On the Ternary Exponential Diophantine Equation Equating a Perfect Power and Sum of Products of Consecutive Integers

Abstract

Consider the Diophantine equation $y^n=x+x(x+1)+\cdots +x(x+1)\cdots (x+k)$, where x, y, n, and k are integers. In 2016, a research article, entitled – ’power values of sums of products of consecutive integers’, primarily proved the inequality $n=$ 19,736 to obtain all solutions $(x,y,n)$ of the equation for the fixed positive integers $k\le 10$. In this paper, we improve the bound as $n\le$ 10,000 for the same case $k\le 10$, and for any fixed general positive integer k, we give an upper bound depending only on k for n.

1. Introduction

In 1976, Tijdeman proved that all integral solutions $(x,y,n)$, $n>0$ and $\left|y\right|>1$, of the equation

$$y^n=f\left(x\right)$$

satisfy $n<c_0$, where $c_0$ is an effectively computable constant depending only on f if $f\left(x\right)$ is an integer polynomial with at least two distinct roots (Shorey-Tijdeman [1], Tijdeman [2], Waldschmidt [3]). In 1987, Brindza in [4] obtained the unconditional form of the result for $f\left(x\right)=f_1^{k_1}\left(x\right)+f_2^{k_2}\left(x\right)+\cdots +f_s^{k_s}\left(x\right)$, where $f_1,f_2,\dots ,f_s$ are integer polynomials and $k_1,k_2,\dots ,k_s$ are positive integers such that $min\{k_i:1\le i\le s\}>s(s-1)$. In 2016, Hajdu, Laishram, and Tengely in [5] proved the above result for $f\left(x\right)=x+x(x+1)+\cdots +x(x+1)\cdots (x+k)$. In 2018, Subburam [6] assured that, for each positive, real $ϵ<1$, there exists an effectively computable constant $c\left(ϵ\right)$ such that

$$max\{x,y,n\}\le c\left(ϵ\right){(\log max\{a,b,c\})}^{2+ϵ},$$

where $(x,y,n)$ is a positive integral solution of the ternary exponential Diophantine equation

$$a^n=b^x+c^y$$

and $a,b,$ and c are fixed positive integers with $gcd(a,b,c)=1$. In 2019, Subburam [7] provided the unconditional form of the first result for $f\left(x\right)={(x+a_1)}^{r_1}+{(x+a_2)}^{r_2}+\cdots +{(x+a_m)}^{r_m}$, where $m\ge 2$; $a_1$, $a_2$, …, $a_m$; $r=r_1,r_2,\dots ,r_m$ are integers such that $r_1\ge r_2\ge \cdots \ge r_m>0$; $gcd(\eta ,\mathrm{cont}(f\left(x\right)\left)\right)=1$; ${\eta }^{1/r}$ is not an integer $>1$; $r_2<r_1-1$ when $r_2<r_1$; $\eta =\left|\right\{r_i:r_1=r_i\left\}\right|$; and $\mathrm{cont}\left(f\right(x\left)\right)$ is the content of $f\left(x\right)$. For further results related to this paper, see Bazsó [8]; Bazsó, Berczes, Hajdu, and Luca [9]; and Tengely and Ulas [10].

In this paper, we consider the Diophantine equation

$$y^n=x+x(x+1)+\cdots +x(x+1)\cdots (x+k)=:f_k\left(x\right)$$

(1)

in integral variables x, y, and n, with $n>0$, where k is a fixed positive integer. In Theorem 2.1 of [5], Hajdu, Laishram, and Tengely proved that there exists an effectively computable constant $c\left(k\right)$ depending only on k such that $(x,y,n)$ satisfy

$$n\le c\left(k\right)$$

if $y\ne 0,-1$. For the case $1\le k\le 10$, they explicitly calculated $c\left(k\right)$ as

$$n\le 19,736.$$

Here, we prove the following theorem. For any positive integers s, $p_1$, $p_2$, …, $p_m$, we denote

$${\displaystyle {\lambda }_s(p_1,\dots ,p_m)=\sum _{\begin{array}{c}{i}_1,\dots ,i_s\\ 1\le i_1<\cdots <i_s\le m\end{array}}{p}_{i_1}{p}_{i_2}\cdots p_{i_s}}$$

and ${\lambda }_0(p_1,\dots ,p_m)=1$. This elementary symmetric polynomial and its upper bound have been studied in Subburam [11].

