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 [email protected] October 18th, 2014 D. Kane (UCSD) ABC Problem October 2014 1 / 27 The ABC Conjecture For integer n 6= 0, let the radical of n be Y Rad(n) := p: pjn For example, Rad(12) = 2 · 3 = 6: The ABC-Conjecture of Masser and Oesterl´estates that Conjecture For any > 0, there are only finitely many triples of relatively prime integers A; B; C so that A + B + C = 0 and max(jAj; jBj; jCj) > Rad(ABC)1+: D. Kane (UCSD) ABC Problem October 2014 2 / 27 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, p2jf (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 2014 3 / 27 Problem History The function field version of the ABC conjecture follows from elementary algebraic geometry. In particular, if f ; g; h 2 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 Teichm¨ullertheory", making it difficult to check. The proof has still not been fully verified. D. Kane (UCSD) ABC Problem October 2014 4 / 27 Experimental Results The ABC@Home project has been searching for ABC triples since 2006. Define the quality of a triple to be log(max jAj; jBj; jCj) q(A; B; C) = : log(Rad(ABC)) The highest quality triples found to date are: 2 + 310 = 235 q = 1:63 112 + 32 · 32 · 56 · 72 = 221 · 23 q = 1:63 19 · 1307 + 2 · 37 = 28 · 322 · 54 q = 1:62 283 + 511 · 132 = 28 · 38 · 173 q = 1:59 As these are all relatively small they seem to provide some support for the conjecture. D. Kane (UCSD) ABC Problem October 2014 5 / 27 Heuristics Why might we expect ABC to be true? Count number of solutions with jAj; jBj; jCj ≈ N and Rad(A) ≤ Na; Rad(B) ≤ Nb; Rad(C) ≤ Nc 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 jnj ≤ N and Rad(n) = m is O(N ). Proof. Qk Let m = i=1 pi Qk ai Pk n = i=1 pi with ai ≥ 1 and i=1 ai log(pi ) ≤ log(N). D. Kane (UCSD) ABC Problem October 2014 6 / 27 Heuristics Why might we expect ABC to be true? Count number of solutions with jAj; jBj; jCj ≈ N and Rad(A) ≤ Na; Rad(B) ≤ Nb; Rad(C) ≤ Nc 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 jnj ≤ N and Rad(n) = m is O(N ). Proof. Qk Let m = i=1 pi Qk ai Pk n = i=1 pi with ai ≥ 1 and i=1 ai log(pi ) ≤ log(N). At most 1 Qk log(N) k! i=1 log(pi ) Largest when pi as small as possible. Optimizing over k gives O(log log log(N)= log log(N)) N = O(N ): D. Kane (UCSD) ABC Problem October 2014 6 / 27 Heuristics Why might we expect ABC to be true? Count number of solutions with jAj; jBj; jCj ≈ N and Rad(A) ≤ Na; Rad(B) ≤ Nb; Rad(C) ≤ Nc 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 jnj ≤ N and Rad(n) = m is O(N ). Na+b+c+ 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 2014 6 / 27 Extended Conjecture However, these heuristics also suggest what happens if a + b + c ≥ 1: Conjecture (Mazur) For a + b + c ≥ 1, let Sa;b;c (N) be the number of relatively prime triples A; B; C with jAj; jBj; jCj ≤ N, Rad(A) ≤ jAja; Rad(B) ≤ jBjb; Rad(C) ≤ jCjc and A + B + C = 0. Then for any > 0 and d = a + b + c − 1, d+ d− N Sa;b;c (N) N : Our goal is to prove this