On the Diophantine equations f (x) = g(y)

TL;DR

This paper investigates conditions under which certain polynomial Diophantine equations have finitely many integer solutions, analyzing specific cases and assuming the ABC Conjecture to establish finiteness criteria.

Contribution

It provides new criteria for finiteness of solutions to equations of the form f(x) = g(y), including cases with monic polynomials and under the ABC Conjecture.

Findings

Finiteness conditions for (y+q1)...(y+qm)=f(x)

Finiteness criteria for f(x)=g(y) under ABC Conjecture

Analysis of polynomial equations with integer solutions

Abstract

The study of finiteness or infiniteness of integer solutions of a Diophantine equation has been considered as a standard problem in the literature. In this paper, for f(x) in Z[x] monic and q1 ,...., qm in Z, we study the conditions for which the Diophantine equatio (y + q1 )(y + q2 ) .... (y + qm ) = f(x) has finitely many solutions in integers. Also assuming ABC Conjecture, we study the conditions for finiteness of integer solutions of the Diophantine equation f(x) = g(y).

Equations

$$(y+q_1)(y+q_2)\cdots (y+q_m)=f(x)$$

$$f(x)=g(y),$$

$$y^2=a_n{x}^n+a_{n-1}{x}^{n-1}+\cdots +a_0,$$

$$(y+q_1)(y+q_2)\cdots (y+q_m)=f(x)$$

$$f(x)=f_{11}(x{)}^{{\alpha }_1}{f}_{12}(x{)}^{{\alpha }_2}\cdots f_{1l}(x{)}^{{\alpha }_l}\pm f_{21}(x{)}^{{\beta }_1}{f}_{22}(x{)}^{{\beta }_2}\cdots f_{2m}(x{)}^{{\beta }_m}$$

$$g(y)=g_1(y{)}^{{\gamma }_1}{g}_2(y{)}^{{\gamma }_2}\cdots g_n(y{)}^{{\gamma }_n},$$

$$f(x)=g(y)$$

$$f(x)=f_{11}(x{)}^{{\alpha }_1}{f}_{12}(x{)}^{{\alpha }_2}\cdots f_{1l}(x{)}^{{\alpha }_l}\pm f_{21}(x{)}^{{\beta }_1}{f}_{22}(x{)}^{{\beta }_2}\cdots f_{2m}(x{)}^{{\beta }_m}$$

$$g(y)=g_1(y{)}^{{\gamma }_1}{g}_2(y{)}^{{\gamma }_2}\cdots g_n(y{)}^{{\gamma }_n},$$

$$f(x)=g(y)$$

$$g(y)=g_1(y{)}^{{\gamma }_1}{g}_2(y{)}^{{\gamma }_2}\cdots g_n(y{)}^{{\gamma }_n},$$

$$f(x)=g(y)$$

$$(y+q_1)\cdots (y+q_m)=f(x)$$

$$c{\ll }_{ϵ}N(a,b,c{)}^{1+ϵ},$$

$$f(x)=g(y)$$

$$a_0-g(c)+\left(a_1-\frac{g^1(c)}{1!}\right)x+\cdots +\left(a_m-\frac{g^m(c)}{m!}\right)x^m+a_{m+1}{x}^{m+1}+\cdots +a_n{x}^n$$

$$A(c)=\left(\begin{array}[]{ccccccc}1&c&c^{2}&\cdots&\cdots&c^{m-1}&c^{m}\\ 0&1&{2\choose 1}c&\cdots&\cdots&{m-1\choose m-2}c^{m-2}&{m\choose m-1}c^{m-1}\\ 0&0&1&\cdots&\cdots&{m-1\choose m-3}c^{m-3}&{m\choose m-2}c^{m-2}\\ \vdots&\vdots&\vdots&\cdots&\cdots&\vdots&\vdots\\ 0&0&0&\cdots&\cdots&1&{m\choose 1}c\\ 0&0&0&\cdots&\cdots&0&1\end{array}\right).$$

$$g(y)=f(x)$$

$$b_0+b_{1}c+b_2{c}^2+\cdots +b_m{c}^m=g(c)=a_0,$$

$$b_1+\left(\genfrac{}{}{0ex}{}{2}{1}\right)b_{2}c+\left(\genfrac{}{}{0ex}{}{3}{2}\right)b_3{c}^2+\cdots +\left(\genfrac{}{}{0ex}{}{m}{m-1}\right)b_m{c}^{m-1}=\frac{g^1(c)}{1!}=a_1,$$

$$b_2+\left(\genfrac{}{}{0ex}{}{3}{1}\right)b_{3}c+\left(\genfrac{}{}{0ex}{}{4}{2}\right)b_4{c}^2+\cdots +\left(\genfrac{}{}{0ex}{}{m}{m-2}\right)b_m{c}^{m-2}=\frac{g^2(c)}{2!}=a_2,$$

$$⋮$$

$$b_{m-2}+\left(\genfrac{}{}{0ex}{}{m-1}{1}\right)b_{m-1}c+\left(\genfrac{}{}{0ex}{}{m}{2}\right)b_m{c}^2=\frac{g^{m-2}(c)}{(m-2)!}=a_{m-2},$$

$$b_{m-1}+\left(\genfrac{}{}{0ex}{}{m}{1}\right)b_{m}c=\frac{g^{m-1}(c)}{(m-1)!}=a_{m-1},$$

$$b_m=\frac{g^m(c)}{m!}=a_m.$$

$$g(y)=f(x)$$

$$(y+q_1)\cdots (y+q_m)=f(x)$$

$$k={\Pi }_{i=1}^m(n+a_i),$$

$$f(x)=x^n+a_1{x}^{n-1}+\cdots +a_n,$$

Full text

[TABLE]

has finitely many solutions in integers. Also assuming $ABC$-Conjecture, we study the conditions for finiteness of integer solutions of the Diophantine equation $f(x)=g(y)$.