Theorem 1.

Let k be any positive integer and

$$\mathfrak{b}=4\left|{\displaystyle \sum _{i=0}^{k-1}{(-1)}^i{A}_{k-i-1}{2}^i}\right|,$$

where $A_0=1$, $A_1=1+{\alpha }_1$, $A_{k-1}=1+{\alpha }_{k-2}$, and $A_j={\alpha }_{j-1}+{\alpha }_j$ for $j=2,3,\dots ,k-2$ and where

$${\displaystyle {\alpha }_m=1+\sum _{i=0}^{m-1}{\lambda }_{i+1}(3,\dots ,k+i-m+1)}$$

for $m=1,2,\dots ,k-2$. Then, all integral solutions $(x,y,n)$, with $y\ne 0,-1,x\ne 1,n\ge 1$, of $\left(1\right)$ satisfy

$$n\le c_2\log \mathfrak{b},$$

where $c_2$ can be bounded using the linear form of the logarithmic method in Laurent, Mignotte, and Nesterenko [12], and an immediate estimation is

$$c_2=\left\{\begin{array}{cc}21,468\hfill & if21>\log n\hfill \\ 26,561{\left(\log \log \mathfrak{b}\right)}^2\hfill & if21\le \log n.\hfill \end{array}\right.$$

If

$$\mathfrak{b}\le 4\times 9\times 11\times 467\times 2,018,957,$$

then all integral solutions $(x,y,n)$, with $y\ne 0,-1,x\ne 1,n\ge 1$, of $\left(1\right)$ satisfy

$$n\le \left\{\begin{array}{cc}max\{1000,824.338\log \mathfrak{b}+0.258\}\hfill & if\mathfrak{b}\le 100\hfill \\ max\{2000,769.218\log \mathfrak{b}+0.258\}\hfill & if100<\mathfrak{b}\le 10,000\hfill \\ max\{10,000,740.683\log \mathfrak{b}+0.234\}\hfill & if\mathfrak{b}>10,000.\hfill \end{array}\right.$$

The result of Hajdu, Laishram, and Tengely in [5] is much stronger than the following corollary. They explicitly obtained all solutions for the values $k\le 10$ using the MAGMA computer program along with two well-known methods (See Subburam [6], Srikanth and Subburam [13], and Subburam and Togbe [14]), after proving that $n\le 19,736$ for $1\le k\le 10$. Here, we have

Corollary 1.

If $1\le k\le 10$, then $n\le$ 10,000.

Hajdu, Laishram, and Tengely studied each of the cases “$(n,k)$ where $n=2$ and k is odd with $1\le k\le 10$” in the proof of Theorem $2.2$ of [5]. Here, we prove the following theorem for any odd k. This can be written as a suitable computer program by considering each step of the following theorem as a sub-program that can be separately and directly run.

Theorem 2.

Let k be odd. Then, we have the following:

(i) 

There uniquely exist rational polynomials $B\left(x\right)$ and $C\left(x\right)$ with $deg\left(C\left(x\right)\right)\le \frac{k-1}{2}$ such that

$$f_k\left(x\right)=B^2\left(x\right)+C\left(x\right).$$

(ii) 

Let l be the least positive integer such that $lB\left(x\right)$ and $l^{2}C\left(x\right)$ have integer coefficients for any nonnegative integer i and $\delta \in \{1,-1\}$

$$P_{i,\delta }\left(x\right)=\delta {(lB\left(x\right)+\delta i)}^2-\delta {\left(lB\left(x\right)\right)}^2-\delta l^{2}C\left(x\right),$$

r is any positive integer,

$$H_1=\{\alpha \in \mathbb{Z}:P_{i,\delta }\left(\alpha \right)=0,\delta \in \{1,-1\},i=0,1,2,\dots ,r-1\},$$

and

$$H_2=\{\alpha \in \mathbb{R}:P_{r,1}\left(\alpha \right)=0\mathrm{or}{P}_{r,-1}\left(\alpha \right)=0\},$$

where $\mathbb{R}$ and $\mathbb{Z}$ are the sets of all real numbers and integers, respectively. If $H_1$ and $H_2$ are empty, then $\left(1\right)$ has no integral solution $(x,y,2)$. Otherwise, all integral solutions $(x,y,2)$ of $\left(1\right)$ satisfy $x\in H_1$ or

$$min{H}_2\le x\le max{H}_2.$$

2. Proofs

Lemma 1.

