2 Preliminaries

It is known by the results of Saradha and Shorey in [Reference Saradha and Shorey10, Reference Saradha and Shorey11] that the only integral solutions of (3) are given by

$$\begin{eqnarray}\frac{6!}{5}=12^{2},\qquad \frac{10!}{7}=720^{2},\qquad \frac{4!}{3}=2^{3}.\end{eqnarray}$$

Write $x=n/s$ and $y=m/t,m\neq 0$ with $s,t$ positive integers and $\gcd (n,s)=\gcd (m,t)=1.$ Then while proving Theorem 1.1, by the above result, we may take $s>1$ and $t>1$ . Further, we may also assume that

(5) $$\begin{eqnarray}\ell >k-1.\end{eqnarray}$$

Equation (3) can be re-written as

$$\begin{eqnarray}(n+s)\cdots (n+(i-1)s)(n+(i+1)s)\cdots (n+ks)=\frac{s^{k-1}m^{\ell }}{t^{\ell }}.\end{eqnarray}$$

Since the left-hand side is an integer and $\gcd (n,s)=\gcd (m,t)=1$ , we get $s^{k-1}=t^{\ell }$ . As $\ell$ is a prime ${>}k-1,$ there is an integer $d>1$ such that $s=d^{\ell }$ and $t=d^{k-1}.$ Thus, (3) gives rise to the equation

(6) $$\begin{eqnarray}\unicode[STIX]{x1D6E5}^{(i)}:=(n+d^{\ell })\cdots (n+(i-1)d^{\ell })(n+(i+1)d^{\ell })\cdots (n+kd^{\ell })=m^{\ell }\end{eqnarray}$$

with $\gcd (n,d)=1.$ Further, we can write each term

$$\begin{eqnarray}n+jd^{\ell }=a_{j}x_{j}^{\ell },\qquad 1\leqslant j\leqslant k,\quad j\neq i\end{eqnarray}$$

with $P(a_{j})<k$ and $a_{j}$ is $\ell$ th power free. Equation (6), when $d^{\ell }$ is replaced by any common difference $D$ , has been the subject of study in the papers [Reference Saradha and Shorey10, Reference Saradha and Shorey12]. This is an equation dealing with perfect powers obtained from a product of consecutive terms in an arithmetic progression with one missing term. It has been shown in [Reference Saradha and Shorey10, Reference Saradha and Shorey12] that such an equation with $D>1$ implies that there exists a prime $\equiv 1~(\text{mod}\,\ell )$ dividing $D.$ In particular, this implies that if a rational solution $(x,y)$ exists for (1), then the denominators of $x$ and $y$ exceed $(2\ell )^{\ell }$ and $(2\ell )^{k-1},$ respectively.

3 Lemmas

As in [Reference Bennett and Siksek2], it is clear that one needs to derive a suitable ternary form from (6). Towards this, we prove the following lemma, which is similar to [Reference Bennett and Siksek2, Lemma 2.1].

Lemma 3.1. Let $k\geqslant 3$ . Suppose that (3) has a rational point $(x,y)$ with $y\neq 0.$ Let $3\leqslant p\leqslant k$ be a prime either dividing $d$ or dividing at least one term and at most two terms of $\unicode[STIX]{x1D6E5}^{(i)}$ and further $k\geqslant 9$ if $p=3$ and $3\nmid d$ . Then there are non-zero integers $a,b,c,u,v,w$ satisfying

$$\begin{eqnarray}au^{\ell }+bv^{\ell }+cw^{\ell }=0\end{eqnarray}$$

such that:

1. (1) $a,b,c$ are $\ell$ th power free integers;

2. (2) $P(abc)<k$ ;

3. (3) $p\nmid abc$ ;

4. (4) $p$ divides precisely one of $u,v,w.$

Proof. Suppose that $p|d.$ Then $p\nmid (n+jd^{\ell })$ for any $1\leqslant j\leqslant k,j\neq i$ since $\gcd (n,d)=1.$ Then the equations

$$\begin{eqnarray}\displaystyle n+2d^{\ell }-(n+d^{\ell }) & = & \displaystyle d^{\ell }\quad \quad ~\text{if }i\neq 2,\nonumber\\ \displaystyle n+d^{\ell }-(n+3d^{\ell }) & = & \displaystyle -2d^{\ell }\quad \text{if }i=2\nonumber\end{eqnarray}$$

satisfy conditions (1)–(4) of the lemma.

