On a variation of the Erdős–Selfridge superelliptic curve

In a recent paper by Das, Laishram and Saradha, it was shown that if there exists a rational solution of yl=(x+1)…(x+i−1)(x+i+1)…(x+k) for i not too close to k/2 and y≠0 , then logl<3k . In this paper, we extend the number of terms that can be missing in the equation and remove the condition on i .


Introduction
The Erdős-Selfridge superelliptic curves are the following family of curves, y l = (x + 1) . . . (x + k). (1) In [4], it is shown to not have any solutions in positive integers x, y, k, l with k, l 2. It has been conjectured by Sander [6] that for l 4 there are no rational solutions to equation (1) with y = 0. In [1], for k 2 a positive integer, there are at most finitely many solutions to (1) with x and y rational numbers, l 2 an integer with (k, l) = (2, 2) and y = 0. Further, it is shown that if l is a prime, then all solutions satisfy log l < 3 k .
In [2], by Das, Laishram and Saradha, they consider the following variation of the Erdős-Selfridge superelliptic curves, for k 2 an integer, l a prime, x and y = 0 rational numbers and i an integer strictly between 1 and k. Letting q be the smallest prime greater than or equal to k/2, they show that if (2) holds and 2 i k − q or q < i < k then log l < 3 k . Further, they show that if (2) holds and 3 k 26, then log l < 3 k . In this paper, we will further the results in [2] by removing the condition on i and also extending the terms that can be missing from the equation. For k 2 an integer, l a prime, i and j integers 1 < i < j < k and t ∈ {0, 1} for i < t < j, we call the following equation the Erdős-Selfridge curve with an incomplete block, We call a solution to (3) with x and y rational numbers and y = 0 a non-trivial rational solution. We note that the case j − i = 2 and i+1 = 0 is the same as (2). Theorem 1. If (x, y) is a non-trivial rational solution to equation (3) for k 27 and j − i − 1 < k/18 − 1, then log l < 3 k . In particular, if j − i = 2, then log l < 3 k holds for k 3. This will be proven by adjusting the proofs in [1,2], by adding in new identities allowing us to consider prime numbers less than k/2 and using a more combinatorial approach.
We will also consider a variation of the Erdős-Selfridge superelliptic curve from which terms in the product have been removed without any specification of their location in the interval [1, k].
for k 2 and L < 0.26 k log k , then log l < 3 k .

Preliminaries
We will assume throughout that l is prime and l > k − 1. We will first prove the existence of primes in the interval [ k 3 , k 2 ]. Following that we will look at the prime decomposition of the factors of equation (3).
Proof. In [5], it is shown that there is always a prime between z and (1 + 1 5 )z, for z 25. Hence, for k 75, the result now follows, and for the other k, it follows from an explicit computation.
Following the work of Bennett and Siksek [1] and of Das, Laishram and Saradha [2], we write the coordinates (x, y) as fractions in lowest common form, x = n/s and y = m/s for m = 0, s and s positive integers. From equation (3), we have As gcd(n, s) = gcd(m, s ) = 1 and l is a prime greater than k − i , it follows there is a positive integer d such that s = d l and s = d k− i .
Hence, equation (3) can be written as for m, n and d integers.
We now write each term in this product as such that x t1 is an integer and a t1 is an lth power free integer. Let p be a prime that divides a t1 , then p must also divide a t2 for some t 2 , hence p divides (t 1 − t 2 )d l . If p divides d, then it must also divide n, contradicting them being co-prime, hence p divides t 1 − t 2 . It now follows that all prime factors of a t are bounded above by k.
We note here that the exact same reasoning applies to equation (4) giving the following equation, for t = 1 if t ∈ S and zero otherwise. Proof. We can assume that no prime p in the range [k/3, k/2] divides d, otherwise the result follows trivially. Such a prime must divide at least two and at most three of the terms n + td l for t ∈ [1, k]. If p does not divide m, then there are at least 2 values of t such that t = 0. We will label these as i p and i p + p. It is then clear that p is in the set of differences of the elements in {t 1 , . . . , t L }. It is easily seen that It is then easily seen that if then there must be such a prime p. For k < 181000, we can explicitly calculate using Magma, the following bound For k 181000, we use the following bounds in [3] x and It is then simple algebraic manipulation to see that for k 181000 It is now seen that with L < 0.26 k log k , inequality (9) is true, completing the Lemma.

Fermat equation
In this section, we will attach a solution to a Fermat equation from a solution of (3) and (4). We will then use what is known about such equations to bound the exponent l. (2) all prime factors of abc are less than or equal to k; (3) p abc; (4) p divides precisely one of u, v, w.
Proof. We first deal with the case of equation (3). Let p be a prime as described and assume that p d, then p must divide m. This follows simply from the following, let j be a value in [1, k] such that n + jd ≡ 0 mod p. Then, if p m, it follows that j − p 0 and j + p k + 1, hence p (k + 1)/2 contradicting our assumption on p. It follows that p either divides d or divides exactly 1, 2 or 3 factors in the Erdős-Selfridge curve.
We first deal with p | d, then it follows that p m, so p a ti x ti . Using (6) we see that choosing a t such that t and t+1 are non-zero that gives the desired result. We now deal with the case that p divides exactly one factor, which we take to be n + td l . We consider the identity, for t a positive integer less than k + 1 such that |t − t| < p. Because L < p − 1, it follows that there exists such a t such that (n + t d l ) appears on the right-hand side of (5). As p must divide n + td l to an lth power, applying (6), we then get an equation satisfying the Lemma, that is, We now consider the case that p divides exactly two factors, n + td l and n + (t + p)d l . We consider a similar identity as before, for α a positive integer less than p.
It is clear that for distinct α and α Hence, as L < p/2 − 1, there exists α such that both n + (t + α)d l and n + (t + p − α)d l appear as factors in (5). Hence, the result now follows from (6) and the same finishing argument as above.
We are left to deal with the case that p divides exactly three factors, n + td l , n + (t + p)d l and n + (t + 2p)d l .
We point out the following identity, (n + td l )(n + (t + p)d l )(n + (t + 2p)d l ) − (n + (t + α)d l )(n + (t + p + α)d l )(n + (t + 2p − 2α)d l ) defined for α a positive integer less than p with α ≡ −p (mod 3). For α and α positive integers either less than p/2, then t + α, t + 2(p + α) This follows from some simple inequalities and calculations mod 3. Hence, it follows that there are more than p 6 − 1 distinct values of α with α ≡ −p (mod 3), such that the terms in (15) involving α do not coincide. So, we see that we have more choices of α than terms deleted, hence at least one α will give us such an equation with all terms defined. We note that as k 26, there will always be a prime greater than or equal to 13 in the permitted interval, meaning we can always take L = 1 for these values of k.
In the case of equation (4), we first apply Lemma 4, then follow the above argument identically.