†† 2010 Mathematics Subject Classification: Primary 11D41, 11D45, Secondary 11D25. Key Words: Diophantine equation, Monic polynomial.

1. Introduction

Consider the equation

[TABLE]

where $f(x)$ and $g(x)$ are polynomials with integral coefficients of degrees $n$ and $m$ respectively. Already we know the following natural problem. Problem 0 Does the equation (1.1) have finitely or infinitely many solutions in integers?

Several mathematicians solved some special cases of this problem. Yuri F. Bilu and Robert F. Tichy in [3] have given a criteria to check whether (1.1) has finitely many solutions in integers with a bounded denominator, when $f(x)$ and $g(x)$ are monic. Baker [1] gives an upper bound for integral solutions of the equation

[TABLE]

where $\displaystyle a_{n},a_{n-1},\cdots,a_{0}$ are rational integers, $a_{n}\neq 0$, when $n\geq 5$ and the polynomial on the right separable. In a major breakthrough, in early seventies, Baker gives an upper bound for integral solutions of the super elliptic equations $y^{m}=f(x)$. Also [2], [4], [5], [9], [10], [11] and [12] give some upper bounds for some equations of the form $y^{m}=f(x)$. Also [8] tells that (1.2) has finitely many solutions in integers, if $a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{0}$ has distinct zeros and $n\geq 3$. Our aim in this paper is to study the conditions for which the equation (1.1) has finitely many solutions in integers. Here without assuming $ABC$ Conjecture, we solve the problem 0 for some special type of $f(x)$ and $g(y)$. When we assume $ABC$ conjecture, we cover much more general types of diophantine equations of the form (1.1) including non-monic polynomials $f(x)$ and $g(y)$. The following three theorems are without assuming $ABC$-Conjecture.

Theorem 1.1**.**

Let $f(x)\in\mathbb{Z}[x]$ be monic, with $\deg f(x)=n>m$ and $\displaystyle q_{1},\cdots,q_{m}\in\mathbb{Z}$. Then the equation

[TABLE]

has finitely many solutions in integers if and only if $|x-y|$ is bounded for all integer solutions $(x,y)$.

Theorem 1.2**.**

Let $f(x)\in\mathbb{Z}[x]$ be monic, $\deg f(x)=m$, at least one of the roots of $f(x)$ be non integral and let $\displaystyle q_{1},q_{2},\cdots,q_{m}\in\mathbb{Z}$. Then the equation

[TABLE]

has finitely many solutions $x$ and $y$ in integers.

Theorem 1.3**.**

Let $f(x)\in\mathbb{Z}[x]$ be monic with $\deg f(x)=n<m$ and $\displaystyle q_{1},\cdots,q_{m}\in\mathbb{Z}$. Then the equation

[TABLE]

has finitely many solutions in integers if and only if $|x-y|$ is bounded for all integer solutios $(x,y)$.

The following three theorems are assuming $ABC$-Conjecture.

Theorem 1.4**.**

Let $\displaystyle f_{11}(x),\cdots,f_{1l}(x),f_{21}(x),\cdots,f_{2m}(x),g_{1}(x),\cdots,g_{n}(x)\in\mathbb{Z}[x]$, $\displaystyle\alpha_{1},\alpha_{2},\cdots,\alpha_{l},\beta_{1},\beta_{2},\cdots,\beta_{m},\gamma_{1},\gamma_{2},\cdots,\gamma_{n}\in\mathbb{Z}^{+}$. Let

[TABLE]

and

[TABLE]

with

1. (1)

$\deg g(y)<\deg f(x)$** 2. (2)

$\deg g(y)>1+\deg(f_{11}(x)f_{12}(x)\cdots f_{1l}(x)f_{21}(x)\cdots f_{2m}(x)g_{1}(x)\cdots g_{n}(x))$** 3. (3)

$f_{11}(x)f_{12}(x)\cdots f_{1l}(x)$* and $f_{21}(x)\cdots f_{2m}(x)$ have no common factor in $\mathbb{Z}[x]$* 4. (4)

leading coefficient of each of $f_{11}(x)\cdots f_{1l}(x)$, $f_{21}(x)\cdots f_{2m}(x)$,

$g_{1}(x)\cdots g_{n}(x)$* is positive.*

If we assume $ABC$-Conjecture, then the equation

[TABLE]

has finitely many solutions $(x,y)$ in integers.

Theorem 1.5**.**

Let $\displaystyle f_{11}(x),\cdots,f_{1l}(x),f_{21}(x),\cdots,f_{2m}(x),g_{1}(x),\cdots,g_{n}(x)\in\mathbb{Z}[x]$, $\displaystyle\alpha_{1},\alpha_{2},\cdots,\alpha_{l},\beta_{1},\beta_{2},\cdots,\beta_{m},\gamma_{1},\gamma_{2},\cdots,\gamma_{n}\in\mathbb{Z}^{+}$. Let

[TABLE]

and

[TABLE]

with

1. (1)

$\deg g(y)>\deg f(x)$** 2. (2)

$\deg f(x)>1+\deg(f_{11}(x)f_{12}(x)\cdots f_{1l}(x)f_{21}(x)\cdots f_{2m}(x)g_{1}(x)\cdots g_{n}(x))$** 3. (3)

$f_{11}(x)f_{12}(x)\cdots f_{1l}(x)$* and $f_{21}(x)\cdots f_{2m}(x)$ have no common factor in $\mathbb{Z}[x]$* 4. (4)

leading coefficient of each of $f_{11}(x)\cdots f_{1l}(x)$, $f_{21}(x)\cdots f_{2m}(x)$,

$g_{1}(x)\cdots g_{n}(x)$* is positive.*

If we assume $ABC$-Conjecture, then the equation

[TABLE]

has finitely many solutions $(x,y)$ in integers.

