A NEW LOWER BOUND IN THE ABC CONJECTURE

. We prove that there exist inﬁnitely many coprime numbers a , b , c with a + b = c and c > rad( abc )exp(6 . 563 √ log c/ log log c ). These are the most extremal examples currently known in the abc conjecture, thereby providing a new lower bound on the tightest possible form of the conjecture. This builds on work of van Frankenhuysen (1999) whom proved the existence of examples satisfying the above bound with the constant 6 . 068 in place of 6 . 563. We show that the constant 6 . 563 may be replaced by 4 (cid:112) 2 δ/e where δ is a constant such that all full-rank unimodular lattices of suﬃciently large dimension n contain a nonzero vector with (cid:96) 1 norm at most n/δ .


Introduction
Three natural numbers , ,  are said to be an  triple if they do not share a common factor and satisfy the equation Informally, the  conjecture says that large  triples cannot be 'very composite' , in the sense of  having a prime factorization containing large powers of small primes.The radical of  is defined to be the product of the primes in the prime factorization of , i.e., rad() |  .
The  conjecture then states that  triples satisfy  =  rad() 1+  (1.1) for every  > 0, where the implied big- constant may depend on .
Presently, the conjecture is far from being proved; not a single  is known for which (1.1) holds. 1 The best known upper bound is due to C. L. Stewart and K. Yu [10] and says that  triples satisfy  =  exp(rad() 1/3 (log rad()) 3 ) .
On the other hand, Stewart and Tijdeman [9] proved in 1986 that there are infinitly many  triples with  > rad() exp  √︁ log /log log  (1.2) for all  < 4. Such  triples are exceptional in the sense that their radical is relatively small in comparison to  and they provide a lower bound on the best possible form of (1.1).In 1997, van Frankenhuysen [3] improved this lower bound by showing that (1.2) holds for  = 4 √ 2, and in 1999 he improved this to  = 6.068 using a sphere-packing idea credited to H. W. Lenstra, Jr.We improve this further by showing that there are infinitely many  triples satisfying (1.2) with  = 6.563.

Preliminaries
Let  be a set of prime numbers.An -unit is defined to be a rational number whose numerator and denominator in lowest terms are divisible by only the primes in .That is, one has This generalizes the notion of units of Z; in particular, the ∅-units are ±1.The height of a rational number / in lowest terms is ℎ( /) max{| |, ||}.This provides a convenient way of measuring the 'size' of an -unit.Finally, if x = ( 1 , . . .,   ) is a vector in R  , we let be its standard ℓ  norm.The existence of exceptional  triples follows from some basic results in the geometry of numbers along with estimates for prime numbers provided by the prime number theorem.In particular, we rely on a result of Rankin [6] guaranteeing the existence of a short nonzero vector in a suitably chosen lattice.

The odd prime number lattice
The result involves in an essential way the odd prime number lattice   generated by the rows b 1 , . . ., b  of the matrix where   denotes the th odd prime number.This lattice has a number of interesting applications.For example, it is used in Schnorr's factoring algorithm [7] and Micciancio's proof that approximating the shortest vector to within a constant factor is NP-hard under a randomized reduction [5].There is an obvious isomorphism between the points of   and the positive { 1 , . . .,   }-units given by Furthermore, this relationship works well with a natural notion of size, as shown in the following lemma.

The kernel sublattice
Let  be the set of positive { 1 , . . .,   }-units, and consider the map  reducing the elements of  modulo 2  .Since each  1 , . . .,   is odd,  :  → (Z/2  Z) * is welldefined.The odd prime number lattice   has an important sublattice that we call the kernel sublattice  , .It consists of those vectors whose associated { 1 , . . .,   }-units lie in the kernel of .Formally, we define Figure 1 plots the first two coordinates of vectors in the kernel sublattice for varying .
Proof Note that  , is discrete and closed under addition and subtraction. , also contains the  linearly independent vectors ord 2  (   )b  for 1 ≤  ≤ , so this demonstrates that  , is a full-rank sublattice of   . ■

