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Article

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

1
Department of Mathematics, Alagappa University, Karaikudi 630004, India
2
Department of Computer and Information Security, Sejong University, Seoul 05006, Korea
3
Ramanujan Centre for Higher Mathematics, Alagappa University, Karaikudi 630003, India
4
Department of Mathematics, School of Advanced Sciences, Kalasalingam Academy of Research and Education, Krishnankoil, Srivilliputhur 626128, India
5
Department of Automotive ICT Convergence Engineering, Daegu Catholic University, Gyeongsan 38430, Korea
6
Department of Computer Science and Engineering, Sejong University, Seoul 05006, Korea
*
Authors to whom correspondence should be addressed.
These authors contributed equally to this work.
Mathematics 2021, 9(15), 1813; https://doi.org/10.3390/math9151813
Submission received: 30 June 2021 / Revised: 22 July 2021 / Accepted: 27 July 2021 / Published: 30 July 2021
(This article belongs to the Special Issue Decidability of Logics and Their Theories and Combinations)

## Abstract

:
Consider the Diophantine equation $y n = x + x ( x + 1 ) + ⋯ + x ( x + 1 ) ⋯ ( x + k )$, 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 $k ≤ 10$. In this paper, we improve the bound as $n ≤$ 10,000 for the same case $k ≤ 10$, and for any fixed general positive integer k, we give an upper bound depending only on k for n.
MSC:
11D61; 11D45

## 1. Introduction

In 1976, Tijdeman proved that all integral solutions $( x , y , n )$, $n > 0$ and $| y | > 1$, of the equation
$y n = f ( x )$
satisfy $n < c 0$, where $c 0$ is an effectively computable constant depending only on f if $f ( x )$ 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 $f ( x ) = f 1 k 1 ( x ) + f 2 k 2 ( x ) + ⋯ + f s k s ( x )$, where $f 1 , f 2 , … , f s$ are integer polynomials and $k 1 , k 2 , … , k s$ are positive integers such that $min { k i : 1 ≤ i ≤ s } > s ( s − 1 )$. In 2016, Hajdu, Laishram, and Tengely in [5] proved the above result for $f ( x ) = x + x ( x + 1 ) + ⋯ + x ( x + 1 ) ⋯ ( x + k )$. In 2018, Subburam [6] assured that, for each positive, real $ϵ < 1$, there exists an effectively computable constant $c ( ϵ )$ such that
$max { x , y , n } ≤ c ( ϵ ) ( log max { a , b , c } ) 2 + ϵ ,$
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 $f ( x ) = ( x + a 1 ) r 1 + ( x + a 2 ) r 2 + ⋯ + ( x + a m ) r m$, where $m ≥ 2$; $a 1$, $a 2$, …, $a m$; $r = r 1 , r 2 , … , r m$ are integers such that $r 1 ≥ r 2 ≥ ⋯ ≥ r m > 0$; $gcd ( η , cont ( f ( x ) ) ) = 1$; $η 1 / r$ is not an integer $> 1$; $r 2 < r 1 − 1$ when $r 2 < r 1$; $η = | { r i : r 1 = r i } |$; and $cont ( f ( x ) )$ is the content of $f ( x )$. 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
$y n = x + x ( x + 1 ) + ⋯ + x ( x + 1 ) ⋯ ( x + k ) = : f k ( x )$
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 $c ( k )$ depending only on k such that $( x , y , n )$ satisfy
$n ≤ c ( k )$
if $y ≠ 0 , − 1$. For the case $1 ≤ k ≤ 10$, they explicitly calculated $c ( k )$ as
$n ≤ 19 , 736 .$
Here, we prove the following theorem. For any positive integers s, $p 1$, $p 2$, …, $p m$, we denote
$λ s ( p 1 , … , p m ) = ∑ i 1 , … , i s 1 ≤ i 1 < ⋯ < i s ≤ m p i 1 p i 2 ⋯ p i s$
and $λ 0 ( p 1 , … , p m ) = 1$. This elementary symmetric polynomial and its upper bound have been studied in Subburam [11].
Theorem 1.
Let k be any positive integer and
$b = 4 ∑ i = 0 k − 1 ( − 1 ) i A k − i − 1 2 i ,$
where $A 0 = 1$, $A 1 = 1 + α 1$, $A k − 1 = 1 + α k − 2$, and $A j = α j − 1 + α j$ for $j = 2 , 3 , … , k − 2$ and where
$α m = 1 + ∑ i = 0 m − 1 λ i + 1 ( 3 , … , k + i − m + 1 )$
for $m = 1 , 2 , … , k − 2$. Then, all integral solutions $( x , y , n )$, with $y ≠ 0 , − 1 , x ≠ 1 , n ≥ 1$, of $( 1 )$ satisfy
$n ≤ c 2 log b ,$
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
$c 2 = 21 , 468 i f 21 > log n 26 , 561 ( log log b ) 2 i f 21 ≤ log n .$
If
$b ≤ 4 × 9 × 11 × 467 × 2 , 018 , 957 ,$
then all integral solutions $( x , y , n )$, with $y ≠ 0 , − 1 , x ≠ 1 , n ≥ 1$, of $( 1 )$ satisfy
$n ≤ max { 1000 , 824.338 log b + 0.258 } i f b ≤ 100 max { 2000 , 769.218 log b + 0.258 } i f 100 < b ≤ 10 , 000 max { 10 , 000 , 740.683 log b + 0.234 } i f b > 10 , 000 .$
The result of Hajdu, Laishram, and Tengely in [5] is much stronger than the following corollary. They explicitly obtained all solutions for the values $k ≤ 10$ 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 $n ≤ 19 , 736$ for $1 ≤ k ≤ 10$. Here, we have
Corollary 1.
If $1 ≤ k ≤ 10$, then $n ≤$ 10,000.
Hajdu, Laishram, and Tengely studied each of the cases “$( n , k )$ where $n = 2$ and k is odd with $1 ≤ k ≤ 10$” 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 $B ( x )$ and $C ( x )$ with $deg ( C ( x ) ) ≤ k − 1 2$ such that
$f k ( x ) = B 2 ( x ) + C ( x ) .$
(ii)
Let l be the least positive integer such that $l B ( x )$ and $l 2 C ( x )$ have integer coefficients for any nonnegative integer i and $δ ∈ { 1 , − 1 }$
$P i , δ ( x ) = δ ( l B ( x ) + δ i ) 2 − δ ( l B ( x ) ) 2 − δ l 2 C ( x ) ,$
r is any positive integer,
$H 1 = { α ∈ Z : P i , δ ( α ) = 0 , δ ∈ { 1 , − 1 } , i = 0 , 1 , 2 , … , r − 1 } ,$
and
$H 2 = { α ∈ R : P r , 1 ( α ) = 0 or P r , − 1 ( α ) = 0 } ,$
where $R$ and $Z$ are the sets of all real numbers and integers, respectively. If $H 1$ and $H 2$ are empty, then $( 1 )$ has no integral solution $( x , y , 2 )$. Otherwise, all integral solutions $( x , y , 2 )$ of $( 1 )$ satisfy $x ∈ H 1$ or
$min H 2 ≤ x ≤ max H 2 .$