Theorem 1.6**.**

Let $f(x),g(x)\in Z[x]$, $f(x)$ be separable with $\deg f(x)>2$ and

[TABLE]

where $\gamma_{1}$, $\gamma_{2}$, …, $\gamma_{n}$ are positive integers $\geq$ 2, $g_{i}(x)\in Z[x]$ for $i=1,2,...,k$. If assume $ABC$ Conjecture, then the equation

[TABLE]

has only finitely many solutions in integers.

As an immediate consequence we have following result.

Corollary 1.7**.**

Assume $ABC$-Conjecture. Let $f(x)\in\mathbb{Z}[x]$ be nonconstant and $\displaystyle q_{1},\cdots,q_{m}\in\mathbb{Z}$.

1. (1)

If $f(x)$ is separable with $\deg f(x)>2$, and for each $j$, $1\leq j\leq m$, $(y+q_{i})^{2}\mid\Pi_{i=1}^{m}(y+q_{i})$, then the equation

[TABLE]

has finitely many integer solutions. 2. (2)

If $m>2$, $\displaystyle q_{1},\cdots,q_{m}$ are distinct and $f(x)=\Pi_{i=1}^{k}f_{i}^{l_{i}}(x)$, where $f_{i}(x)\in\mathbb{Z}[x]$ and $l_{i}\geq 2$ for $1\leq i\leq k$, then the equation

[TABLE]

has finitely many integer solutions.

Our paper contains five sections. In the second section we state $ABC$-Conjecture. Here We also establish some results to equation (1.1). In the third Section we study the conditions in Cases $m<n$ and $m>n$, for which the equation (1.1) has finitely many solutions in integers. In fourth section, we study Case $m=n$. In the fifth section, we study the conditions assuming $ABC$-Conjecture for which the equation (1.1) has finitely many solutions in integers.

2. Preliminaries

The $ABC$-Conjecture. (Oesterl$\acute{e}$, Masser, Szpiro): Fix $\epsilon>0$. If $\displaystyle a,b,c$ are coprime positive integers satisfying $a+b=c$ then

[TABLE]

where $N(a,b,c)$ is the product of distinct prime divisors of $abc$. The following result is a consequence of $truth$ of $ABC$-conjecture and is a part of Theorem 6 in [6].

Lemma 2.1**.**

Assume that the $ABC$-Conjecture is true. Fix $\epsilon>0$. If $g(x)\in\mathbb{Z}[x]$ has no repeated roots and $q^{2}\mid g(m)$, then $q\ll_{\epsilon}|m|^{1+\epsilon}$.

An integer is called powerful if $p^{2}$ divides $n$ for every prime $p$ dividing $n$.

Remark 2.2 :

Lemma 2.1 implies that if $g(x)\in\mathbb{Z}[x]$ has degree $>2$ and is separable, then $g(m)$ is powerful for only finitely many integers $m$.

The following results are without assuming $ABC$-Conjecture.

Lemma 2.3**.**

Let $\displaystyle f(x),g(x)\in\mathbb{Z}[x]$ with $\displaystyle\deg f(x)>\deg g(x)$. Then the equation

[TABLE]

has finitely many integer solutions if and only if $|x-y|$ is bounded for all integral solutions $(x,y)$.

Proof.

Let $f(x)=a_{0}+\cdots+a_{n}x^{n}$ and $g(x)=b_{0}+\cdots+b_{m}x^{m}$ with $a_{n}\neq 0\neq b_{m}$. Given $n>m$. If possible assume that the equation $g(y)=f(x)$ has infinitely many integer solutions $x=a$, $y=b$, with $|a-b|$ bounded, i.e. $a-b$ takes values in a finite subset $S$ (say) of $\mathbb{Z}$. There should be an integer $c\in S$, such that the equation $g(b)=f(a)$ is satisfied by infinitely many $\displaystyle a,b\in\mathbb{Z},b-a=c$. As $b=a+c$, we have $g(a+c)=f(a)$ for infinitely many integers $a$. i.e., $f(a)=g(c)+\frac{g^{1}(c)}{1!}a+\frac{g^{2}(c)}{2!}a^{2}+\cdots+\frac{g^{m}(c)}{m!}a^{m}$ for infinitely many integers $a$. This implies that the polynomial

[TABLE]

over $\mathbb{Z}$ has infinitely many integers roots $x=a$. Therefore $a_{0}-g(c)=a_{1}-g^{1}(c)=\cdots=a_{m}-\frac{g^{m}(c)}{m!}=a_{m+1}=\cdots=a_{n}=0$, which is a contradiction to the hypothesis. ∎

For $c$ an integer Let us define a matrix $A(c)$ as follows:

[TABLE]

Lemma 2.4**.**

Let $\displaystyle a_{0}+\cdots+a_{m}x^{m}=f(x),b_{0}+\cdots+b_{m}x^{m}=g(x)\in\mathbb{Z}[x]$ with $\deg f(x)=\deg g(x)=m\geq 1$. Then the equation

[TABLE]

has finitely many solutions $(x,y)$ in integers if and only if $|x-y|$ is bounded and $\displaystyle A(y-x)(b_{0},b_{1},\cdots,b_{m})^{t}\neq(a_{0},a_{1},\cdots,a_{m})^{t}$.

