fermat s Last Theorem for all sufficiently large exponents Implies that for every irreducible, integer polynomial f, f (n) is square-free for a constant fraction of n unless for some prime p, p 2 f (n) for all n A uniform version of ABC over number fields implies that the Dirichlet L-functions do not have Siegel zeroes. D. Kane (UCSD) ABC Problem October / 27

4 Problem History The function field version of the ABC conjecture follows from elementary algebraic geometry. In particular, if f, g, h F[t] are relatively prime and f + g + h = 0 then max(deg(f ), deg(g), deg(h)) deg(rad(fgh)) 1. The version for number fields though seems to be much more difficult. In 2012, Shinichi Mochizuki submitted a purported proof of the conjecture. The proof uses Mochizuki s inter-universal Teichmüller theory, making it difficult to check. The proof has still not been fully verified. D. Kane (UCSD) ABC Problem October / 27

5 Experimental Results The project has been searching for ABC triples since Define the quality of a triple to be q(a, B, C) = log(max A, B, C ). log(rad(abc)) The highest quality triples found to date are: = 23 5 q = = q = = q = = q = 1.59 As these are all relatively small they seem to provide some support for the conjecture. D. Kane (UCSD) ABC Problem October / 27

6 Heuristics Why might we expect ABC to be true? Count number of solutions with A, B, C N and Rad(A) N a, Rad(B) N b, Rad(C) N c for fixed a, b, c with a + b + c < 1. How many such A, B, C are there? Lemma For any m, the number of n with n N and Rad(n) = m is O ɛ (N ɛ ). Proof. Let m = k i=1 p i n = k i=1 pa i i with a i 1 and k i=1 a i log(p i ) log(n). D. Kane (UCSD) ABC Problem October / 27

7 Heuristics Why might we expect ABC to be true? Count number of solutions with A, B, C N and Rad(A) N a, Rad(B) N b, Rad(C) N c for fixed a, b, c with a + b + c < 1. How many such A, B, C are there? Lemma For any m, the number of n with n N and Rad(n) = m is O ɛ (N ɛ ). Proof. Let m = k i=1 p i n = k i=1 pa i i with a i 1 and k i=1 a i log(p i ) log(n). log(n) log(p i ) At most 1 k k! i=1 Largest when p i as small as possible. Optimizing over k gives N O(log log log(n)/ log log(n)) = O ɛ (N ɛ ). D. Kane (UCSD) ABC Problem October / 27

8 Heuristics Why might we expect ABC to be true? Count number of solutions with A, B, C N and Rad(A) N a, Rad(B) N b, Rad(C) N c for fixed a, b, c with a + b + c < 1. How many such A, B, C are there? Lemma For any m, the number of n with n N and Rad(n) = m is O ɛ (N ɛ ). N a+b+c+ɛ many triples with small radicals. About 1/N of them should have A + B + C = 0. Expect finitely many solutions if a + b + c < 1 D. Kane (UCSD) ABC Problem October / 27

9 Extended Conjecture However, these heuristics also suggest what happens if a + b + c 1. Conjecture (Mazur) For a + b + c 1, let S a,b,c (N) be the number of relatively prime triples A, B, C with A, B, C N, Rad(A) A a, Rad(B) B b, Rad(C) C c and A + B + C = 0. Then for any ɛ > 0 and d = a + b + c 1, N d+ɛ ɛ S a,b,c (N) ɛ N d ɛ. Our goal is to prove this conjecture for a + b + c 2. D. Kane (UCSD) ABC Problem October / 27

10 Lower Bounds Theorem For 0 < a, b, c 1, with a + b + c > 1, S a,b,c (N) = Ω(N d log(n) 2 ). Basic idea: Force A, B, C to have small radicals by making them divisible by 2 x, 3 y, 5 z. Lattice problem Difficulties Lattice too skew Need A, B, C relatively prime D. Kane (UCSD) ABC Problem October / 27

11 The Lattice Pick x, y, z as small as possible so that 2 x 1 N 1 a, 3 y 1 N 1 b, 5 z 1 N 1 c. Let α = 2 x, β = 3 y, γ = 5 z If A N and α A, A /Rad(A) α/rad(α) N 1 a A 1 a Thus Rad(A) A a Similar argument for B and C. Consider triples (A, B, C) so that A + B + C = 0 and α A, β B, γ C This is a 2-dimensional lattice L Let P be the polygon given by the set of (x 1, x 2, x 3 ) with x 1 + x 2 + x 3 = 0 and x i N for all i Points of P L whose coordinates are relatively prime give us solutions D. Kane (UCSD) ABC Problem October / 27

12 Geometry of Numbers Count points in L P Draw fundamental domain of L around each point in intersection Roughly approximates P Compare areas D. Kane (UCSD) ABC Problem October / 27

