On Some Properties of ABC Triples and Radicals of an Integer

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, in other words, 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 [1,2]. The statement of the abc conjecture is as follows:

 Conjecture 1.

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

So far there is no widely accepted proof for the abc conjecture and the conjecture is still regarded as an open problem.

So far, a proven upper bound for c 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 [3].

Besides, Stewart and Tijdeman proved that the following lower bound holds for infinitely many abc triples in 1986 [4]:

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

(2)

for all $k<4$, and van Frankenhuysen improved k to $6.068$ in 2000 [5].

Moreover, Granville and Tucker ([6]) has proposed the following conjecture:

 Conjecture 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’s a stronger conjecture proposed by Baker in 2004 [7], stating that

 Conjecture 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, [8]) 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 3 implies that $c<rad{\left(abc\right)}^{{\displaystyle \frac{7}{4}}}$.

In this paper, we intend to discuss some properties of the radical of integers and the triples involved in the abc conjecture, improving the upper bound for the c in an abc triple, and apply our result to the Nagell-Ljunggren equation as an example.

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 $p_1,\dots ,p_{\omega \left(n\right)}$ 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. Properties of Radicals

In this section, we will discuss the general properties of the radical of a positive integer.

Definition 1. 

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

Theorem 1. 

If $gcd(a,b)=1$, then $R\left(ab\right)=R\left(a\right)R\left(b\right)$

Proof. 

Since $rad\left(n\right)$ is multiplicative, we have $R\left(ab\right)={\displaystyle \frac{rad\left(ab\right)}{ab}}={\displaystyle \frac{rad\left(a\right)}{a}}{\displaystyle \frac{rad\left(b\right)}{b}}=R\left(a\right)R\left(b\right)$ □

Theorem 2. 

$R\left(n\right)=1$ if and only if n is square-free, $R\left(n\right)\le {\displaystyle \frac{1}{2}}$ if $R\left(n\right)$ is not square-free.

Proof. 

By definition of $rad\left(n\right)$, we have $rad\left(n\right)\le n$, and if n is square-free, then $rad\left(n\right)=n$ and vice versa.

If n is not square-free, then there’s a prime factor $p\mid n$ such that $p^2\mid n$, which implies that $p·rad\left(n\right)\le n$.

Therefore we have

$$R\left(n\right)={\displaystyle \frac{rad\left(n\right)}{n}}\le {\displaystyle \frac{rad\left(n\right)}{p·rad\left(n\right)}}={\displaystyle \frac{1}{p}}\le {\displaystyle \frac{1}{2}}.$$

(3)

And this completes the proof. □

Remark 1. 

Let $\mu \left(n\right)$ be the Möbius function of n, and $\omega \left(n\right)$ be the number of distinct prime factors of n, then we have $\mu \left(n\right)={(-1)}^{\omega \left(n\right)}⌊R\left(n\right)⌋$.

Theorem 3. 

If $gcd(a,b)=1$, then $R\left(a\right)+R\left(b\right)\le R\left(ab\right)+1$

Proof. 

Since $R\left(a\right)\le 1$ and $R\left(b\right)\le 1$, we have $R\left(a\right)-1\le 0\wedge R\left(b\right)-1\le 0$. Therefore we have

$$\begin{array}{cc}\hfill & 0\le \left(R\right(a)-1)\left(R\right(b)-1)\hfill \\ \hfill & \Rightarrow 0\le R\left(ab\right)-\left(R\right(a)+R(b\left)\right)+1\hfill \\ \hfill & \Rightarrow R\left(a\right)+R\left(b\right)\le R\left(ab\right)+1.\hfill \end{array}$$

(4)

And this completes the proof. □

3. Properties of ABC Triples

In this section, we will discuss the properties of abc triples i.e. positive integers $a,b,c$ such that $a+b=c$, $gcd(a,b,c)=1$ and $rad\left(abc\right)<c$.

Definition 2. 