## 2. Proofs

Lemma 1.
Let $k ≥ 3$. Then, all integral solutions $( x , y , n )$, $n > 0$ and $y ≠ 0$, of $( 1 )$ satisfy the equation
$a 2 b 1 y 2 n − b 2 a 1 y 1 n = 2 b 1 a 1 ,$
where $a 1 , a 2 , b 1$, and $b 2$ are positive integers such that
$a 1 a 2 b 1 b 2 ∣ 4 ∑ i = 0 k − 1 ( − 1 ) i A k − i − 1 2 i ,$
$A i$ is the coefficient of $x k − i − 1$ in the polynomial $f k ( x ) / x ( x + 2 )$,
$x = b 2 b 1 y 1 n , and x + 2 = a 2 a 1 y 2 n$
for some nonzero integers $y 1$ and $y 2$.
Proof.
Let $k ≥ 3$. Let $( x , y , n )$, with $n > 0$ and $y ≠ 0$, be any integral solution of the Diophantine equation
$y n = x + x ( x + 1 ) + ⋯ + x ( x + 1 ) ⋯ ( x + k ) .$
This can be written as
$y n = x ( x + 2 ) g k ( x )$
for some integer polynomial $g k ( x )$, which is not divided by x and $x + 2$, since $k ≥ 3$. Let d and q be positive integers such that
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$, $gcd ( d 2 2 , ( x / d ) ) = gcd ( d 1 2 , ( ( x + 2 ) g k ( x ) / d ) ) = 1$, $q 1 q 2 = q$, and $gcd ( q 1 , q 2 ) = 1 = gcd ( q 2 2 , ( ( x + 2 ) / q ) ) = gcd ( q 1 2 , ( x g k ( x ) / q ) ) = 1 .$ Then,
$d 1 2 d x = y 1 n a n d q 1 2 q ( x + 2 ) = y 2 n$
for some nonzero integers $y 1$ and $y 2$, since $y ≠ 0$ and $n ≥ 1$. From this, we have
$q d 1 2 y 2 n − d q 1 2 y 1 n = 2 q 1 2 d 1 2 and so q 2 d 1 y 2 n − d 2 q 1 y 1 n = 2 q 1 d 1 .$
Let
$g k ( x ) = f k ( x ) / ( x ( x + 2 ) ) = x k − 1 + A 1 x k − 2 + ⋯ + A k − 1$
and
$g ( x ) = x 2 + 2 x .$
Then, for each integer l with $0 ≤ l ≤ k − 1$,
$h l ( x ) = ∑ i = 0 l ( − 1 ) i A l − i 2 i x k − l − 1 + A l + 1 x k − l − 2 + ⋯ + A k − 1 .$
In particular,
$h k − 1 ( x ) = ∑ i = 0 k − 1 ( − 1 ) i A k − i − 1 2 i .$
This implies that
$gcd ( g ( x ) , g k ( x ) ) ∣ ∑ i = 0 k − 1 ( − 1 ) i A k − i − 1 2 i ,$
where $A i$ is the coefficient of $x k − i − 1$ in the polynomial $g k ( x )$.
If x is odd, then $d ∣ x$, $d ∣ g k ( x )$, $q ∣ ( x + 2 )$, $q ∣ g k ( x )$ and so $d q ∣ gcd ( g ( x ) , g k ( x ) )$. Suppose that x is even. Then,
$d q 4 ∣ x ( x + 2 ) 4 and d q 4 ∣ g k ( x ) .$
Hence, we have
$d q ∣ 4 gcd ( g ( x ) , g k ( x ) ) and so d q ∣ 4 ∑ i = 0 k − 1 ( − 1 ) i A k − i − 1 2 i .$
This proves the lemma. □
Lemma 2
(Hajdu, Laishram, and Tengely [5]). Let a, b, and c be positive integers with $a < b ≤ 4 × 2 , 018 , 957 × 99 × 467$ and $c ≤ 2 a b$. Then, the Diophantine equation
$a u n − b v n = ± c ,$
in integral variables $u > v > 1$, implies
$n ≤ max { 1000 , 824.338 log b + 0.258 } i f b ≤ 100 max { 2000 , 769.218 log b + 0.258 } i f 100 < b ≤ 10 , 000 max { 10 , 000 , 740.683 log b + 0.234 } i f b > 10 , 000$
Lemma 3
(Szalay [15]). Suppose that $p ≥ 2$ and $r ≥ 1$ are integers and that
$F ( x ) = x r p + a r p − 1 x r p − 1 + ⋯ + a 0$