Hermite's constant
The Hermite constant   is defined to be the smallest positive number such that every lattice of dimension  and volume det() contains a nonzero vector x with ∥x∥ 2 2 ≤   det() 2/ .We are interested in the "Manhattan distance" ℓ 1 norm instead of the usual Euclidean norm, so we define the related constants   by the smallest positive number such that every full-rank lattice of dimension  contains a nonzero vector x with ∥x∥ 1 ≤   det() 1/ .By Minkowski's theorem [2] applied to a generalized octahedron (a 'sphere' in the ℓ 1 norm), every full-rank lattice of dimension  contains a nonzero lattice point x with ∥x∥ 1 ≤ (! det()) 1/ .It follows that   ≤ (!) 1/ ∼ /, but better bounds on   are known.Blichfeldt [1] showed that , where ! Γ( + 1).Improving this, Rankin [6] showed the following.

A full-rank kernel sublattice
Since  , ∈ R +1 is of dimension  (i.e., not full-rank) it is awkward to use Rankin's result on  , directly.The basis matrix of  , cannot simply be rotated to embed it in R  , since rotation does not preserve the ℓ 1 norm.To circumvent this and work with a full-rank lattice we adjoin the new basis vector b +1 = [0, . . ., 0,  3 ] to   to form a full-rank lattice   (and similarly a full-rank lattice  , ).
Proof The basis matrix of   adjoined with b +1 is an upper-triangular matrix, so det(  ) =  3  =1 log   .The index of  , in   is 2 −1 when  ≥ 2 by the same argument as in Lemma 2.2, so det( , ) = 2 −1 det(  ).■ Our choice of  will ultimately be asymptotic to  log 2 , and in this case det( , ) 1/(+1) grows slightly more than linearly in .Finally, we will require the fact that any vector in   including a nontrivial coefficient on b +1 must be sufficiently large (have length at least  3 in the ℓ 1 norm).
Without loss of generality suppose that  +1 > 0 and for contradiction suppose and this is nonsensical since the left-hand side is nonnegative.■

Asymptotic formulae
Let    and let () be the prime counting function, so that  = () − 1.The prime number theorem [4] states that () ∼ li() where li() is the logarithmic integral

Exceptional 𝒂𝒃𝒄 triples
For our purposes the importance of the kernel sublattice is that it lets us show the existence of  triples in which  is large relative to rad().The following lemma shows how this may be done.
Lemma 3.1 For all  ≲  log 2  and sufficiently large , there exists an  triple satisfying .
Proof By the definition of  from Corollary 2.4, for all sufficiently large  there exists a nonzero x ∈  , with Say x = +1 =1   b  .For sufficiently large  we must have  +1 = 0, since by Lemma 2.7 if  +1 ≠ 0 then ∥x∥ 1 ≥  3 .This would contradict (3.1) since by Lemma 2.6 the right-hand side is  ( 2+ ).

Optimal choice of 𝒎
The first bound in Lemma 3.1 allows us to show the existence of infinitely many  triples whose ratio of  to rad() grows arbitrarily large.Using the second bound, we can even show that this ratio grows faster than a function of .It is not immediately clear how to choose  optimally, i.e., to maximize the ratio /rad().
For convenience, let  denote the right-hand side of the second inequality in Lemma 3.1 with    (log ).Then 2 and our goal becomes to choose  as a function of  to maximize  log(/ 2  ).Choosing  as asymptotically slow-growing as possible in terms of  will maximize this in terms of .We must take  >  2   /() for the logarithm to be positive, so we take    2  for some constant .Note that with this choice  ∼  log 2 , so Lemma 3.1 applies.We have that  log(/ Using the permissible value for  derived by Rankin's bound in Corollary 2.4, the constant in the exponent becomes approximately 6.56338.As mentioned in Section 2.3, the best known upper bound on  is 2, meaning that the constant in the exponent would become 8 if this upper bound was shown to be tight.