On ABC Triples

1. Introduction

The radical of a positive integer n, often denoted as $rad\left(n\right)$, is the product of all prime factors of n, that is, the largest square-free factor of n. The abc conjecture is a conjecture in number theory first proposed by Joseph Oesterlé and David Masser in 1985 that is related to the radicals of positive integers [4,5]. The statement of the abc conjecture is as follows:

Conjecture 1.1.

Let $a,b,c$ be positive integers such that $c=a+b$ and $gcd(a,b,c)=1$, then for each positive real number $\epsilon >0$, there are only finite triples $(a,b,c)$ such that $rad{\left(abc\right)}^{1+\epsilon }<c$.

A set of three positive integers $(a,b,c)$ that is a subject of Conjecture 1 is called an abc triple. So far there is no widely accepted proof for the abc conjecture and the conjecture is still regarded as an open problem.

A proven upper bound for c in an abc triple is

$$logc<K·rad{\left(abc\right)}^{{\displaystyle \frac{1}{3}}}{(lograd\left(abc\right))}^3$$

(1)

with K being a constant independent of a, b and c. This upper bound was proven by Stewart and Yu in 2001 [6].

In addition, Stewart and Tijdeman proved that the following lower bound holds for infinitely many abc triples in 1986 [7]:

$$logc>lograd\left(abc\right)+k{\displaystyle \frac{\sqrt{logc}}{loglogc}}$$

(2)

for all $k<4$, and van Frankenhuysen improved k to $6.068$ in 2000.[8]

Moreover, Granville and Tucker ([2]) has proposed the following conjecture:

Conjecture 1.2.

Let $a,b,c$ be positive integers such that $c=a+b$ and $gcd(a,b,c)=1$, then $c<rad{\left(abc\right)}^2$.

Besides, there is a stronger conjecture proposed by Baker in 2004 [1], stating that

Conjecture 1.3.

Let $a,b,c$ be positive integers such that $c=a+b$ and $gcd(a,b,c)=1$, then $c<{\displaystyle \frac{6}{5}}rad\left(abc\right){\displaystyle \frac{{(lograd\left(abc\right))}^{\omega \left(abc\right)}}{\left(\omega \right(abc\left)\right)!}}$

where $\omega \left(abc\right)$ is the number of distinct prime factors of $abc$.

Since it has been shown by Laishram and Shorey (2011,[3]) that ${\displaystyle \frac{6}{5}}{\displaystyle \frac{{(lograd\left(abc\right))}^{\omega \left(abc\right)}}{\left(\omega \right(abc\left)\right)!}}<rad{\left(abc\right)}^{{\displaystyle \frac{3}{4}}}$, Conjecture 1.3 implies that $c<rad{\left(abc\right)}^{{\displaystyle \frac{7}{4}}}$.

In this paper, we intend to discuss some properties of triples involved in the abc conjecture, improving the upper bound for c in an abc triple. Unless otherwise specified, $rad\left(n\right)={\prod }_{i=1}^{\omega \left(n\right)}{p}_i$ indicates the radical of a positive integer $n={\prod }_{i=1}^{\omega \left(n\right)}{p}_i^{a_i}$ with $p_1,\cdots ,p_{\omega \left(n\right)}$ being prime numbers and $1\le a_1,\dots ,a_{\omega \left(n\right)}$ being positive integers, and $logn$ indicates the natural logarithm of n.

2. Main Results

Definition 1.

$R\left(n\right)={\displaystyle \frac{rad\left(n\right)}{n}}$

Theorem 1.

If $c=a+b$ and $gcd(a,b,c)=1$, then $c<rad{\left(abc\right)}^2$

Proof. 

W.l.o.g. assume that $a<b$. First, we cannot have $rad{\left(abc\right)}^2=c$, because if $rad{\left(a\right)}^{2}rad{\left(b\right)}^{2}rad{\left(c\right)}^2=rad{\left(abc\right)}^2=c$, then we have $rad\left(a\right)rad\left(b\right)\mid c\Rightarrow gcd(a,b,c)>1$, which is a contradiction, thus we either have $rad{\left(abc\right)}^2>c$ or $rad{\left(abc\right)}^2<c$.