Let $k\ge 3$. Then, all integral solutions $(x,y,n)$, $n>0$ and $y\ne 0$, of $\left(1\right)$ satisfy the equation

$$a_2{b}_1{y}_2^n-b_2{a}_1{y}_1^n=2b_1{a}_1,$$

where $a_1,a_2,b_1$, and $b_2$ are positive integers such that

$${\displaystyle a_1{a}_2{b}_1{b}_2\mid 4\sum _{i=0}^{k-1}{(-1)}^i{A}_{k-i-1}{2}^i,}$$

$A_i$ is the coefficient of $x^{k-i-1}$ in the polynomial $f_k\left(x\right)/x(x+2)$,

$$x=\left(\frac{b_2}{b_1}\right)y_1^n,\mathrm{and}x+2=\left(\frac{a_2}{a_1}\right)y_2^n$$

for some nonzero integers $y_1$ and $y_2$.

Proof. 

Let $k\ge 3$. Let $(x,y,n)$, with $n>0$ and $y\ne 0$, be any integral solution of the Diophantine equation

$$y^n=x+x(x+1)+\cdots +x(x+1)\cdots (x+k).$$

This can be written as

$$y^n=x(x+2)g_k\left(x\right)$$

for some integer polynomial $g_k\left(x\right)$, which is not divided by x and $x+2$, since $k\ge 3$. Let d and q be positive integers such that

$$\gcd (x,(x+2)g_k\left(x\right))=d\mathrm{and}\gcd ((x+2),x{g}_k\left(x\right))=q.$$

Let $d_1$, $d_2$, $q_1$, and $q_2$ be positive integers such that $d_1{d}_2=d$$gcd(d_1,d_2)=1$, $gcd(d_2^2,(x/d))=gcd(d_1^2,((x+2)g_k\left(x\right)/d))=1$, $q_1{q}_2=q$, and $gcd(q_1,q_2)=1=gcd(q_2^2,((x+2)/q))=gcd(q_1^2,(x{g}_k\left(x\right)/q))=1.$ Then,

$$\left(\frac{d_1^2}{d}\right)x=y_1^{n}and\left(\frac{q_1^2}{q}\right)(x+2)=y_2^n$$

for some nonzero integers $y_1$ and $y_2$, since $y\ne 0$ and $n\ge 1$. From this, we have

$$q{d}_1^2{y}_2^n-d{q}_1^2{y}_1^n=2q_1^2{d}_1^2\mathrm{and}\mathrm{so}{q}_2{d}_1{y}_2^n-d_2{q}_1{y}_1^n=2q_1{d}_1.$$

Let

$$g_k\left(x\right)=f_k\left(x\right)/\left(x(x+2)\right)=x^{k-1}+A_1{x}^{k-2}+\cdots +A_{k-1}$$

and

$$g\left(x\right)=x^2+2x.$$

Then, for each integer l with $0\le l\le k-1$,

$$h_l\left(x\right)=\left({\displaystyle \sum _{i=0}^l{(-1)}^i{A}_{l-i}{2}^i}\right)x^{k-l-1}+A_{l+1}{x}^{k-l-2}+\cdots +A_{k-1}.$$

In particular,

$${\displaystyle h_{k-1}\left(x\right)=\sum _{i=0}^{k-1}{(-1)}^i{A}_{k-i-1}{2}^i.}$$

This implies that

$${\displaystyle gcd(g\left(x\right),g_k\left(x\right))\mid \sum _{i=0}^{k-1}{(-1)}^i{A}_{k-i-1}{2}^i,}$$

where $A_i$ is the coefficient of $x^{k-i-1}$ in the polynomial $g_k\left(x\right)$.