Proof.

If possible let the equation $g(y)=f(x)$ has infinitely many integer solutions $x=a$, $y=b$, $|a-b|$ bounded and for $c=b-a$,

$\displaystyle A(c)(b_{0},b_{1},\cdots,b_{m})^{t}\neq(a_{0},a_{1},\cdots,a_{m})^{t}$. This implies that there should be at least one integer $c$ such that $b-a=c$ and $g(b)=f(a)$ are satisfied by infinitely many integers $a,\,b$. Thus we have $f(a)=g(a+c)=g(c)+\frac{g^{1}(c)}{1!}a+\frac{g^{2}(c)}{2!}a^{2}+\cdots+\frac{g^{m}(c)}{m!}a^{m}$ for infinitely many integers $a$. This implies

[TABLE]

Which is same as $\displaystyle A(c)(b_{0},b_{1},\cdots,b_{m})^{t}=(a_{0},a_{1},\cdots,a_{m})^{t}$ a contradiction to our assumption. ∎

Corollary 2.5**.**

Let $f(x)$, $g(x)\in\mathbb{Z}[x]$ be as in the lemma, with $a_{m}\neq b_{m}$. Then the equation

[TABLE]

has finitely many integer solutions if and only if $|x-y|$ is bounded for all integer solutions $(x,y)$.

Proof.

Immediate from the Lemma. ∎

Corollary 2.6**.**

Let $f(x)$, $g(x)\in\mathbb{Z}[x]$ be as in the lemma, $\deg f(x)=\deg g(x)=m\geq 2$ and $a_{m-1}-b_{m-1}$ not divisible by $mb_{m}$. Then the equation

[TABLE]

has finitely many integer solutions if and only if $|x-y|$ is bounded for all integer solutions $(x,y)$.

Proof.

Let us assume that the equation $g(y)=f(x)$ is satisfied by infinitely many integers $x=a$, $y=b$ with $|a-b|$ bounded. So by the lemma we have $b_{m-1}+mb_{m}c=a_{m-1}$, hence $mb_{m}$ divides $b_{m-1}-a_{m-1}$, which is a contradiction. ∎

3. Cases $m<n$ and $m>n$

Proof of theorem 1.1: As $\displaystyle\deg f(x)=n>m=\deg((y+q_{1})\cdots(y+q_{m}))$ by Lemma 2.3 the theorem immediately follows.$\hfill\square$ Proof of theorem 1.3: As $\displaystyle\deg((y+q_{1})\cdots(y+q_{m}))=m>n=\deg f(x)$, immediate from Lemma 2.3.$\hfill\square$

4. Case $m=n$

In this section, for $f(x)\in\mathbb{Z}[x]$, monic with $\deg f(x)=m$, we study the conditions for which the equation

[TABLE]

has finitely many solutions.

4.1. THEOREMS ON MONIC POLYNOMIALS

In this subsection, we give the definition for a complete composite number of a monic polynomial with positive integral coefficients. Using this definition we prove one main theorem on polynomials. This theorem is needed to prove the all theorems of the next subsection.

Definition 4.1 :

Let $f(x)$ be a polynomial of degree $m$ with positive integral coefficients. Then a positive integer $k$ is said to be complete composite of $f(x)$, if $k=f(n)$, for some positive integer $n$ and

[TABLE]

where $\displaystyle a_{1},\cdots,a_{m}$ are positive integers. If $f(a)$ is complete composite of $f(x)$, then $a$ is called a complete number of $f(x)$.

Lemma 4.2**.**

Let $f(x)$ be a monic polynomial of degree $n$ with positive integral coefficients. If there are infinitely many complete composites of $f(x)$, then $f(x)$ has a factor $x+c$ with positive integer $c$.

Proof.

Let $f(x)$ be a monic polynomial of degree $n$ with positive integral coefficients. Let

[TABLE]

where $\displaystyle a_{1},a_{2},\cdots,a_{n}$ are positive integers. Let there be infinitely many complete composites of $f(x)$. Therefore there are infinitely many positive integers $m$ such that

[TABLE]

where $\displaystyle b_{1m},b_{2m},\cdots,b_{nm}$ are positive integers. If $b_{im}=c$, for infinitely many pairs $(i,m)$, where $c$ is a positive integer, then $m+c$ divides $f(m)$, for infinitely many positive integers $m$. For $f(x)$ and $x+c$, there exist two polynomials $g(x)$ and $h(x)$ with integral coefficients such that

[TABLE]

where $h(x)$ is a constant polynomial (see [7]). Since $\displaystyle f(n),n+c$ and $g(n)$ are integers for any integer $n$, we have

[TABLE]

where $h(n)$ is an integer. Since $n+c$ divides $f(n)$, for infinitely many positive integers $n$, $n+c$ divides $h(n)$, for infinitely many positive integers $n$. Therefore $h(n)=0$, because $h(x)$ is a constant polynomial. So

[TABLE]

This gives that $f(x)$ has a factor $x+c$ with positive integer $c$. Suppose that $b_{im}$ is not a constant for infinitely many pairs $(i,m)$. From this, it is clear that for each positive integer $k$, there exist a positive integer $N$ such that