Claim: Assume that $z={\displaystyle \frac{1}{4rad\left(abc\right)}}$ and $rad{\left(abc\right)}^2<c$, then for all nonnegative integers $0\le m$, ${(1+{\displaystyle \frac{z}{2b}})}^{m}rad{\left(abc\right)}^2<c$ implies that ${(1+{\displaystyle \frac{z}{2b}})}^{m+1}rad{\left(abc\right)}^2<c$.

Proof of the claim:

We prove the claim by mathematical induction.

First, since $rad{\left(abc\right)}^2<c$ and since $c=a+b<2b$, we have $rad{\left(abc\right)}^2<c<2b$

Therefore, we have

$$R\left(ab\right)={\displaystyle \frac{rad\left(ab\right)}{ab}}<{\displaystyle \frac{rad\left(ab\right)}{a\left(\frac{rad{\left(abc\right)}^2}{2}\right)}}={\displaystyle \frac{2rad\left(ab\right)}{a·rad{\left(abc\right)}^2}}$$

(3)

and

$$R\left(c\right)={\displaystyle \frac{rad\left(c\right)}{c}}<{\displaystyle \frac{rad\left(c\right)}{rad{\left(abc\right)}^2}}$$

(4)

This implies that $R\left(ab\right)-{\displaystyle \frac{2rad\left(ab\right)}{a·rad{\left(abc\right)}^2}}<0$ and $R\left(c\right)-{\displaystyle \frac{rad\left(c\right)}{rad{\left(abc\right)}^2}}<0$

Therefore we have

$$0<(R\left(ab\right)-{\displaystyle \frac{2rad\left(ab\right)}{a·rad{\left(abc\right)}^2}})(R\left(c\right)-{\displaystyle \frac{rad\left(c\right)}{rad{\left(abc\right)}^2}})$$

(5)

Which implies that

$${\displaystyle \frac{2rad\left(ab\right)}{a·rad{\left(abc\right)}^2}}R\left(c\right)+{\displaystyle \frac{rad\left(c\right)}{rad{\left(abc\right)}^2}}R\left(ab\right)<R\left(abc\right)+\left({\displaystyle \frac{2}{a·rad{\left(abc\right)}^3}}\right)$$

(6)

Multiplying a on both sides, we have

$$\begin{array}{cc}\hfill & {\displaystyle \frac{2}{c·rad\left(abc\right)}}+{\displaystyle \frac{1}{b·rad\left(abc\right)}}\hfill \\ \hfill & <a·R\left(abc\right)+\left({\displaystyle \frac{2}{rad{\left(abc\right)}^3}}\right)=a·{\displaystyle \frac{rad\left(abc\right)}{abc}}+\left({\displaystyle \frac{2}{rad{\left(abc\right)}^3}}\right).\hfill \end{array}$$

(7)

Since $rad{\left(abc\right)}^2<c$, (6) implies that

$${\displaystyle \frac{2}{c·rad\left(abc\right)}}+{\displaystyle \frac{1}{b·rad\left(abc\right)}}<{\displaystyle \frac{1}{b\sqrt{c}}}+{\displaystyle \frac{2}{rad{\left(abc\right)}^3}}$$

(8)

Let $\rho ={\displaystyle \frac{1}{rad\left(abc\right)}}$, then by multiplying $bc$ on both sides, we have

$$(2b+c)\rho <\sqrt{c}+2{\rho }^{3}bc$$

(9)

Which implies that

$${\displaystyle \frac{2b+c-\frac{\sqrt{c}}{\rho }}{2bc}}<{\rho }^2\Rightarrow rad{\left(abc\right)}^2<{\displaystyle \frac{2b}{2b+c-\frac{\sqrt{c}}{\rho }}}c$$

(10)