If x is odd, then $d\mid x$, $d\mid g_k\left(x\right)$, $q\mid (x+2)$, $q\mid g_k\left(x\right)$ and so $dq\mid gcd(g\left(x\right),g_k\left(x\right))$. Suppose that x is even. Then,

$$\frac{dq}{4}\mid \frac{x(x+2)}{4}\mathrm{and}\frac{dq}{4}\mid g_k\left(x\right).$$

Hence, we have

$${\displaystyle dq\mid 4gcd(g\left(x\right),g_k\left(x\right))\mathrm{and}\mathrm{so}dq\mid 4\sum _{i=0}^{k-1}{(-1)}^i{A}_{k-i-1}{2}^i.}$$

This proves the lemma. □

Lemma 2

(Hajdu, Laishram, and Tengely [5]). Let a, b, and c be positive integers with $a<b\le 4\times 2,018,957\times 99\times 467$ and $c\le 2ab$. Then, the Diophantine equation

$$a{u}^n-b{v}^n=\pm c,$$

in integral variables $u>v>1$, implies

$$n\le \left\{\begin{array}{cc}max\{1000,824.338\log b+0.258\}\hfill & ifb\le 100\hfill \\ max\{2000,769.218\log b+0.258\}\hfill & if100<b\le 10,000\hfill \\ max\{10,000,740.683\log b+0.234\}\hfill & ifb>10,000\hfill \end{array}\right.$$

Lemma 3

(Szalay [15]). Suppose that $p\ge 2$ and $r\ge 1$ are integers and that

$$F\left(x\right)=x^{rp}+a_{rp-1}{x}^{rp-1}+\cdots +a_0$$

is a polynomial with integer coefficients. Then, rational polynomials

$$B\left(x\right)=x^r+b_{r-1}{x}^{r-1}+\cdots +b_0$$

and $C\left(x\right)$ with $deg\left(C\right(x\left)\right)\le rp-r-1$ uniquely exist for which

$$F\left(x\right)=B^p\left(x\right)+C\left(x\right).$$

Lemma 4

(Srikanth and Subburam [13]). Let p be a prime number, $B\left(x\right)$ and $C\left(x\right)$ be nonzero rational polynomials with $deg\left(C\right(x\left)\right)<(p-1)deg\left(B\right(x\left)\right)$, l be a positive integer such that $lB\left(x\right)$ and $l^{p}C\left(x\right)$ have integer coefficients for any nonnegative integer i and $\delta \in \{1,-1\}$:

$$P_{i,\delta }\left(x\right)=\delta {(lB\left(x\right)+\delta i)}^p-\delta {\left(lB\left(x\right)\right)}^p-\delta l^{p}C\left(x\right),$$

r be any positive integer,

$$H_1=\{\alpha \in \mathbb{Z}:P_{i,\delta }\left(\alpha \right)=0,\delta \in \{1,-1\},i=0,1,2,\dots ,r-1\},$$

and

$$H_2=\{\alpha \in \mathbb{R}:P_{r,1}\left(\alpha \right)=0\mathrm{or}{P}_{r,-1}\left(\alpha \right)=0\}.$$

If $H_1$ and $H_2$ are empty, then the Diophantine equation

$$y^p=B{\left(x\right)}^p+C\left(x\right)$$

has no integral solution $(x,y)$. Otherwise, all integral solutions $(x,y)$ of the equation satisfy $x\in H_1$ or

$$min{H}_2\le x\le max{H}_2.$$

In some other new way as per Note 2, using Laurent’s result leads to a better result. For our present purpose, the following lemma is enough.

Lemma 5

(Laurent, Mignotte, and Nesterenko [12]). Let l, m, ${\alpha }_1$, ${\alpha }_2$, ${\beta }_1$, and ${\beta }_2$ be positive integers such that $l\log ({\alpha }_1/{\alpha }_2)-m\log ({\beta }_1/{\beta }_2)\ne 0$. Let

$$\mathsf{\Gamma }=\left|{\left(\frac{{\alpha }_1}{{\alpha }_2}\right)}^l{\left(\frac{{\beta }_1}{{\beta }_2}\right)}^m-1\right|.$$

Then, we have

$$\left|\mathsf{\Gamma }\right|>0.5exp\left\{-24.34\log \alpha \log \beta {\left(max\left\{\gamma +0.14,21\right\}\right)}^2\right\},$$

where $\alpha =max\{3,{\alpha }_1,{\alpha }_2\}$, $\beta =max\{3,{\beta }_1,{\beta }_2\}$ and $\gamma =\log \left(\frac{l}{\log \beta }+\frac{m}{\log \alpha }\right)$.