A set of three positive integers $(a,b,c)$ is called an abc triple if $c=a+b$, $gcd(a,b,c)=1$ and $rad\left(abc\right)<c$.

Theorem 4. 

If $(a,b,c)$ is an abc triple, then c is not square-free

Proof. 

If c is square-free, then we have $c=rad\left(c\right)$, therefore, we have

$$c=rad\left(c\right)\le rad\left(a\right)rad\left(b\right)rad\left(c\right)=rad\left(abc\right)$$

(5)

□

Theorem 5. 

If $(a,b,c)$ is an abc triple, then at most one of a or b is square-free.

Proof. 

First, by Theorem 3, we have

$$R\left(a\right)+R\left(b\right)+R\left(c\right)-2=\left(R\right(a)+R(b)-1)+R\left(c\right)-1\le R\left(ab\right)+R\left(c\right)-1\le R\left(abc\right).$$

(6)

Since $(a,b,c)$ is an abc triple, we have

$$R\left(a\right)+R\left(b\right)+R\left(c\right)\le 2+R\left(abc\right)=2+{\displaystyle \frac{rad\left(abc\right)}{abc}}<2+{\displaystyle \frac{c}{abc}}=2+{\displaystyle \frac{1}{ab}}.$$

(7)

If both of a and b are square-free, then we have

$$\begin{array}{cc}\hfill & 2+R\left(c\right)=R\left(a\right)+R\left(b\right)+R\left(c\right)\le 2+{\displaystyle \frac{1}{ab}}\hfill \\ \hfill & \Rightarrow rad\left(c\right)\le {\displaystyle \frac{c}{ab}}={\displaystyle \frac{a+b}{ab}}<2.\hfill \end{array}$$

(8)

But this leads to a contradiction.

Therefore, if $(a,b,c)$ is an abc triple, at most one of a or b is square-free. □

Theorem 6. 

If $(a,b,c)$ is an abc triple with a being square-free, then $a<b<c$

Proof. 

$b<c$ follows from the fact that $c=a+b$

Assume that a is square-free and $a>b$, then we have

$$a·rad\left(b\right)rad\left(c\right)=rad\left(a\right)rad\left(b\right)rad\left(c\right)=rad\left(abc\right)<c=a+b.$$

(9)

Since b and c are not square-free by Theorem 5, we have $4\le b<c$, which implies that $2\le rad\left(b\right)$ and $2\le rad\left(c\right)$, thus we have

$$4=2\times 2\le rad\left(b\right)rad\left(c\right)<1+{\displaystyle \frac{b}{a}}<2.$$

(10)

Which is a contradiction.

Therefore, if $(a,b,c)$ is an abc triple with a being square-free, then $a<b<c$. □

Theorem 7. 

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

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

(11)

and

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

(12)

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

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

(13)

Which implies that

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

(14)

Multiplying a on both sides, we have

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

(15)

Since $rad{\left(abc\right)}^2<c$, (14) implies that

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

(16)

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

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

(17)

Which implies that

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

(18)

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

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

(19)

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

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

(20)

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 (18) we have

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

(21)

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

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

(22)

and

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

(23)

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

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

(24)

Which implies that

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

(25)

Since $\zeta rad{\left(abc\right)}^2<c$, (25) implies that

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

(26)

Multiplying a on both sides, we have

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

(27)

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

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

(28)

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

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

(29)

Which implies that

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

(30)

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

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

(31)

which implies that

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

(32)

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

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

(33)

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

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

(34)

Therefore, (33) implies that

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

(35)

However, since $⌊\zeta rad{\left(abc\right)}^2⌋\le \zeta rad{\left(abc\right)}^2$, (35) 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 (29), we have

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

(36)

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, (36) 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 (30), we have

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

(37)

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 $c<rad{\left(abc\right)}^2$. □

