2.1 The odd prime number lattice

The result involves in an essential way the odd prime number lattice $L_n$ generated by the rows

, $\dotsc $ ,

of the matrix

where $p_i$ denotes the ith odd prime number. This lattice has a number of interesting applications. For example, it is used in Schnorr’s factoring algorithm [Reference Schnorr7] and Micciancio’s proof that approximating the shortest vector to within a constant factor is NP-hard under a randomized reduction [Reference Micciancio5]. There is an obvious isomorphism between the points of $L_n$ and the positive $ \left \lbrace {p_1,\dotsc ,p_n} \right \rbrace $ -units given by

Furthermore, this relationship works well with a natural notion of size, as shown in the following lemma.

Proof Without loss of generality, suppose $p\geq q$ . Then

as required, since $h(p/q)=p$ by assumption.

2.2 The kernel sublattice

Let P be the set of positive $ \left \lbrace {p_1,\dotsc ,p_n} \right \rbrace $ -units, and consider the map $\phi $ reducing the elements of P modulo $2^m$ . Since each $p_1$ , $\dotsc $ , $p_n$ is odd, $\phi \colon P\to (\mathbb {Z}/2^m\mathbb {Z})^*$ is well defined. The odd prime number lattice $L_n$ has an important sublattice that we call the kernel sublattice $L_{n,m}$ . It consists of those vectors whose associated $ \left \lbrace {p_1,\dotsc ,p_n} \right \rbrace $ -units lie in the kernel of $\phi $ . Formally, we define

Figure 1 plots the first two coordinates of vectors in the kernel sublattice for varying m.

Figure 1 Plots of $\{\ (x,y) : (x,y,z)\in L_{2,m}\ \}$ for $1\leq m\leq 8$ .

2.3 Hermite’s constant

The Hermite constant $\gamma _n$ is defined to be the smallest positive number such that every lattice of dimension n and volume $\det (L)$ contains a nonzero vector

with

We are interested in the “Manhattan distance” $\ell _1$ -norm instead of the usual Euclidean norm, so we define the related constants $\delta _n$ by the smallest positive number such that every full-rank lattice of dimension n contains a nonzero vector

with

By Minkowski’s theorem [Reference Cassels2] applied to a generalized octahedron (a “sphere” in the $\ell _1$ -norm), every full-rank lattice of dimension n contains a nonzero lattice point

with

. It follows that $\delta _n \leq (n!)^{1/n} \sim n/e$ , but better bounds on $\delta _n$ are known. Blichfeldt [Reference Blichfeldt1] showed that

$$\begin{align*}\delta_n \leq \sqrt{\frac{4(n+1)(n+2)}{3\pi(n+3)}}\bigg({\frac{2(n+1)}{n+3}\Big({\frac{n}{2}+1}\Big)!}\bigg)^{1/n} \sim \frac{n}{ \sqrt{1.5\pi e}} , \end{align*}$$

where

. Improving this, Rankin [Reference Rankin6] showed the following.

Lemma 2.3 For all integer n and real $x\in [1/2,1]$ , we have

$$\begin{align*}\delta_n \leq \Bigg({\frac{2-x}{1-x}}\Bigg)^{x-1}\Bigg({\frac{1+x n}{x}(x n)!}\Bigg)^{1/n}\frac{n^{1-x}}{x!} \sim \Bigg({\frac{2-x}{1-x}}\Bigg)^{x-1}\Bigg({\frac{x}{e}}\Bigg)^x \frac{n}{x!}. \end{align*}$$

Proof Note that $((1+xn)/x)^{1/n}=1+O((\log n)/n)$ and

$$\begin{align*}(x n)!^{1/n} = \Big({\sqrt{2\pi xn}\mspace{1.5mu}\Big({\frac{xn}{e}}\Big)^{xn} \big({1+O(n^{-1})}\big)}\Big)^{1/n} = \Big({\frac{xn}{e}}\Big)^x \big( {1+O \big({\tfrac{\log n}{n}}\big)}\big). \end{align*}$$

Then, by Lemma 2.3, it follows that

$$\begin{align*}\delta_n \leq \Bigg({\frac{2-x}{1-x}}\Bigg)^{x-1} \Bigg({\frac{x}{e}}\Bigg)^x \frac{n}{x!} + O(\log n), \end{align*}$$

and the function $x\mapsto \big( {\frac {1-x}{2-x}}\big)^{x-1} \big( {\frac {e}{x}}\big)^x x!$ for $1/2\leq x\leq 1$ reaches a maximum of approximately $3.65931$ at $x\approx 0.645467$ .

The best possible value $\delta $ can achieve in Corollary 2.4 is unknown, but the Minkowski–Hlawka theorem [Reference Cassels2] applied to a generalized octahedron shows that in any dimension n, there is always a full-rank lattice L with all of its nonzero lattice points having ; here, $\zeta $ is the Riemann zeta function. It follows that $\delta _n>(\zeta (n)\mspace {1.5mu}n!)^{1/n}\mspace {-1.5mu}/2\sim n/(2e)$ , so we must have $\delta \leq 2e$ .