is a polynomial with integer coefficients. Then, rational polynomials
$B ( x ) = x r + b r − 1 x r − 1 + ⋯ + b 0$
and $C ( x )$ with $deg ( C ( x ) ) ≤ r p − r − 1$ uniquely exist for which
$F ( x ) = B p ( x ) + C ( x ) .$
Lemma 4
(Srikanth and Subburam [13]). Let p be a prime number, $B ( x )$ and $C ( x )$ be nonzero rational polynomials with $deg ( C ( x ) ) < ( p − 1 ) deg ( B ( x ) )$, l be a positive integer such that $l B ( x )$ and $l p C ( x )$ have integer coefficients for any nonnegative integer i and $δ ∈ { 1 , − 1 }$:
$P i , δ ( x ) = δ ( l B ( x ) + δ i ) p − δ ( l B ( x ) ) p − δ l p C ( x ) ,$
r be any positive integer,
$H 1 = { α ∈ Z : P i , δ ( α ) = 0 , δ ∈ { 1 , − 1 } , i = 0 , 1 , 2 , … , r − 1 } ,$
and
$H 2 = { α ∈ R : P r , 1 ( α ) = 0 or P r , − 1 ( α ) = 0 } .$
If $H 1$ and $H 2$ are empty, then the Diophantine equation
$y p = B ( x ) p + C ( x )$
has no integral solution $( x , y )$. Otherwise, all integral solutions $( x , y )$ of the equation satisfy $x ∈ H 1$ or
$min H 2 ≤ x ≤ max H 2 .$
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, $α 1$, $α 2$, $β 1$, and $β 2$ be positive integers such that $l log ( α 1 / α 2 ) − m log ( β 1 / β 2 ) ≠ 0$. Let
$Γ = α 1 α 2 l β 1 β 2 m − 1 .$
Then, we have
$| Γ | > 0.5 exp − 24.34 log α log β max γ + 0.14 , 21 2 ,$
where $α = max { 3 , α 1 , α 2 }$, $β = max { 3 , β 1 , β 2 }$ and $γ = log l log β + m log α$.
Proof of Theorem 1.
Assume that $k ≥ 3$. Then, by Lemma 1, all integral solutions $( x , y , n )$, $y ≠ 0 , − 1$ and $n ≥ 1$, of $( 1 )$ satisfy the equation
$a y 2 n − b y 1 n = c ,$
where $y 1$ and $y 2$ are nonzero integers, a and b are positive integers such that $c ≤ 2 a b$,
$a b ∣ 4 ∑ i = 0 k − 1 ( − 1 ) i A k − i − 1 2 i ,$
and $A i$ is the coefficient of $x k − i − 1$ in the polynomial $f k ( x ) / x ( x + 2 )$. Without loss of generality, we can take $y 1 > y 2$ to prove the result. From (2), we write
$1 − a b y 2 y 1 n = c b y 1 n .$
Next, take $α 1 = a$, $α 2 = b$, $β 1 = y 2$, $β 2 = y 1$, $l = 1$, and $m = n$ in Lemma 5. Then, by the lemma, we obtain
$c b y 1 n ≥ exp { − 24.3414 ( log max { 3 , a , b } ) ( log max { 3 , y 1 } ) max { 21 , ( log n ) } 2 .$
From this, we obtain the required bound. Next, assume that $1 ≤ k ≤ 2$. Then, we can write Equation (1) as
$y 1 n = c 1 x$
and
$y 2 2 = c 2 ( x + 2 ) i ,$
where
$c 1 , c 2 ∈ 1 / 4 , 1 / 2 , 1 , 2 , 4$
and $i ∈ { 1 , 2 }$. 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
$g k ( x ) = 1 + ( x + 1 ) 1 + ( x + 3 ) + ⋯ + ( x + 3 ) ( x + 4 ) ⋯ ( x + k ) .$
and
$g k ( x ) = x k − 1 + A 1 x k − 2 + ⋯ + A k − 1 .$
Then, we obtain $A 0 = 1$, $A 1 = 1 + α 1$, $A k − 1 = 1 + α k − 2$, and $A j = α j − 1 + α j$ for $j = 2 , 3 , … , k − 2$ and
$α m = 1 + ∑ i = 0 m − 1 λ i + 1 ( 3 , … , k + i − m + 1 )$