13 Point Count Lemma Let L be a lattice in a two dimensional vector space V and P a convex polygon in V. Let m be the minimum separation between points of L. Then Proof (sketch). L P = Volume(P) CoVolume(L) + O ( ) Perimeter(P) + 1. m Make L a square lattice (this can be done without increasing Perimeter(P) by too much) Put fundamental domain around each point of intersection Covers P exactly up to points within m of the perimeter Comparing area to area of P yields result D. Kane (UCSD) ABC Problem October / 27

14 Application Recall: L = {(A, B, C) Z 3, A + B + C = 0, α A, β B, γ C} α = 2 x N 1 a, β = 3 y N 1 b, γ = 5 z N 1 c P = {(x 1, x 2, x 3 ) R 3 : x 1 + x 2 + x 3 = 0, x i N for all i} By Lemma ( L P = Θ N 2 N 3 a b c ( = Θ N a+b+c 1) + O ) ( ) Perimeter(P) + O + 1 m ( ) Perimeter(P) + 1 m Problem: m might be very small D. Kane (UCSD) ABC Problem October / 27

15 Modified Lattice Suppose L has short vector (αr, βs, γt) Idea: replace L by some similar L for which we do not have an exceptionally short vector (want shortest vector CoVol) Technique Let 2 h gcd(αr, βs, γt) Let 2 k N (3 a b c)/2 /m Let q be a prime q > 5, q r and q small O(log(N)) Let q w 1 > 2 h+k Replace α by α := 2 x h k q w, define L = {(A, B, C) Z 3, A + B + C = 0, α A, β b, γ C} Analysis α /Rad(α ) = α/rad(α)2 h k q w 1 > N 1 a, good enough L has short vector v = (α 2 k r, β2 h q w s, γ2 h q w t) v N (3 a b c)/2 v not a multiple of other vectors in L For any w in L either w is a multiple of v or w v CoVolume(L ) = Θ(α βγ) Shortest vector in L has length at least Ω(N (3 a b c)/2 / log(n)) D. Kane (UCSD) ABC Problem October / 27

16 Summary Have lattice L and polygon P Have intersection L P = Vol(P) CoVol(L ) + O ( Perim(P) m ) + 1 = Ω(N d / log(n)) + O(N d/2 log(n) + 1) If (A, B, C) L P then: A + B + C = 0 A, B, C N A/Rad(A) N 1 a A 1 a so Rad(A) A a Similarly, have Rad(B) B b, Rad(C) C c Still need: A, B, C relatively prime. D. Kane (UCSD) ABC Problem October / 27

17 Sieving Need to count just the relatively prime triples Let L n = {(A, B, C) L : n A, n B, n C} Want = (A,B,C) L P { 1 if gcd(a, B, C) = 1 0 otherwise µ(n) (A,B,C) L P n gcd(a,b,c) = n µ(n) L n P D. Kane (UCSD) ABC Problem October / 27

18 Approximation Deal with sum µ(n) L n P n Truncate sum at n = N d/2 log 2 (N) Use Lemma to approximate inner sum Õ(N d/2 ) = = =Θ n=1 Õ(N d/2 ) n=1 ( ) µ(n)vol(p) Perimeter(P) CoVol(L n) + O ShortVec(L n) + 1 ( ) Vol(P) µ(n) gcd(30q, n) CoVol(L ) n 2 + Õ(N d/2 /n) ) + Õ(Nd/2 ) = Ω(N a+b+c 1 log 2 (N)) ( Vol(P) CoVol(L ) This proves the lower bound. D. Kane (UCSD) ABC Problem October / 27

19 Lattice Upper Bounds Idea: A, B, C must have large repeated parts in their factorization with few ways to select them. Consider first fixing the repeated parts and then use lattice techniques to bound the number of solutions. First we need some definitions: u(n) := p n p e(n) := p α n,α>1 p α = n /u(n) v(n) := p 2 n p = Rad(e(n)) = Rad(n)/u(n) D. Kane (UCSD) ABC Problem October / 27

20 Basic Technique Fix the values of A, B, C, v(a), v(b), v(c) each to within a factor of 2 (there are only log 6 (N) ways to do this) Fix the exact values of v(a), v(b), v(c) (there are O(v(A)v(B)v(C)) ways to do this) Fix the exact values of e(a), e(b), e(c) (there are O(N ɛ ) ways to do this since Rad(e(X )) = v(x ) and e(x ) N) Count the number of A, B, C D. Kane (UCSD) ABC Problem October / 27

21 Conditions on A, B, C 1 A + B + C = 0 2 e(a) A, e(b) B, e(c) C 3 A, B, C N and are of at most the prespecified sizes. 4 We also need that the selected values e(a), e(b), e(c) are the appropriate quantities for A, B, C, but we will ignore this condition as we only need an upper bound Note that conditions 1 and 2 above define a lattice, while condition 3 defines a polygon. D. Kane (UCSD) ABC Problem October / 27

