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. p|n

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(|A|, |B|, |C|) > 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, p2|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 2014 3 / 27 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 Teichmüllertheory”, 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 |A|, |B|, |C|) 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 |A|, |B|, |C| ≈ 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 |n| ≤ 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 |A|, |B|, |C| ≈ 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 |n| ≤ 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 ).

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 |A|, |B|, |C| ≤ N, Rad(A) ≤ |A|a, Rad(B) ≤ |B|b, Rad(C) ≤ |C|c 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 |A| ≤ N and α|A, 1−a 1−a I |A|/Rad(A) ≥ α/Rad(α) ≥ N ≥ |A| a I Thus Rad(A) ≤ |A| I 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 (x1, x2, x3) with x1 + x2 + x3 = 0 and |xi | ≤ 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) |L ∩ P| = + 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 = {(A, B, C) ∈ Z , A + B + C = 0, α|A, β|B, γ|C} x 1−a y 1−b z 1−c I α = 2 ≈ N , β = 3 ≈ N , γ = 5 ≈ N 3 I P = {(x1, x2, x3) ∈ R : x1 + x2 + x3 = 0, |xi | ≤ N for all i} By Lemma

N2 Perimeter(P) |L ∩ P| = 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 do√ not have an exceptionally short vector (want shortest vector ≈ CoVol) Technique h I Let 2 || 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 |r 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 = {(A, B, C) ∈ Z3, A + B + C = 0, α0|A, β|b, γ|C} 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 |v| ≈ 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 |w||v| ≥ 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) |L0 ∩ P| = + O + 1 CoVol(L0) m = Ω(Nd / log(N)) + O(Nd/2 log(N) + 1)

If (A, B, C) ∈ L0 ∩ P then:

I A + B + C = 0 I |A|, |B|, |C| ≤ N 1−a 1−a a I A/Rad(A) ≥ N ≥ |A| so Rad(A) ≤ |A| b c I Similarly, have Rad(B) ≤ |B| , Rad(C) ≤ |C| 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 = {(A, B, C) ∈ L : n|A, n|B, n|C} Want ( X 1 if gcd(A, B, C) = 1 0 otherwise (A,B,C)∈L0∩P X X = µ(n) (A,B,C)∈L0∩P n| gcd(A,B,C) X 0 = µ(n)|Ln ∩ P| n

D. Kane (UCSD) ABC Problem October 2014 15 / 27 Approximation Deal with sum X 0 µ(n)|Ln ∩ P| n

Truncate sum at n = Nd/2 log2(N) Use Lemma to approximate inner sum

O˜(Nd/2) X µ(n)Vol(P) Perimeter(P) = + O + 1 CoVol(L0 ) ShortVec(L0 ) n=1 n n O˜(Nd/2) X Vol(P) µ(n) gcd(30q, n) = + O˜(Nd/2/n) CoVol(L0) n2 n=1 Vol(P) =Θ + O˜(Nd/2) = Ω(Na+b+c−1 log−2(N)) CoVol(L0)

This proves the lower bound.

D. Kane (UCSD) ABC Problem October 2014 16 / 27 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: Y u(n) := p p||n Y e(n) := pα = |n|/u(n) pα||n,α>1 Y v(n) := p = Rad(e(n)) = Rad(n)/u(n) p2|n

D. Kane (UCSD) ABC Problem October 2014 17 / 27 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 log6(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) 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 2014 18 / 27 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 2014 19 / 27 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 + 1 . CoVolume(L)

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 2014 20 / 27 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) 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)NO + 1 CoVolume(L) |A||B|v(A)v(B)v(C)N ≤ O + v(A)v(B)v(C)N e(A)e(B)e(C) NRad(A)Rad(B)Rad(C) ≤ O + v(A)v(B)v(C)N |C| ≤ O(N|A|a|B|b|C|c−1 + v(A)v(B)v(C)N) The above is at most O(Na+b+c−1+ + v(A)v(B)v(C)N)

D. Kane (UCSD) ABC Problem October 2014 21 / 27 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(Na+b+c−1+ + v(A)v(B)v(C)N).

This bound works well when v(A)v(B)v(C) Na+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 2014 22 / 27 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 2014 23 / 27 Solutions to the Equation Thus, we are left with trying to bound the number of integer solutions to the equation 2 2 2 x1y1 + x2y2 + x3y3 = 0

with xi , yi within certain bounds. Fortunately, this problem is well studied Lemma (Pierre Le Boudec)

The number of solutions to the above equation with |xi | ≤ Ui ≤ N and |yi | ≤ Vi ≤ N and x1y1, x2y2, x3y3 pairwise relatively prime is at most

2/3 1/3 O(N (U1U2U3) (V1V2V3) ).

2 Thus, letting V1 = v(A), V2 = v(B), V3 = v(C), Ui = N/Vi , 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(N2+(v(A)v(B)v(C))−1).

D. Kane (UCSD) ABC Problem October 2014 24 / 27 Putting it Together Theorem For any a, b, c, N, > 0 we have that

a+b+c−1+ 1+ Sa,b,c (N) = O(N + N ).

Proof. Fix the values of v(A), v(B), v(C) to within factors of 2 (there are only log3(N) ways to do this) If v(A)v(B)v(C) N, then the lattice upper bound yields a+b+c−1+ 1+ O(N + N ) If v(A)v(B)v(C) N, then the cubic solutions upper bound yields 1+ O(N ) Summing over all ranges for v(A), v(B), v(C) yields the appropriate bound.

D. Kane (UCSD) ABC Problem October 2014 25 / 27 Further Improvements

We suspect that the lower bounds are essentially optimal These techniques cannot easily improve the upper bound

I Case when v(A)v(B)v(C) ≈ N I Lattice problem cannot exclude one solution per choice of v(A), v(B), v(C) I We expect that cubic in question actually has N solutions in the appropriate range I Need something new Perhaps bound solutions to equation

2 3 2 3 2 3 x1y1 z1 + x2y2 z2 + x3y3 z3 = 0

for xi , yi , zi in some appropriate ranges.

D. Kane (UCSD) ABC Problem October 2014 26 / 27 Acknowledgements

This work was done with the support of an NSF graduate fellowship.

D. Kane (UCSD) ABC Problem October 2014 27 / 27 E. Bombieri and W. Gubler, Heights in Diophantine Geometry, Cambridge Univ. Press; Cambridge 2006. Pierre Le Boudec Density of Rational Points on a Certain Smooth Bihomogeneous Threefold, preprint http://arxiv.org/abs/1308.0033 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, 195-202.

D. Kane (UCSD) ABC Problem October 2014 27 / 27