2.4 A full-rank kernel sublattice

Since $L_{n,m}\in \mathbb {R}^{n+1}$ is of dimension n (i.e., not full-rank), it is awkward to use Rankin’s result on $L_{n,m}$ directly. The basis matrix of $L_{n,m}$ cannot simply be rotated to embed it in $\mathbb {R}^n$ , since rotation does not preserve the $\ell _1$ -norm. To circumvent this and work with a full-rank lattice, we adjoin the new basis vector to $L_n$ to form a full-rank lattice $\overline {L}_n$ (and similarly a full-rank lattice $\overline {L}_{n,m}$ ).

Our choice of m will ultimately be asymptotic to $n\log _2 n$ , and in this case, $\det (\overline {L}_{n,m})^{1/(n+1)}$ grows slightly more than linearly in n.

Finally, we will require the fact that any vector in $\overline {L}_n$ including a nontrivial coefficient on must be sufficiently large (have length at least $n^3$ in the $\ell _1$ -norm).

Proof We have .

Without loss of generality, suppose that $e_{n+1}>0$ , and for contradiction, suppose . Then

$$\begin{align*}\sum_{i=1}^n e_i\log p_i+e_{n+1}n^3 \leq \Bigg\lvert{\sum_{i=1}^n e_i\log p_i+e_{n+1}n^3}\Bigg\rvert < n^3 e_{n+1} - \sum_{i=1}^n \left\lvert {e_i} \right\rvert \log p_i \end{align*}$$

implies $\sum _{i=1}^n (e_i+ \left \lvert {e_i} \right \rvert ) \log p_i < 0$ , and this is nonsensical since the left-hand side is nonnegative.

2.5 Asymptotic formulae

Let , and let $\pi (x)$ be the prime counting function, so that $n=\pi (x)-1$ . The prime number theorem [Reference Ingham4] states that $\pi (x)\sim \operatorname {\mathrm {li}}(x)$ where $\operatorname {\mathrm {li}}(x)$ is the logarithmic integral $\int _0^x\frac {{\,\mathrm {d}} t}{\log t}$ with asymptotic expansion

(2.1) $$ \begin{align} \operatorname{\mathrm{li}}(x) = \frac{x}{\log x} + \frac{x}{\log^2 x} + \frac{2x}{\log^3 x} + O \Bigg({\frac{x}{\log^4 x}}\Bigg). \end{align} $$

In fact, the error term $\pi (x)-\operatorname {\mathrm {li}}(x)$ is $O(x/\exp (C\log ^{1/2} x))$ for some constant $C>0$ . The following estimates are consequences of this (cf. [Reference Stewart and Tijdeman9, Lemma 2]). For the convenience of the reader, proofs are given in the Appendix.

3.1 Optimal choice of m

The first bound in Lemma 3.1 allows us to show the existence of infinitely many $abc$ triples whose ratio of c to $\operatorname {\mathrm {rad}}(abc)$ grows arbitrarily large. Using the second bound, we can even show that this ratio grows faster than a function of c. It is not immediately clear how to choose m optimally, i.e., to maximize the ratio $c/\operatorname {\mathrm {rad}}(abc)$ .

For convenience, let R denote the right-hand side of the second inequality in Lemma 3.1 with . Then $2^{m-1}= \big( {\frac {\delta R}{n+l_n}}\big)^{n+1}/(n^3\prod _{i=1}^n\log p_i)$ , so the bounds of Lemma 3.1 can be rewritten in terms of R:

(3.2) $$ \begin{align} \frac{(\delta R/(n+l_n))^{n+1}}{n^3\prod_{i=1}^n p_i\log p_i}\operatorname{\mathrm{rad}}(abc) \leq c \quad\text{and}\quad 2\log c \leq R. \end{align} $$

The question now becomes how to choose R in terms of n so that $c/\operatorname {\mathrm {rad}}(abc)$ is maximized.

Taking the logarithm of the first inequality in (3.2) gives

$$\begin{align*}(n+1)\log \Bigg({\frac{\delta R}{n+l_n}}\Bigg) - 3\log n - \sum_{i=1}^n \log p_i - \sum_{i=1}^n \log\log p_i + \log\operatorname{\mathrm{rad}}(abc) \leq \log c. \end{align*}$$

Using the asymptotic formulae in Lemmas 2.8 and 2.9 with $\log (n+l_n)=\log n+O(l_n/n)$ , this becomes

(3.3) $$ \begin{align} n\log \Bigg({\frac{e\delta R}{np_n\log p_n}}\Bigg) + \frac{2p_n}{\log^2 p_n} + O \Bigg({\frac{p_n}{\log^3 p_n}}\Bigg) + \log\operatorname{\mathrm{rad}}(abc) \leq \log c. \end{align} $$