If $2b+c-{\displaystyle \frac{\sqrt{c}}{\rho }}\le 2b+z$, then we have

$$\begin{array}{cc}\hfill & c-{\displaystyle \frac{\sqrt{c}}{\rho }}-z\le 0\Rightarrow \sqrt{c}\le {\displaystyle \frac{1}{2}}({\displaystyle \frac{1}{\rho }}+\sqrt{{\displaystyle \frac{1}{{\rho }^2}}+4z})\hfill \\ \hfill & \Rightarrow c\le {\displaystyle \frac{1}{4}}({\displaystyle \frac{2}{{\rho }^2}}+{\displaystyle \frac{2}{\rho }}\sqrt{{\displaystyle \frac{1}{{\rho }^2}}+4z})\le {\displaystyle \frac{1}{{\rho }^2}}+z\hfill \end{array}$$

(11)

By Bernoulli’s inequality, we have $\sqrt{{\displaystyle \frac{1}{{\rho }^2}}+4z}\le {\displaystyle \frac{1}{\rho }}(1+2z{\rho }^2)={\displaystyle \frac{1}{\rho }}+2z\rho$, therefore, we have

$$c\le {\displaystyle \frac{1}{4}}({\displaystyle \frac{2}{{\rho }^2}}+{\displaystyle \frac{2}{\rho }}\sqrt{{\displaystyle \frac{1}{{\rho }^2}}+4z})\le {\displaystyle \frac{1}{{\rho }^2}}+z=rad{\left(abc\right)}^2+{\displaystyle \frac{1}{4rad\left(abc\right)}}$$

(12)

but since $rad\left(abc\right)$ and c are positive integers with $6\le rad\left(abc\right)$, we have ${\displaystyle \frac{1}{4rad\left(abc\right)}}<1$, this implies that $c\le rad{\left(abc\right)}^2$, which contradicts with our assumption.

Therefore we have $2b+z<2b+c-{\displaystyle \frac{\sqrt{c}}{\rho }}$, and by (10) we have

$$\begin{array}{c}\hfill rad{\left(abc\right)}^2<{\displaystyle \frac{2b}{2b+c-\frac{\sqrt{c}}{\rho }}}c<{\displaystyle \frac{2b}{2b+z}}c\Rightarrow (1+{\displaystyle \frac{z}{2b}})rad{\left(abc\right)}^2<c\end{array}$$

(13)

Thus the claim holds for $m=0$.

Now assume that the proposition holds for all positive integers $1\le m\le k-1$, then for the case $m=k$, since ${(1+{\displaystyle \frac{z}{2b}})}^{k}rad{\left(abc\right)}^2<c$ and since $c=a+b<2b$, we have ${(1+{\displaystyle \frac{z}{2b}})}^{k}rad{\left(abc\right)}^2<c<2b$. For the sake of brevity, we may have $\zeta ={(1+{\displaystyle \frac{z}{2b}})}^k$ in the following paragraphs.

Therefore, we have

$$R\left(ab\right)={\displaystyle \frac{rad\left(ab\right)}{ab}}<{\displaystyle \frac{rad\left(ab\right)}{a\left(\zeta \frac{rad{\left(abc\right)}^2}{2}\right)}}={\displaystyle \frac{2rad\left(ab\right)}{\zeta a·rad{\left(abc\right)}^2}}$$

(14)

and

$$R\left(c\right)={\displaystyle \frac{rad\left(c\right)}{c}}<{\displaystyle \frac{rad\left(c\right)}{\zeta rad{\left(abc\right)}^2}}$$

(15)

This implies that $R\left(ab\right)-{\displaystyle \frac{2rad\left(ab\right)}{\zeta a·rad{\left(abc\right)}^2}}<0$ and $R\left(c\right)-{\displaystyle \frac{rad\left(c\right)}{\zeta rad{\left(abc\right)}^2}}<0$

Therefore we have

$$0<(R\left(ab\right)-{\displaystyle \frac{2rad\left(ab\right)}{\zeta a·rad{\left(abc\right)}^2}})(R\left(c\right)-{\displaystyle \frac{rad\left(c\right)}{\zeta rad{\left(abc\right)}^2}})$$