for $m = 1 , 2 , … , k − 2$. □
Next, we consider the case that
$b ≤ 4 × 9 × 11 × 467 × 2 , 018 , 957 .$
If $y 1 = 1$, $y 2 = 1$, or $y 1 = y 2$, then we have
$x = d 2 d 1 = 1 , x = q 2 q 1 − 2 = − 1 , x = 2 q 1 d 2 d 1 q 2 − q 1 d 2 ,$
where $d 1 , d 2 , q 1$ and $q 2$ are positive integers such that $d 1 d 2 q 1 q 2 = a b$. 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
$740.683 log b ≤ 8982.9 .$
In a similar way, for the case $k < 10$, we have
$max { 10 , 000 , 740.683 log b + 0.23 } ≤ 10 , 000 .$
Hence, Lemma 2 confirms the result. □
Proof of Theorem 2.
Take $F ( x ) = x + x ( x + 1 ) + ⋯ + x ( x + 1 ) ⋯ ( x + k )$ in Lemma 3. Since k is odd, so $2 ∣ deg ( F ( x ) )$, $p = 2$, and $r = k + 1 2$. Then, by Lemma 3, there uniquely exist rational polynomials $B ( x )$ and $C ( x )$ with $deg ( C ( x ) ) ≤ k − 1 2$ such that
$F ( x ) = B 2 ( x ) + C ( x ) .$
Now, by Lemma 4, we have the theorem. □
Note 1.
First, find the values of the elementary symmetric forms$λ i + 1 ( 3 , … , k + i − m + 1 )$for$i = 0 , … , m − 1$and$m = 1 , 2 , … , k − 2$. Next, obtain$α 1$, $α 2$, …, $α k − 2$and so$A 0$, $A 1$, …, $A k − 1$. Using this, calculate$| A k − i − 1 − 2 A k − i − 2 |$and so
$2 i | A k − i − 1 − 2 A k − i − 2 | = | A k − i − 1 2 i − A k − i − 2 2 i + 1 |$
for$i = 0 , 2 , 4 , …$. In this way, for any positive integer k, we can find the exact value of$b$in Theorem 1. Therefore, it is not so hard to decide for which k is
$b ≤ 4 × 9 × 11 × 467 × 2 , 018 , 957$
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$C ≤ 2 A B$, $B > A$and$B ≤ 4 × 9 × 11 × 467 × 2 , 018 , 957$. Then, the equation
$A u n − B v n = ± C$
in integer variables$u > v > 1 , n > 3$implies
$n ≤ C m ( max { m , h n } ) 2 ( log B ) 2 + ( τ − 1 ) q 0 log u 0 + 1 log u 0 + log 4 log u 0 ,$
where
$h n = log n ( τ + 1 ) log B + 1 2 log u + ( τ − 1 ) q 0 + ϵ m ,$
in which$q 0$, $u 0$, $C m , m , τ$, and$ϵ m$are positive real numbers such that$u ≥ u 0$, $log ( u / v ) ≤ q 0$, $C m > 1$, $ϵ m > 1$, and$τ > 1$.
If we use the above observation in Lemma 1 of this paper, then we obtain the bound
$n ≤ c 2 ′ ( log n − log log b ) 2 log b$
and so an immediate estimation is
$n ≤ c 2 log b ,$
where $c 2$ is as in Theorem 1 and $c 2 ′$ is a positive real number depending on $u 0 , q 0 , C m , m , τ$, 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 $( 1 )$ 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]),
$y n = a 0 x + a 1 x ( x + 1 ) + ⋯ + a k x ( x + 1 ) ⋯ ( x + k ) ,$
where $k , a 0 , a 1 , ⋯ , a k$ are fixed integers and $x , y , n$ are integral variables in obtaining a better upper bound (depending only on $k , a 0 , a 1 , ⋯ , a k$) for all $max { x , y , n }$, where $( x , y , n )$ is an integral solution of the general equation.