[TABLE]

for all $\displaystyle i=1,2,\cdots,n$ and for all $m\geq N$, where $m$ is a complete number of $f(x)$. From equation (4.3), we get

[TABLE]

Since for each positive integer $k$, there exist a positive integer $N$ such that

[TABLE]

for all $\displaystyle i=1,2,\cdots,n$ and for all $m\geq N$, where $m$ is a complete number of $f(m)$, then we can choose a complete number $d$ of $f(x)$ such that

[TABLE]

Therefore

[TABLE]

which is a contradiction to (4.4). So our assumption is wrong. This proves the theorem. ∎

Lemma 4.3**.**

Let $f(x)$ be a monic polynomial of degree $n$ with positive integral coefficients. Then there are infinitely many complete composites of $f(x)$ if and only if $f(x)$ is the product of $n$ one degree monic polynomials with positive integral coefficients.

Proof.

Let $f(x)$ be a monic polynomial of degree $n$ with positive integral coefficients. Let there be infinitely many complete composites of $f(x)$. We prove this result using Mathematical induction on $n$. Clearly the result is true for $n=1$. Let us assume that the result is true for all positive integers $<n$. We prove this result for $n$. Since there are infinitely many complete composites of $f(x)$, by Lemma 4.2, $f(x)$ has a factor $x+c$ with positive integer $c$. Therefore

[TABLE]

where $g(x)$ is a polynomial of degree $n-1$ with positive integral coefficients. Since there are infinitely many complete composites of $f(x)$, there are infinitely many complete composites of $g(x)$. By our induction hypothesis, $g(x)$ is the product of $n-1$ monic polynomials of degree one with positive integral coefficients. Therefore $f(x)$ is the product of $n$ monic polynomials of degree one with positive integral coefficients. Converse part is obvious. ∎

Lemma 4.4**.**

Let $f(x)$ be a monic polynomial of degree $n$ with integral coefficients. Then there are infinitely many complete composites of $g(x)=f(x+h)$, (where $h$ is a positive integer such that the coefficients of $g(x)$ are positive) if and only if $f(x)$ is the product of $n$ monic polynomials of one degree with integral coefficients.

Proof.

Let $f(x)$ be a monic polynomial of degree $n$ with integral coefficients. Let there be infinitely many complete composites of $g(x)=f(x+h)$, where $h$ is a positive integer such that $g(x)$ has positive coefficients. By Lemma 4.3, $g(x)$ is the product of $n$ monic polynomials of degree one with positive integral coefficients. Let

[TABLE]

Therefore

[TABLE]

Therefore $f(x)$ is the product of $n$ one degree monic polynomials with integer coefficients. Converse part is obvious. ∎

4.2. Finiteness of integral solutions

In this subsection, we study the finiteness of integral solutions of equation (1.1) using the results from above subsection.

Lemma 4.5**.**

Let $f(x)$ be a monic polynomial of degree $m$ with positive integral coefficients, all roots of $f(x)$ be not integers and let $\displaystyle q_{1},q_{2},\cdots,q_{m}$ be positive integers. Then the equation

[TABLE]

has finitely many solutions $x$ and $y$ in positive integers.

Proof.

Let $f(x)$ be a monic polynomial of degree $m$ with positive integral coefficients, all roots of $f(x)$ be not integers and let $\displaystyle q_{1},q_{2},\cdots,q_{m}$ be positive integers. Let

[TABLE]

Suppose that there is a solution $x=a$ and $y=b$ in positive integers for (4.5) such that

[TABLE]

Therefore

[TABLE]

Since $a^{m}<f(a)$,

[TABLE]

Which is a contradiction, because (4.5) has the solution $x=a$ and $y=b$. Therefore

[TABLE]

This implies that $a\in\{\displaystyle b,b+1,b+2,\cdots,b+q_{i}-1,b+q_{i}\}$ or $b>a$. Suppose that for (4.5), there are infinitely many solutions $x=a$ and $y=b$ in positive integers such that $a=b+c$, where $c$ is a positive integer such that $0\leq c\leq q_{i}$. Then $a-c+q_{j}$ divides $f(a)$, for $\displaystyle j=1,2,\cdots,m$ and infinitely many positive integers $a$. This implies that all roots of $f(x)$ are integers. Which is a contradiction to our assumption. Therefore there are infinitely many solutions $x=a$ and $y=b$ in positive integers for (4.5) such that

[TABLE]

This implies that there are infinitely many complete composites of $f(x)$. By Lemma (4.3), $f(x)$ has only integral roots. Which is a contradiction to our assumption. This proves the lemma. ∎

Lemma 4.6**.**

Let $f(x)$ be a monic polynomial of degree $m$ with integral coefficients, all the roots of $f(x)$ be not integers and let $\displaystyle q_{1},q_{2},\cdots,q_{m}$ be integers. Then the equation (1.1),

[TABLE]

has finitely many solutions $x$ and $y$ in positive integers.

Proof.

