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

Consider the Diophantine equation yn=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.

In this paper, we consider the Diophantine equation 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 if y = 0, −1. For the case 1 ≤ k ≤ 10, they explicitly calculated c(k) as n ≤ 19,736.
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: (ii) Let l be the least positive integer such that lB(x) and l 2 C(x) have integer coefficients for any nonnegative integer i and δ ∈ {1, −1} r is any positive integer, 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 .

Proofs
Then, all integral solutions (x, y, n), n > 0 and y = 0, of (1) satisfy the equation a 2 b 1 y n 2 − b 2 a 1 y n 1 = 2b 1 a 1 , where a 1 , a 2 , b 1 , and b 2 are positive integers such that , and x + 2 = a 2 a 1 y 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 This can be written as 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 for some nonzero integers y 1 and y 2 , since y = 0 and n ≥ 1. From this, we have and so q 2 d 1 y n 2 − d 2 q 1 y n 1 = 2q 1 d 1 . Let Then, for each integer l with 0 ≤ l ≤ k − 1, In particular, This implies that , q | g k (x) and so dq | gcd(g(x), g k (x)). Suppose that x is even. Then, Hence, we have dq | 4 gcd(g(x), g k (x)) and so dq | 4 This proves the lemma.
If H 1 and H 2 are empty, then the Diophantine equation has no integral solution (x, y). Otherwise, all integral solutions (x, y) of the equation satisfy 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.
If y 1 = 1, y 2 = 1, or y 1 = y 2 , then we have 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.
Hence, Lemma 2 confirms the result.
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 ≤ 2AB, B > A and B ≤ 4 × 9 × 11 × 467 × 2,018,957. Then, the equation in which q 0 , u 0 , C m , m, τ, and m are positive real numbers such that u ≥ u 0 , log(u/v) ≤ q 0 , C m > 1, 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 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 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.

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.