(16)

Which implies that

$$\begin{array}{cc}\hfill & {\displaystyle \frac{2rad\left(ab\right)}{\zeta a·rad{\left(abc\right)}^2}}R\left(c\right)+{\displaystyle \frac{rad\left(c\right)}{\zeta rad{\left(abc\right)}^2}}R\left(ab\right)<R\left(abc\right)+\left({\displaystyle \frac{2}{{\zeta }^{2}a·rad{\left(abc\right)}^3}}\right)\hfill \end{array}$$

(17)

Since $\zeta rad{\left(abc\right)}^2<c$, (17) implies that

$$\begin{array}{cc}\hfill & {\displaystyle \frac{2rad\left(abc\right)}{\zeta ac·rad{\left(abc\right)}^2}}+{\displaystyle \frac{rad\left(abc\right)}{\zeta ab·rad{\left(abc\right)}^2}}={\displaystyle \frac{2rad\left(ab\right)}{\zeta a·rad{\left(abc\right)}^2}}R\left(c\right)+{\displaystyle \frac{rad\left(c\right)}{\zeta rad{\left(abc\right)}^2}}R\left(ab\right)\hfill \\ \hfill & <R\left(abc\right)+\left({\displaystyle \frac{2}{{\zeta }^{2}a·rad{\left(abc\right)}^3}}\right).\hfill \end{array}$$

(18)

Multiplying a on both sides, we have

$$\begin{array}{cc}\hfill & {\displaystyle \frac{2}{\zeta c·rad\left(abc\right)}}+{\displaystyle \frac{1}{\zeta b·rad\left(abc\right)}}\hfill \\ \hfill & <a·R\left(abc\right)+\left({\displaystyle \frac{2}{{\zeta }^{2}rad{\left(abc\right)}^3}}\right)=a·{\displaystyle \frac{rad\left(abc\right)}{abc}}+\left({\displaystyle \frac{2}{{\zeta }^{2}rad{\left(abc\right)}^3}}\right).\hfill \end{array}$$

(19)

Since $\zeta rad{\left(abc\right)}^2<c$, we have

$$\begin{array}{cc}\hfill & {\displaystyle \frac{2}{\zeta c·rad\left(abc\right)}}+{\displaystyle \frac{1}{\zeta b·rad\left(abc\right)}}<{\displaystyle \frac{rad\left(abc\right)}{bc}}+\left({\displaystyle \frac{2}{{\zeta }^{2}rad{\left(abc\right)}^3}}\right)\hfill \\ \hfill & <{\displaystyle \frac{\sqrt{\frac{c}{\zeta }}}{bc}}+{\displaystyle \frac{2}{{\zeta }^{2}rad{\left(abc\right)}^3}}={\displaystyle \frac{1}{b\sqrt{\zeta c}}}+{\displaystyle \frac{2}{{\zeta }^{2}rad{\left(abc\right)}^3}}\hfill \end{array}$$

(20)

Let $\rho ={\displaystyle \frac{1}{\sqrt{\zeta }rad\left(abc\right)}}$, then by multiplying $\sqrt{\zeta }bc$ on both sides, we have

$$(2b+c)\rho <\sqrt{c}+2{\rho }^{3}bc$$

(21)

Which implies that

$${\displaystyle \frac{2b+c-\frac{\sqrt{c}}{\rho }}{2bc}}<{\rho }^2\Rightarrow \zeta rad{\left(abc\right)}^2<{\displaystyle \frac{2b}{2b+c-\frac{\sqrt{c}}{\rho }}}c$$

(22)