By the prime number theorem $n=\operatorname {\mathrm {li}}(p_n)+O(p_n/\log ^2 p_n)$ and (2.1), the leftmost term becomes

$$\begin{align*}n\log \Bigg({\frac{e\delta R}{p_n^2 \big({1+1/\log p_n+O(1/\log^2 p_n)}\big)}}\Bigg) , \end{align*}$$

and with $\log (1+1/x)=1/x+O(1/x^2)$ as $x\to \infty $ , this is

$$\begin{align*}n\log \Bigg({\frac{e\delta R}{p_n^2}}\Bigg) - \frac{n}{\log p_n} + O \Bigg({\frac{n}{\log^2 p_n}}\Bigg). \end{align*}$$

Using (2.1) again on the last two terms and putting this back into (3.3), we get

(3.4) $$ \begin{align} n\log \Bigg({\frac{e\delta R}{p_n^2}}\Bigg) + \frac{p_n}{\log^2 p_n} + O \Bigg({\frac{p_n}{\log^3 p_n}}\Bigg) + \log\operatorname{\mathrm{rad}}(abc) \leq \log c , \end{align} $$

and our goal becomes to choose R as a function of n to maximize $n\log (e\delta R/p_n^2)$ . Choosing R as asymptotically slow-growing as possible in terms of n will maximize this in terms of R. We must take $R>p_n^2/(e\delta )$ for the logarithm to be positive, so we take for some constant k. Note that with this choice $m\sim n\log _2 n$ , so Lemma 3.1 applies. We have that $n\log (e\delta R/p_n^2)$ simplifies to

$$\begin{align*}n\log(e\delta k) \sim \frac{p_n}{\log p_n}\log(e\delta k) = \frac{\sqrt{R/k}}{\log\sqrt{R/k}}\log(e\delta k) \sim \frac{2\sqrt{R/k}}{\log R}\log(e\delta k). \end{align*}$$

For fixed R, this is maximized when . Using $R=ep_n^2/\delta $ in (3.4),

$$\begin{align*}2n + \frac{p_n}{\log^2 p_n} + O \Bigg({\frac{p_n}{\log^3 p_n}}\Bigg) + \log\operatorname{\mathrm{rad}}(abc) \leq \log c. \end{align*}$$

By the prime number theorem and (2.1) again,

$$\begin{align*}\frac{2p_n}{\log p_n} + \frac{3p_n}{\log^2 p_n} + O \Bigg({\frac{p_n}{\log^3 p_n}}\Bigg) + \log\operatorname{\mathrm{rad}}(abc) \leq \log c. \end{align*}$$

Rewriting in terms of R,

$$\begin{align*}\frac{2\sqrt{\delta R/e}}{\log \sqrt{\delta R/e}} + \frac{3\sqrt{\delta R/e}}{\log^2 \sqrt{\delta R/e}} + O \Bigg({\frac{\sqrt{R}}{\log^3 R}}\Bigg) + \log\operatorname{\mathrm{rad}}(abc) \leq \log c. \end{align*}$$

Simplifying,

$$\begin{align*}\frac{4\sqrt{\delta R/e}}{\log(\delta R/e)} + \frac{12\sqrt{\delta R/e}}{\log^2(\delta R/e)} + O \Bigg({\frac{\sqrt{R}}{\log^3 R}}\Bigg) + \log\operatorname{\mathrm{rad}}(abc) \leq \log c. \end{align*}$$

Using $1/(x+y)=1/x-y/x^2+O(x^{-3})$ as $x\to \infty $ , this gives

$$\begin{align*}\frac{4\sqrt{\delta R/e}}{\log (R/2)} + \frac{(12-4\log(2\delta/e))\sqrt{\delta R/e}}{\log^2 R} + O \Bigg({\frac{\sqrt{R}}{\log^3 R}}\Bigg) + \log\operatorname{\mathrm{rad}}(abc) \leq \log c. \end{align*}$$

Using that $2\delta <e^4$ , the second term on the left is positive, and so for sufficiently large R, the middle two terms are necessarily positive. Therefore, for sufficiently large R, this can be simplified to

$$\begin{align*}\frac{4\sqrt{\delta R/e}}{\log(R/2)} + \log\operatorname{\mathrm{rad}}(abc) \leq \log c. \end{align*}$$

Using that $2\log c\leq R$ from (3.2) and the increasing monotonicity of $\sqrt {R}/\log (R/2)$ for sufficiently large R, we finally achieve that

$$\begin{align*}\frac{4\sqrt{2(\delta/e)\log c}}{\log\log c} + \log\operatorname{\mathrm{rad}}(abc) \leq \log c. \end{align*}$$

Taking the exponential, this proves the following theorem.

Theorem 3.1 There are infinitely many $abc$ triples satisfying

$$\begin{align*}\exp \Bigg({\frac{4\sqrt{2(\delta/e)\log c}}{\log\log c}}\Bigg) \operatorname{\mathrm{rad}}(abc) \leq c. \end{align*}$$

Using the permissible value for $\delta $ 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 $\delta $ is $2e$ , meaning that the constant in the exponent would become $8$ if this upper bound was shown to be tight.