Proof of Theorem 1. 

Assume that $k\ge 3$. Then, by Lemma 1, all integral solutions $(x,y,n)$, $y\ne 0,-1$ and $n\ge 1$, of $\left(1\right)$ satisfy the equation

$$a{y}_2^n-b{y}_1^n=c,$$

(2)

where $y_1$ and $y_2$ are nonzero integers, a and b are positive integers such that $c\le 2ab$,

$${\displaystyle ab\mid 4\sum _{i=0}^{k-1}{(-1)}^i{A}_{k-i-1}{2}^i,}$$

and $A_i$ is the coefficient of $x^{k-i-1}$ in the polynomial $f_k\left(x\right)/x(x+2)$. Without loss of generality, we can take $y_1>y_2$ to prove the result. From (2), we write

$$\left|1-\left(\frac{a}{b}\right){\left(\frac{y_2}{y_1}\right)}^n\right|=\frac{c}{b{y}_1^n}.$$

Next, take ${\alpha }_1=a$, ${\alpha }_2=b$, ${\beta }_1=y_2$, ${\beta }_2=y_1$, $l=1$, and $m=n$ in Lemma 5. Then, by the lemma, we obtain

$$\frac{c}{b{y}_1^n}\ge exp\{-24.3414(\log max\{3,a,b\})(\log max\{3,y_1\})max{\{21,\left(\log n\right)\}}^2.$$

From this, we obtain the required bound. Next, assume that $1\le k\le 2$. Then, we can write Equation (1) as

$$y_1^n=c_{1}x$$

and

$$y_2^2=c_2{(x+2)}^i,$$

where

$$c_1,c_2\in \left\{1/4,1/2,1,2,4\right\}$$

and $i\in \{1,2\}$. In the same way, we can obtain the required bound. To find the exact values of $A_0$, $A_1$, …, $A_{k-1}$, equate the coefficients of the polynomials

$$g_k\left(x\right)=1+(x+1)\left(1+(x+3)+\cdots +(x+3)(x+4)\cdots (x+k)\right).$$

and

$$g_k\left(x\right)=x^{k-1}+A_1{x}^{k-2}+\cdots +A_{k-1}.$$

Then, we obtain $A_0=1$, $A_1=1+{\alpha }_1$, $A_{k-1}=1+{\alpha }_{k-2}$, and $A_j={\alpha }_{j-1}+{\alpha }_j$ for $j=2,3,\dots ,k-2$ and

$${\displaystyle {\alpha }_m=1+\sum _{i=0}^{m-1}{\lambda }_{i+1}(3,\dots ,k+i-m+1)}$$

for $m=1,2,\dots ,k-2$. □

Next, we consider the case that

$$\mathfrak{b}\le 4\times 9\times 11\times 467\times 2,018,957.$$

If $y_1=1$, $y_2=1$, or $y_1=y_2$, then we have

$$x=\frac{d_2}{d_1}=1,x=\frac{q_2}{q_1}-2=-1,x=\frac{2q_1{d}_2}{d_1{q}_2-q_1{d}_2},$$

where $d_1,d_2,q_1$ and $q_2$ are positive integers such that $d_1{d}_2{q}_1{q}_2=ab$. These three equations give the required upper bound. Hence, Lemma 2 completes the theorem.

Proof of Corollary 1. 

Take $k=10$ in Theorem 1. Then, $A_0=1$, $A_1=54$, $A_2=1258$, $A_3=16,541$, $A_4=134,716$, $A_5=700,776$, $A_6=2,309,303$, $A_7=4,589,458$, $A_8=4,880,507$, $A_9=2,018,957$, and $b/4=46,233$ and so

$$740.683\log \mathfrak{b}\le 8982.9.$$

In a similar way, for the case $k<10$, we have

$$max\{10,000,740.683\log \mathfrak{b}+0.23\}\le 10,000.$$

Hence, Lemma 2 confirms the result. □

Proof of Theorem 2. 