## Author Contributions

Conceptualization, S.S.; data curation, S.A.; formal analysis, S.S., N.A., and M.K.; methodology, N.A. and S.A.; project administration, W.C. and G.P.J.; resources, W.C. and G.P.J.; software, M.K.; supervision, W.C. and G.P.J.; validation, L.N.; visualization, L.N.; writing—original draft, S.S. and N.A.; writing—review and editing, G.P.J. All authors have read and agreed to the published version of the manuscript.

## Funding

Anbazhagan and Amutha thank the RUSA grant sanctioned vide letter No. F 24-51/2014-U, Policy (TN Multi-Gen), Dept. of Edn. Govt. of India, Dt. 9 October 2018; the DST-PURSE 2nd Phase programme vide letter No. SR/PURSE Phase 2/38 (G) Dt. 21 February 2017; and the DST (FST—level I) 657876570 vide letter No. SR/FIST/MS-I/2018/17 Dt. 20 December 2018. S. Subburam’s research has been honored by the National Board of Higher Mathematics (NBHM), Department of Atomic Energy, Government of India (IN).

Not applicable.

Not applicable.

Not applicable.

## Conflicts of Interest

The authors declare no conflict of interest.

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Subburam, S.; Nkenyereye, L.; Anbazhagan, N.; Amutha, S.; Kameswari, M.; Cho, W.; Joshi, G.P. On the Ternary Exponential Diophantine Equation Equating a Perfect Power and Sum of Products of Consecutive Integers. Mathematics 2021, 9, 1813. https://doi.org/10.3390/math9151813

AMA Style

Subburam S, Nkenyereye L, Anbazhagan N, Amutha S, Kameswari M, Cho W, Joshi GP. On the Ternary Exponential Diophantine Equation Equating a Perfect Power and Sum of Products of Consecutive Integers. Mathematics. 2021; 9(15):1813. https://doi.org/10.3390/math9151813

Chicago/Turabian Style

Subburam, S., Lewis Nkenyereye, N. Anbazhagan, S. Amutha, M. Kameswari, Woong Cho, and Gyanendra Prasad Joshi. 2021. "On the Ternary Exponential Diophantine Equation Equating a Perfect Power and Sum of Products of Consecutive Integers" Mathematics 9, no. 15: 1813. https://doi.org/10.3390/math9151813

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