conjecture for a + b + c ≥ 2: D. Kane (UCSD) ABC Problem October 2014 7 / 27 Lower Bounds Theorem d −2 For 0 < a; b; c ≤ 1; with a + b + c > 1, Sa;b;c (N) = Ω(N log(N) ). Basic idea: Force A; B; C to have small radicals by making them divisible by 2x ; 3y ; 5z . Lattice problem Difficulties I Lattice too skew I Need A; B; C relatively prime D. Kane (UCSD) ABC Problem October 2014 8 / 27 The Lattice Pick x; y; z as small as possible so that 2x−1 ≥ N1−a; 3y−1 ≥ N1−b; 5z−1 ≥ N1−c . Let α = 2x ; β = 3y ; γ = 5z If jAj ≤ N and αjA, 1−a 1−a I jAj=Rad(A) ≥ α=Rad(α) ≥ N ≥ jAj a I Thus Rad(A) ≤ jAj I Similar argument for B and C. Consider triples (A; B; C) so that A + B + C = 0 and αjA; βjB; γjC This is a 2-dimensional lattice L Let P be the polygon given by the set of (x1; x2; x3) with x1 + x2 + x3 = 0 and jxi j ≤ N for all i Points of P \ L whose coordinates are relatively prime give us solutions D. Kane (UCSD) ABC Problem October 2014 9 / 27 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 2014 10 / 27 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 Volume(P) Perimeter(P) jL \ Pj = + O + 1 : CoVolume(L) m Proof (sketch). 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 2014 11 / 27 Application Recall: 3 I L = f(A; B; C) 2 Z ; A + B + C = 0; αjA; βjB; γjCg x 1−a y 1−b z 1−c I α = 2 ≈ N ; β = 3 ≈ N ; γ = 5 ≈ N 3 I P = f(x1; x2; x3) 2 R : x1 + x2 + x3 = 0; jxi j ≤ N for all ig By Lemma N2 Perimeter(P) jL \ Pj = Θ + O + 1 N3−a−b−c m Perimeter(P) = Θ Na+b+c−1 + O + 1 m Problem: m might be very small D. Kane (UCSD) ABC Problem October 2014 12 / 27 Modified Lattice Suppose L has short vector (αr; βs; γt) 0 Idea: replace L by some similar L for which we dop not have an exceptionally short vector (want shortest vector ≈ CoVol) Technique h I Let 2 jj gcd(αr; βs; γt) k (3−a−b−c)=2 I Let 2 ≈ N =m I Let q be a prime q > 5; q 6 jr and q small O(log(N)) w−1 h+k I Let q > 2 0 x−h−k w I Replace α by α := 2 q , define L0 = f(A; B; C) 2 Z3; A + B + C = 0; α0jA; βjb; γjCg Analysis 0 0 −h−k w−1 1−a I α =Rad(α ) = α=Rad(α)2 q > N , good enough 0 0 k −h w −h w I L has short vector v = (α 2 r; β2 q s; γ2 q t) (3−a−b−c)=2 I jvj ≈ N 0 I v not a multiple of other vectors in L 0 I For any w in L either w is a multiple of v or jwjjvj ≥ CoVolume(L0) = Θ(α0βγ) 0 (3−a−b−c)=2 I Shortest vector in L has length at least Ω(N = log(N)) D. Kane (UCSD) ABC Problem October 2014 13 / 27 Summary Have lattice L0 and polygon P Have intersection Vol(P) Perim(P) jL0 \ Pj = + O + 1 CoVol(L0) m = Ω(Nd = log(N)) + O(Nd=2 log(N) + 1) If (A; B; C) 2 L0 \ P then: I A + B + C = 0 I jAj; jBj; jCj ≤ N 1−a 1−a a I A=Rad(A) ≥ N ≥ jAj so Rad(A) ≤ jAj b c I Similarly, have Rad(B) ≤ jBj ; Rad(C) ≤ jCj I Still need: A; B; C relatively prime. D. Kane (UCSD) ABC Problem October 2014 14 / 27 Sieving Need to count just the relatively prime triples 0 0 Let Ln = f(A; B; C) 2 L : njA; njB; njCg Want ( X 1 if gcd(A; B; C) = 1 0 otherwise (A;B;C)2L0\P X X = µ(n) (A;B;C)2L0\P nj gcd(A;B;C) X 0 = µ(n)jLn \ Pj n D.