Let $f(x)$ be a monic polynomial of degree $m$ with integral coefficients, all roots of $f(x)$ be not integers and let $\displaystyle q_{1},q_{2},\cdots,q_{m}$ be integers. Let

[TABLE]

where $h$ is a positive integer such that the coefficients of $g(x)$ are positive integers. Let

[TABLE]

Clearly $k+1+q_{i}$ is a positive integer for all $\displaystyle i=1,2,\cdots,m$. Let $p_{i}=k+1+qi$. Since all roots of $f(x)$ are not integers, by Lemma 4.5, the equation

[TABLE]

has finitely many solutions in positive integers. This proves the theorem. ∎

Lemma 4.7**.**

Let $f(x)$ be a monic polynomial of degree $m$ with integral coefficients, all the roots of $f(x)$ be not integers and let $\displaystyle q_{1},q_{2},\cdots,q_{m}$ be integers. Then the equation (1.1),

[TABLE]

has finitely many solutions $x$ and $y$ in negative integers.

Proof.

Let $f(x)$ be a monic polynomial of degree $m$ with integral coefficients, all the roots of $f(x)$ be not integers and let $\displaystyle q_{1},q_{2},\cdots,q_{m}$ be integers. Suppose that there are infinitely many solutions $x=-a$ and $y=-b$ for (1.1), where $a$ and $b$ are positive integers. Therefore there are infinitely many solutions $x=a$ and $y=b$ in positive integers for the equation

[TABLE]

If $m$ is even, then (4.6) implies

[TABLE]

where $f(-x)$ is a monic polynomial with integral coefficients. Therefore (4.7) has finitely many solutions $a$ and $b$ in positive integers, by Lemma 4.6.This gives a contradiction to that (4.6) has infinitely many solutions $a$ and $b$. If $m$ is odd, then (4.6) implies

[TABLE]

where $-f(-x)$ is a monic polynomial with integral coefficients. Therefore the equation (4.8) has finitely many solutions in positive integers, by Lemma 4.6. This gives a contradiction to that (4.6) has infinitely many solutions $a$ and $b$. Therefore (1.1) has finitely many solutions $x$ and $y$ in negative integers. ∎

Proof of Theorem 1.2: Let $f(x)$ be a monic polynomial of degree $m$ with integral coefficients, all roots of $f(x)$ be not integers and let $\displaystyle q_{1},q_{2},\cdots,q_{m}$ be integers. Suppose that there are infinitely many solutions $x=-a$ and $y=b$ for the equation (1.1) such that $a$ and $b$ are positive integers. Therefore there are infinitely many solutions $x=a$ and $y=b$ in positive integers for the equation

[TABLE]

from (1.1). If $m$ is even, by Lemma 4.6, the equation (4.9) has finitely many solutions in positive integers. Which is a contradiction to our assumption. If $m$ is odd, then $f(-k)$ is negative for all positive integers $k>l$, where $l$ is some positive integer. Which is a contradiction to $b+q_{i}$ is positive for all $\displaystyle i=1,2,\cdots,m$ and for infinitely many positive integers $b$. Suppose that there are infinitely many solutions $x=a$ and $y=-b$ for the equation (1.1) such that $a$ and $b$ are positive integers. Therefore

[TABLE]

If $m$ is even, by Lemma 4.6, there are finitely many solutions in positive integers. Which is a contradiction to our assumption. If $m$ is odd, then we get a contradiction, because left hand side of (4.10) is negative for infinitely many positive integers $b$, but right hand side of (4.10) is positive for all positive integers $a$. From the above two cases, we get that the equation (1.1) has infinitely many solutions $x=a$ and $y=b$ in positive integers or the equation (1.1) has infinitely many solutions $x=-a$ and $y=-b$, where $a$ and $b$ are positive integers. This contradicts the two Lemmas (4.6) and (4.7). This proves the theorem. $\square$

5. Diophantine Applications of $ABC$ Conjecture

Proof of Theorem 1.4: Suppose that there are infinitely many solutions $(x,y)\in\mathbb{Z}\times\mathbb{Z}$ for the equation $(\ref{e1})$. First we assume that the equation $(\ref{e1})$ has infinitely many solutions $(a,b)$ in positive integers. Then

[TABLE]

for infinitely many $(a,b)\in\mathbb{N}\times\mathbb{N}$. Therefore

[TABLE]

for infinitely many $(a,b)\in\mathbb{N}\times\mathbb{N}$. Now we shall prove that for infinitely many $(a,b)\in\mathbb{N}\times\mathbb{N}$,

[TABLE]

Suppose for infinitely many $(a,b)$,

[TABLE]

then by division algorithm, we will have a non constant common factor of the polynomials $f_{11}(x)\cdots f_{1l}(x)$ and $f_{21}(x)\cdots f_{2m}(x)$ which is a contradiction to the hypothesis (3) of the theorem. This establishes equation (5.12). Since $f_{11}(x)\cdots f_{1l}(x)$, $f_{21}(x)\cdots f_{2m}(x)$ and $g_{1}(y)\cdots g_{n}(y)$ have positive leading coefficients, we get positive integers $N_{1}$ and $N_{2}$ such that for all integers $x\geq N_{1}$ and $y\geq N_{2}$, $f_{11}(x)\cdots f_{1l}(x)$, $f_{21}(x)\cdots f_{2m}(x)$ and $g_{1}(y)\cdots g_{n}(y)$ are positive integers. Let $\epsilon>0$ be a real number such that

[TABLE]

Case i. Assume that

[TABLE]

for infinitely many $(a,b)\in\mathbb{N}\times\mathbb{N}$ with

1. (1)

$a\geq N_{1}$ and $b\geq N_{2}$ 2. (2)

$f_{11}(a)^{\alpha_{1}}f_{12}(a)^{\alpha_{2}}\cdots f_{1l}(a)^{\alpha_{l}},f_{21}(a)^{\beta_{1}}f_{22}(a)^{\beta_{2}}\cdots f_{2m}(a)^{\beta_{m}},g(b)\in\mathbb{N}$ 3. (3)

$f_{11}(a)^{\alpha_{1}}f_{12}(a)^{\alpha_{2}}\cdots f_{1l}(a)^{\alpha_{l}},f_{21}(a)^{\beta_{1}}f_{22}(a)^{\beta_{2}}\cdots f_{2m}(a)^{\beta_{m}},g(b)$ are coprimes.