22 Another Lattice Lemma Lemma Let L be a 2 dimensional lattice and P a centrally symmetric convex polygon. Then the number of vectors in L P which are not positive integer multiples of other vectors in L is ( ) Volume(P) O CoVolume(L) + 1. Proof (sketch). If P only contains, multiples of a single point, we have at most 2 and are done. Otherwise, P contains a fundamental domain of L Thus, 2P contains a fundamental domain around each point of L P Thus, 4Volume(P) = Volume(2P) L P CoVolume(L) D. Kane (UCSD) ABC Problem October / 27

23 Upper Bound Fix A, B, C, v(a), v(b), v(c) to within factors of 2. Assume C A, B At most O(v(A)v(B)v(C)N ɛ ) ways to select e(a), e(b), e(c) Number of triples is at most number of non-multiple points in L P, bound with Lemma ( ) Volume(P) #Solutions v(a)v(b)v(c)n ɛ O CoVolume(L) + 1 ( ) A B v(a)v(b)v(c)n ɛ O + v(a)v(b)v(c)n ɛ e(a)e(b)e(c) ( N ɛ ) Rad(A)Rad(B)Rad(C) O + v(a)v(b)v(c)n ɛ C The above is at most O(N ɛ A a B b C c 1 + v(a)v(b)v(c)n ɛ ) O(N a+b+c 1+ɛ + v(a)v(b)v(c)n ɛ ) D. Kane (UCSD) ABC Problem October / 27

24 Upper Bound Proposition Fix 0 < a, b, c 1. The number of solutions to the ABC problem with parameters (a, b, c, N) and with v(a), v(b), v(c) lying in fixed dyadic intervals is O(N a+b+c 1+ɛ + v(a)v(b)v(c)n ɛ ). This bound works well when v(a)v(b)v(c) N a+b+c 1. In particular, if a + b + c 5/2, the second term can be ignored. On the other hand, we could run into trouble otherwise. We need a new upper bound technique to cover this case. D. Kane (UCSD) ABC Problem October / 27

25 Large Squares Basic idea: If v(a) is large, then A is divisible by a large square Let T (n) := n/v(n) 2 Bound solutions to T (A)v(A) 2 + T (B)v(B) 2 + T (C)v(C) 2 = 0 D. Kane (UCSD) ABC Problem October / 27

26 Solutions to the Equation Thus, we are left with trying to bound the number of integer solutions to the equation x 1 y x 2 y x 3 y 2 3 = 0 with x i, y i within certain bounds. Fortunately, this problem is well studied Lemma (Pierre Le Boudec) The number of solutions to the above equation with x i U i N and y i V i N and x 1 y 1, x 2 y 2, x 3 y 3 pairwise relatively prime is at most O(N ɛ (U 1 U 2 U 3 ) 2/3 (V 1 V 2 V 3 ) 1/3 ). Thus, letting V 1 = v(a), V 2 = v(b), V 3 = v(c), U i = N/V 2 i, we find that the number of solutions to the ABC problem with v(a), v(b), v(c) in given ranges is at most O(N 2+ɛ (v(a)v(b)v(c)) 1 ). D. Kane (UCSD) ABC Problem October / 27

27 Putting it Together Theorem For any a, b, c, N, ɛ > 0 we have that S a,b,c (N) = O ɛ (N a+b+c 1+ɛ + N 1+ɛ ). Proof. Fix the values of v(a), v(b), v(c) to within factors of 2 (there are only log 3 (N) ways to do this) If v(a)v(b)v(c) N, then the lattice upper bound yields O ɛ (N a+b+c 1+ɛ + N 1+ɛ ) If v(a)v(b)v(c) N, then the cubic solutions upper bound yields O ɛ (N 1+ɛ ) Summing over all ranges for v(a), v(b), v(c) yields the appropriate bound. D. Kane (UCSD) ABC Problem October / 27

28 Further Improvements We suspect that the lower bounds are essentially optimal These techniques cannot easily improve the upper bound Case when v(a)v(b)v(c) N Lattice problem cannot exclude one solution per choice of v(a), v(b), v(c) We expect that cubic in question actually has N solutions in the appropriate range Need something new Perhaps bound solutions to equation x 1 y1 2 z1 3 + x 2 y2 2 z2 3 + x 3 y3 2 z3 3 = 0 for x i, y i, z i in some appropriate ranges. D. Kane (UCSD) ABC Problem October / 27

29 Acknowledgements This work was done with the support of an NSF graduate fellowship. D. Kane (UCSD) ABC Problem October / 27

30 E. Bombieri and W. Gubler, Heights in Diophantine Geometry, Cambridge Univ. Press; Cambridge Pierre Le Boudec Density of Rational Points on a Certain Smooth Bihomogeneous Threefold, preprint J. W. S. Cassels, An Introduction to the Geometry of Numbers Springer Classics in Mathematics, Springer-Verlag; 1997 (reprint of 1959 and 1971 Springer-Verlag editions). Andrew Granville Notes on Mazur s Questions about Powers of Numbers, unpublished. Barry Mazur Questions about Powers of Numbers Notices of the AMS, 47 (2000), no. 2, D. Kane (UCSD) ABC Problem October / 27