Take $F\left(x\right)=x+x(x+1)+\cdots +x(x+1)\cdots (x+k)$ in Lemma 3. Since k is odd, so $2\mid deg\left(F\right(x\left)\right)$, $p=2$, and $r=\frac{k+1}{2}$. Then, by Lemma 3, there uniquely exist rational polynomials $B\left(x\right)$ and $C\left(x\right)$ with $deg\left(C\left(x\right)\right)\le \frac{k-1}{2}$ such that

$$F\left(x\right)=B^2\left(x\right)+C\left(x\right).$$

Now, by Lemma 4, we have the theorem. □

Note 1. 

First, find the values of the elementary symmetric forms${\lambda }_{i+1}(3,\dots ,k+i-m+1)$for$i=0,\dots ,m-1$and$m=1,2,\dots ,k-2$. Next, obtain${\alpha }_1$, ${\alpha }_2$, …, ${\alpha }_{k-2}$and so$A_0$, $A_1$, …, $A_{k-1}$. Using this, calculate$|A_{k-i-1}-2A_{k-i-2}|$and so

$$2^i|A_{k-i-1}-2A_{k-i-2}|=|A_{k-i-1}{2}^i-A_{k-i-2}{2}^{i+1}|$$

for$i=0,2,4,\dots$. In this way, for any positive integer k, we can find the exact value of$\mathfrak{b}$in Theorem 1. Therefore, it is not so hard to decide for which k is

$$\mathfrak{b}\le 4\times 9\times 11\times 467\times 2,018,957$$

as in Theorem 1. For this work, we can use a suitable computer program.

Note 2. 

The result of Laurent [16] is an improvement on the result of Laurent, Mignotte, and Nesterenko [12]. From the proof, using the result of Laurent [16] and Proposition 4.1 in Hajdu, Laishram, and Tengely [5], we write the following:

Let A, B, and C be positive integers with$C\le 2AB$, $B>A$and$B\le 4\times 9\times 11\times 467\times 2,018,957$. Then, the equation

$$A{u}^n-B{v}^n=\pm C$$

in integer variables$u>v>1,n>3$implies

$$n\le C_m{(max\{m,h_n\})}^2\left(\log B\right)\left(2+\frac{(\tau -1)q_0}{\log u_0}+\frac{1}{\log u_0}\right)+\frac{\log 4}{\log u_0},$$

where

$$h_n=\log \left(\frac{n}{(\tau +1)\log B}+\frac{1}{2\log u+(\tau -1)q_0}\right)+{ϵ}_m,$$

in which$q_0$, $u_0$, $C_m,m,\tau$, and${ϵ}_m$are positive real numbers such that$u\ge u_0$, $\log (u/v)\le q_0$, $C_m>1$, ${ϵ}_m>1$, and$\tau >1$.

If we use the above observation in Lemma 1 of this paper, then we obtain the bound

$$n\le c_2^{\prime }{(\log n-\log \log \mathfrak{b})}^2\log \mathfrak{b}$$

and so an immediate estimation is

$$n\le c_2\log \mathfrak{b},$$

where $c_2$ is as in Theorem 1 and $c_2^{\prime }$ is a positive real number depending on $u_0,q_0,C_m,m,\tau$, and ${ϵ}_m$. Though there are better bounds in the literature than what the linear form of the logarithmic method in Laurent, Mignotte, and Nesterenko [12] gives, it is sufficient to obtain an explicit bound only in terms of k using our method, which simplifies the arguments in Section 5 of [5] as well.

3. Conclusions

This article implied a method to obtain an upper bound for all n where $(x,y,n)$ is an integral solution of $\left(1\right)$ and to improve the method and algorithm of [4]. The same method can be applied to study the general Diophantine equation (see [8,9,10]),

$$y^n=a_{0}x+a_{1}x(x+1)+\cdots +a_{k}x(x+1)\cdots (x+k),$$

where $k,a_0,a_1,\cdots ,a_k$ are fixed integers and $x,y,n$ are integral variables in obtaining a better upper bound (depending only on $k,a_0,a_1,\cdots ,a_k$) for all $\max \{x,y,n\}$, where $(x,y,n)$ is an integral solution of the general equation.