If $2b+c-{\displaystyle \frac{\sqrt{c}}{\rho }}\le 2b+z$, then since $⌊\zeta rad{\left(abc\right)}^2⌋+1\le c$, $2b+c-{\displaystyle \frac{\sqrt{c}}{\rho }}\le 2b+z$ imply that

$$\begin{array}{cc}\hfill & c\le \left(\sqrt{\zeta }rad\left(abc\right)\right)\sqrt{c}+z\hfill \\ \hfill & \Rightarrow \sqrt{c}(\sqrt{c}-\sqrt{\zeta }rad\left(abc\right))\le z={\displaystyle \frac{1}{4rad\left(abc\right)}}\hfill \\ \hfill & \Rightarrow \sqrt{c}\le \sqrt{\zeta }rad\left(abc\right)+{\displaystyle \frac{1}{4rad\left(abc\right)\sqrt{c}}}.\hfill \end{array}$$

(23)

which implies that

$$\begin{array}{cc}\hfill & c\le \zeta rad{\left(abc\right)}^2+{\displaystyle \frac{2\sqrt{\zeta }rad\left(abc\right)}{4rad\left(abc\right)\sqrt{c}}}+{\displaystyle \frac{1}{16c·rad{\left(abc\right)}^2}}\hfill \\ \hfill & \le \zeta rad{\left(abc\right)}^2+{\displaystyle \frac{2\sqrt{\zeta }rad\left(abc\right)}{4rad\left(abc\right)\sqrt{\zeta }rad\left(abc\right)}}+{\displaystyle \frac{1}{16c·rad{\left(abc\right)}^2}}\hfill \\ \hfill & =\zeta rad{\left(abc\right)}^2+{\displaystyle \frac{1}{2rad\left(abc\right)}}+{\displaystyle \frac{1}{16c·rad{\left(abc\right)}^2}}\hfill \\ \hfill & \le ⌊\zeta rad{\left(abc\right)}^2⌋+1+{\displaystyle \frac{1}{2rad\left(abc\right)}}+{\displaystyle \frac{1}{16c·rad{\left(abc\right)}^2}}\hfill \end{array}$$

(24)

But since $3\le c$ is a positive integer and since $gcd(a,b,c)=1$ and $a+b=c$, we have $6\le rad\left(abc\right)$, thus ${\displaystyle \frac{1}{2rad\left(abc\right)}}+{\displaystyle \frac{1}{16c·rad{\left(abc\right)}^2}}<{\displaystyle \frac{1}{12}}+{\displaystyle \frac{1}{16\times 6^2}}<1$, therefore, (24) implies that $⌊\zeta rad{\left(abc\right)}^2⌋+1\le c\le ⌊\zeta rad{\left(abc\right)}^2⌋+1$, which indicates that $c=⌊\zeta rad{\left(abc\right)}^2⌋+1$.

By (21) and $c=⌊\zeta rad{\left(abc\right)}^2⌋+1$, we have

$$2b<{\displaystyle \frac{\sqrt{c}}{\rho }}+(2{\rho }^{2}b-1)c=\sqrt{{\displaystyle \frac{⌊\zeta rad{\left(abc\right)}^2⌋+1}{\zeta rad{\left(abc\right)}^2}}}+({\displaystyle \frac{2b}{\zeta rad{\left(abc\right)}^2}}-1)(⌊\zeta rad{\left(abc\right)}^2⌋+1)$$

(25)

Here we must have ${\displaystyle \frac{2b}{\zeta rad{\left(abc\right)}^2}}-1\le 1$, this is because if $1<{\displaystyle \frac{2b}{\zeta rad{\left(abc\right)}^2}}-1$, then we have $\zeta rad{\left(abc\right)}^2<b\le (c-1)$, which implies that $\zeta rad{\left(abc\right)}^2+1<c\le ⌊\zeta rad{\left(abc\right)}^2⌋+1$, a contradiction.