So by $ABC$ Conjecture, for a fixed $\epsilon>0$, there exists a constant $c_{1}$ such that

[TABLE]

[TABLE]

So

[TABLE]

Therefore

[TABLE]

Case ii. Assume that

[TABLE]

for infinitely many $(a,b)\in\mathbb{N}\times\mathbb{N}$ with

1. (1)

$a\geq N_{1}$ and $b\geq N_{2}$ 2. (2)

$f_{11}(a)^{\alpha_{1}}f_{12}(a)^{\alpha_{2}}\cdots f_{1l}(a)^{\alpha_{l}},f_{21}(a)^{\beta_{1}}f_{22}(a)^{\beta_{2}}\cdots f_{2m}(a)^{\beta_{m}},g(b)\in\mathbb{N}$ 3. (3)

$f_{11}(a)^{\alpha_{1}}f_{12}(a)^{\alpha_{2}}\cdots f_{1l}(a)^{\alpha_{l}},f_{21}(a)^{\beta_{1}}f_{22}(a)^{\beta_{2}}\cdots f_{2m}(a)^{\beta_{m}},g(b)$ are coprimes.

So by $ABC$ Conjecture, for a fixed $\epsilon>0$, there exists a constant $c_{2}$ such that

[TABLE]

[TABLE]

Also as by equation (5.13), $g(b)\leq f_{11}(a)^{\alpha_{1}}\cdots f_{1l}(a)^{\alpha_{l}}$, we have

[TABLE]

[TABLE]

So

[TABLE]

Therefore

[TABLE]

So from Case i and Case ii, we get

[TABLE]

where $c=max\left\{c_{1},c_{2}\right\}$. If $b<a$ for infinitely many $(a,b)\in\mathbb{N}\times\mathbb{N}$, then $f(b)<f(a)$, since leading coefficients of $f(x)$ is positive. Therefore

[TABLE]

so

[TABLE]

This implies that $\deg g(x)>\deg f(x)$. This contradicts to $(i)$. So $b\geq a$ for infinitely many $(a,b)\in\mathbb{N}\times\mathbb{N}$. Therefore from (5), we can write

[TABLE]

for infinitely many positive integers $b$, since $f_{11}(a)f_{12}(a)\cdots f_{1l}(a)$ and

$f_{21}(a)f_{22}(a)\cdots f_{2m}(a)$ have positive leading coefficients. So

[TABLE]

Also as for $b\geq M$ for some positive integer $M$,

[TABLE]

we have

[TABLE]

Since $\epsilon\deg\left(f_{11}(x)f_{12}(x)\cdots f_{1l}(x)f_{21}(x)f_{22}(x)\cdots f_{2m}(x)g_{1}(x)\cdots g_{n}(x)\right)<1$, $\deg g(y)<1+\deg(f_{11}(x)f_{12}(x)\cdots f_{1l}(x)f_{21}(x)\cdots f_{2m}(x)g_{1}(x)\cdots g_{n}(x))$. This is a contradiction to (ii). So our assumption is wrong. There are only finitely many positive integral solutions $(x,y)$ for (1.1). Second we shall prove that there are only finitely many integral solutions $(x,y)=(a,b)$ for (1.1). Let $(a,b)$ be any integral solutions for (1.1). Then $(a,b)$ can be written in the form $(\pm a_{1},\pm b_{1})$, where $a_{1},b_{1}\in\mathbb{N}$. Now we replace $(x,y)$ by $(\pm x_{1},\pm y_{1})$ in (1.1). Then we get that

[TABLE]

That is,

[TABLE]

We can reduce this equation as either

[TABLE]

or

[TABLE]

It is clear that the above two equations satisfy the conditions (1), (2), (3) and (4). So the above two equations have only finitely many integral solutions. This proves the theorem.$\hfill\square$ Proof of Theorem 1.5: Suppose that there are infinitely many solutions $(x,y)\in\mathbb{Z}\times\mathbb{Z}$ for equation $(\ref{e1})$. First we assume that the equation $(\ref{e1})$ has infinitely many solutions $(a,b)$ in positive integers. Then

[TABLE]

for infinitely many $(a,b)\in\mathbb{N}\times\mathbb{N}$. Therefore

[TABLE]

for infinitely many $(a,b)\in\mathbb{N}\times\mathbb{N}$. Now we shall prove that for infinitely many $(a,b)\in\mathbb{N}\times\mathbb{N}$,

[TABLE]

Suppose for infinitely many $(a,b)$,

[TABLE]

then by division algorithm, we will have a non constant common factor of the polynomials $f_{11}(x)\cdots f_{1l}(x)$ and $f_{21}(x)\cdots f_{2m}(x)$ which is a contradiction to the hypothesis $(3)$ of the theorem. This establishes equation $(\ref{subbugcd})$. Since $f_{11}(x)\cdots f_{1l}(x)$, $f_{21}(x)\cdots f_{2m}(x)$ and $g_{1}(y)\cdots g_{n}(y)$ have positive leading coefficients, we get positive integers $N_{1}$ and $N_{2}$ such that for all integers $x\geq N_{1}$ and $y\geq N_{2}$, $f_{11}(x)\cdots f_{1l}(x)$, $f_{21}(x)\cdots f_{2m}(x)$ and $g_{1}(y)\cdots g_{n}(y)$ are positive integers. Let $\epsilon>0$ be a real number such that