4. Additional Result

In this section, we will apply our result on the Nagell-Ljunggren equation as an example of the application of our results. The Nagell-Ljunggren equation is a Diophantine equation of the form ${\displaystyle \frac{x^n-1}{x-1}}=y^q$, with $x,y,n,q$ being positive integers and $3\le n,2\le q$. It has been conjectured that the following solutions are all positive solutions of the Nagell-Ljunggren equation:

$${\displaystyle \frac{3^5-1}{3-1}}={11}^2,{\displaystyle \frac{7^4-1}{7-1}}={20}^2\mathrm{and}{\displaystyle \frac{{18}^3-1}{18-1}}=7^3$$

(38)

In this section, we will show that the above solutions are all positive solutions of the Nagell-Ljunggren equation by applying Theorem 7.

Theorem 8. 

The solutions listed in (38) are all positive solutions of the Nagell-Ljunggren equation.

Proof. 

Assume that $(x,y,n,q)$ is a set of positive integers such that ${\displaystyle \frac{x^n-1}{x-1}}=y^q$, then since 29 is the smallest prime factor of n for this solution, we have $29\le n$ and $3\le q$ ([9])

By moving terms, we get $(x-1)y^q+1=x^n$, then by Theorem 7, we have

$$\begin{array}{cc}\hfill & (x-1)y^q+1=x^n<rad{(1·(x-1)y^q·x^n)}^2\le x^2{(x-1)}^2{y}^2<x^4{y}^2\hfill \\ \hfill & \Rightarrow y^{q-2}<x^2(x-1)<x^3\hfill \\ \hfill & \Rightarrow x^n<x^4{y}^2<x^{4+{\displaystyle \frac{6}{q-2}}}\Rightarrow n<4+{\displaystyle \frac{6}{q-2}}.\hfill \end{array}$$

(39)

Since $29\le n$, we have

$$29\le n<4+{\displaystyle \frac{6}{q-2}}\Rightarrow 25q-50<6\Rightarrow q<{\displaystyle \frac{56}{25}}\Rightarrow q=2.$$

(40)

But since we have $3\le q$ as well, this leads to a contradiction.

Therefore, the solutions listed in (38) are all positive solutions of the Nagell-Ljunggren equation. □

5. Discussions

In this paper, we showed that if $(a,b,c)$ is an abc triple, then at most one of a and b is square-free, and c can’t be a square-free number, which is consistent with a casual observation on computational results like ones from ABC@home [10,11].

Also, note that in general $a+b=c$ does not imply $rad\left(a\right)+rad\left(b\right)\le rad\left(c\right)$ or $rad\left(c\right)\le rad\left(a\right)+rad\left(b\right)$, and this is one major barrier in the research of relevant topics. To see this, both of $(a,b,c)=(2,3^{10}\times 109,{23}^5)$ and $(a,b,c)=(1,2\times 3^7,5^4\times 7)$ satisfy the condition $a+b=c$,10] but the former one has $rad\left(c\right)\le rad\left(a\right)+rad\left(b\right)$ and the latter one has $rad\left(a\right)+rad\left(b\right)\le rad\left(c\right)$. There’s no fixed relationship for the logarithm of radicals of integers either, since again $(a,b,c)=(2,3^{10}\times 109,{23}^5)$ gives $lograd\left(c\right)<lograd\left(a\right)+lograd\left(b\right)$ and $(a,b,c)=(1,2\times 3^7,5^4\times 7)$ gives $lograd\left(a\right)+lograd\left(b\right)<lograd\left(c\right)$.

Regarding the abc conjecture specifically, it has been shown from the data of ABC@Home [11] that there are abc triples with the square roots of its members also form a Pythagorean triple like $(a,b,c)$=$(2^4\times {11}^2,3^4\times {13}^2,5^6)$, and also abc triples like $(a,b,c)=(2^7,{17}^3,{71}^2)$ with all its members being perfect powers; however, such triples do not seem common among all known pairs. A question is, are there only finitely many abc triples such that both of a, b and c are perfect powers? More specifically, are there only finitely many abc triples that the square roots of its members also form a Pythagorean triple?