Also, we must have $b-a=1$, this is because if $2\le b-a$, then we have

$$2\le b-a\Rightarrow a+2\le b\Rightarrow a\le b-2\Rightarrow c=a+b\le 2b-2\Rightarrow c+2\le 2b$$

(26)

Therefore, (25) implies that

$$\begin{array}{cc}\hfill & ⌊\zeta rad{\left(abc\right)}^2⌋+3=c+2\le 2b<⌊\zeta rad{\left(abc\right)}^2⌋+1+\sqrt{{\displaystyle \frac{⌊\zeta rad{\left(abc\right)}^2⌋+1}{\zeta rad{\left(abc\right)}^2}}}\hfill \\ \hfill & \Rightarrow 4<{\displaystyle \frac{⌊\zeta rad{\left(abc\right)}^2⌋+1}{\zeta rad{\left(abc\right)}^2}}\Rightarrow 4\zeta rad{\left(abc\right)}^2<⌊\zeta rad{\left(abc\right)}^2⌋+1\hfill \end{array}$$

(27)

However, since $⌊\zeta rad{\left(abc\right)}^2⌋\le \zeta rad{\left(abc\right)}^2$, (27) implies that $4\zeta rad{\left(abc\right)}^2<\zeta rad{\left(abc\right)}^2+1$, which further implies that $3rad{\left(abc\right)}^2\le 3\zeta rad{\left(abc\right)}^2<1$, a contradiction.

Therefore, we have $b-a=1$, and by $b-a=1$ we have $a=b-1$, which implies that $c=a+b=2b-1$.

Again, by (21), we have

$$\begin{array}{cc}\hfill & 2b=c+1<{\displaystyle \frac{\sqrt{c}}{\rho }}+(2{\rho }^{2}b-1)c\hfill \\ \hfill & \Rightarrow 2c+1<{\displaystyle \frac{\sqrt{c}}{\rho }}+{\rho }^2(c+1)\hfill \\ \hfill & \Rightarrow 2c+1<c+{\displaystyle \frac{⌊\zeta rad{\left(abc\right)}^2⌋+2}{\zeta rad{\left(abc\right)}^2}}\le c+1+{\displaystyle \frac{2}{\zeta rad{\left(abc\right)}^2}}\hfill \\ \hfill & \Rightarrow c<{\displaystyle \frac{2}{\zeta rad{\left(abc\right)}^2}}\hfill \end{array}$$

(28)

But since $6\le rad\left(abc\right)$ and $1<\zeta ={(1+{\displaystyle \frac{z}{2b}})}^k$, we have ${\displaystyle \frac{2}{\zeta rad{\left(abc\right)}^2}}<1$, therefore, (28) implies that $c<1$, which is a contradiction.

Therefore we have $2b+z<2b+c-{\displaystyle \frac{\sqrt{c}}{\rho }}$, and by $2b+z<2b+c-{\displaystyle \frac{\sqrt{c}}{\rho }}$ and (22), we have

$$\begin{array}{cc}\hfill & {(1+{\displaystyle \frac{z}{2b}})}^{k}rad{\left(abc\right)}^2<{\displaystyle \frac{2b}{2b+c-\frac{\sqrt{c}}{\rho }}}c<{\displaystyle \frac{2b}{2b+z}}c\hfill \\ \hfill & \Rightarrow {(1+{\displaystyle \frac{z}{2b}})}^{k+1}rad{\left(abc\right)}^2={(1+{\displaystyle \frac{z}{2b}})}^k(1+{\displaystyle \frac{z}{2b}})rad{\left(abc\right)}^2<c\hfill \end{array}$$

(29)

Thus the claim holds for $m=k$ as well.

Therefore, by mathematical induction, the claim holds for all $0\le m$, which implies that if $rad{\left(abc\right)}^2<c$, then ${(1+{\displaystyle \frac{z}{2b}})}^{m}rad{\left(abc\right)}^2<c$ holds for all non-negative integers $0\le m$.

However, since c is finite and $1<1+{\displaystyle \frac{z}{2b}}$, by the Archimedean property, there is a positive integer $\tau$ such that $c<{(1+{\displaystyle \frac{z}{2b}})}^{\tau }rad{\left(abc\right)}^2$, therefore, the claim leads to a contradiction, thus $rad{\left(abc\right)}^2<c$ is false.

Therefore $c<rad{\left(abc\right)}^2$. □