[TABLE]

Case i. Assume that

[TABLE]

for infinitely many $(a,b)\in\mathbb{N}\times\mathbb{N}$ with

1. (1)

$a\geq N_{1}$ and $b\geq N_{2}$ 2. (2)

$f_{11}(a)^{\alpha_{1}}\cdots f_{1l}(a)^{\alpha_{l}},f_{21}(a)^{\beta_{1}}\cdots f_{2m}(a)^{\beta_{m}},g(b)\in\mathbb{N}$ 3. (3)

$f_{11}(a)^{\alpha_{1}}\cdots f_{1l}(a)^{\alpha_{l}},f_{21}(a)^{\beta_{1}}\cdots f_{2m}(a)^{\beta_{m}},g(b)$ are coprimes.

So by $ABC$ Conjecture, for any fixed $\epsilon>0$, there exists a constant $c_{1}$ such that

[TABLE]

So

[TABLE]

Therefore

[TABLE]

Case ii. Assume that

[TABLE]

for infinitely many $(a,b)\in\mathbb{N}\times\mathbb{N}$ with

1. (1)

$a\geq N_{1}$ and $b\geq N_{2}$ 2. (2)

$f_{11}(a)^{\alpha_{1}}\cdots f_{1l}(a)^{\alpha_{l}},f_{21}(a)^{\beta_{1}}\cdots f_{2m}(a)^{\beta_{m}},g(b)\in\mathbb{N}$ 3. (3)

$f_{11}(a)^{\alpha_{1}}\cdots f_{1l}(a)^{\alpha_{l}},f_{21}(a)^{\beta_{1}}\cdots f_{2m}(a)^{\beta_{m}},g(b)$ are coprimes.

So by $ABC$ Conjecture, for any fixed $\epsilon>0$, there exists a constant $c_{2}$ such that

[TABLE]

[TABLE]

Also as by equation (5.16) $g(b)\leq f_{11}(a)^{\alpha_{1}}\cdots f_{1l}(a)^{\alpha_{l}}$, we have

[TABLE]

So

[TABLE]

Therefore

[TABLE]

So from Case i and Case ii, we get

[TABLE]

where $c=\max\left\{c_{1},c_{2}\right\}$.

[TABLE]

since

[TABLE]

If $b>a$ for infinitely many $(a,b)\in\mathbb{N}\times\mathbb{N}$, then $f(b)>f(a)$, since leading coefficients of $f(x)$ is positive. Therefore

[TABLE]

So

[TABLE]

This implies that $\deg g(x)<\deg f(x)$. This contradicts to $(i)$. So $b\leq a$ for infinitely many $(a,b)\in\mathbb{N}\times\mathbb{N}$. Also since $f_{11}(x)\cdots f_{1l}(x)$ and $f_{21}(x)\cdots f_{2m}(x)$ have positive leading coefficients, by equation (5.17), we get

[TABLE]

Thus as for $a\geq M$ for some positive integer $M$,

[TABLE]

we have

[TABLE]

Since $\epsilon\deg\left(f_{11}(x)\cdots f_{1l}(x)f_{21}(x)\cdots f_{2m}(x)g_{1}(x)\cdots g_{n}(x)\right)<1$, $\deg f(x)<1+\deg(f_{11}(x)\cdots f_{1l}(x)f_{21}(x)\cdots f_{2m}(x)g_{1}(x)\cdots g_{n}(x))$. This is a contradiction to (ii). So there can be only finitely many positive integral solutions $(x,y)$ for (1.1). Second we shall prove that there are only finitely many integral solutions $(x,y)=(a,b)$ for (1.1). Let $(a,b)$ be any integral solutions for (1.1). Then $(a,b)$ can be written as the form $(\pm a_{1},\pm b_{1})$, where $a_{1},b_{1}\in\mathbb{N}$. Now we replace (x, y) by $(\pm x_{1},\pm y_{1})$ in (1.1). Then we get that

[TABLE]

That is,

[TABLE]

We can reduce this equation as either

[TABLE]

or

[TABLE]

It is clear that the above two equations satisfy the conditions (1), (2), (3) and (4). So the above two equations have only finitely many integral solutions. This proves theorem.$\hfill\square$

Proof of Theorem 1.6: Suppose there are infinitely many solutions $(x,y)=(a,b)$ in integers. Then

[TABLE]

for infinitely many $(a,b)\in\mathbb{Z}\times\mathbb{Z}$. That is for infinitely many integers $a,b\in\mathbb{Z}$, we have

[TABLE]

Since $\gamma_{1}$, …, $\gamma_{n}$ are positive integers $\geq 2$. So there exists a positive integer $N$ such that for every integer $b>N$, $g_{1}(b)^{\gamma_{1}}g_{2}(b)^{\gamma_{2}}\cdots g_{n}(b)^{\gamma_{n}}$ is a powerful number, while $f(x)$ being separable of $\deg>2$, by Remark 2.2, $f(x)$ can be powerful numbers only for finitely many integers $x$. Which is a contradiction. So our assumption is wrong. $\hfill\square$ Acknowledgments: The second author would like to thank the Harish Chandra Research Institute, Allahabad for financial support during his research study.