Let $p\nmid d.$ Suppose that $p$ divides only one term of $\unicode[STIX]{x1D6E5}^{(i)}$ , say, $n+jd^{\ell }.$ Then $p$ occurs to an $\ell$ th power in this term. Hence, $p\nmid a_{j}$ and $p|x_{j}.$ Note that if $j=i+1$ or $i-1$ , then $n+(j-2)d^{\ell }$ or $n+(j+2)d^{\ell }$ occur as terms of $\unicode[STIX]{x1D6E5}^{(i)}$ , respectively, since $2\leqslant i\leqslant k-1$ . Form equations as follows.

$$\begin{eqnarray}\displaystyle n+jd^{\ell }-(n+(j-1)d^{\ell }) & = & \displaystyle d^{\ell }\quad \quad ~~\text{if }i\leqslant j-2,\nonumber\\ \displaystyle n+jd^{\ell }-(n+(j+1)d^{\ell }) & = & \displaystyle -d^{\ell }\quad ~~~\text{if }i\geqslant j+2,\nonumber\\ \displaystyle n+jd^{\ell }-(n+(j-2)d^{\ell }) & = & \displaystyle 2d^{\ell }\quad \quad \text{if }i=j-1,\nonumber\\ \displaystyle n+jd^{\ell }-(n+(j+2)d^{\ell }) & = & \displaystyle -2d^{\ell }~\quad \text{if }i=j+1.\nonumber\end{eqnarray}$$

These equations satisfy conditions (1)–(4) of the lemma.

Suppose that $p$ divides exactly two terms. Let $j$ be the smallest index such that $n+jd^{\ell }$ is a term of $\unicode[STIX]{x1D6E5}^{(i)}$ divisible by $p$ . Suppose that $j+p=i$ . Then $n+jd^{\ell }$ and $n+(j+2p)d^{\ell }$ are the only terms of $\unicode[STIX]{x1D6E5}^{(i)}$ divisible by $p$ and $i\notin \{j+1,j+2p-1\}$ . Consider the equation

$$\begin{eqnarray}\displaystyle & & \displaystyle (n+jd^{\ell })(n+(j+2p)d^{\ell })-(n+(j+1)d^{\ell })(n+(j+2p-1)d^{\ell })\nonumber\\ \displaystyle & & \displaystyle \quad =-(2p-1)d^{\ell }.\nonumber\end{eqnarray}$$

This satisfies conditions (1)–(4) of the lemma.

Hence, we may assume that $n+jd^{\ell }$ and $n+(j+p)d^{\ell }$ are the only terms of $\unicode[STIX]{x1D6E5}^{(i)}$ divisible by $p$ . Then form the equation

$$\begin{eqnarray}(n+jd^{\ell })(n+(j+p)d^{\ell })-(n+(j+1)d^{\ell })(n+(j+p-1)d^{\ell })=-(p-1)d^{2\ell }\end{eqnarray}$$

if $i\not \in \{j+1,j+p-1\}$ . This satisfies conditions (1)–(4) of the lemma. Thus, we need to consider $i\in \{j+1,j+p-1\}$ . For $p\geqslant 5$ , we take

$$\begin{eqnarray}(n+jd^{\ell })(n+(j+p)d^{\ell })-(n+(j+2)d^{\ell })(n+(j+p-2)d^{\ell })=-2(p-2)d^{2\ell },\end{eqnarray}$$

which satisfies conditions (1)–(4) of the lemma. Let $p=3$ . Since $i\in \{j+1,j+2\}$ , we see that $1\leqslant j\leqslant 3$ and hence $k<j+6\leqslant 9$ , contradicting the hypothesis.◻

From Lemma 3.1, the work of Kraus [Reference Kraus6] and the discussions in the proof of Theorem 1 of [Reference Bennett and Siksek2], we obtain the following lemma as an easy consequence.

Lemma 3.2. Suppose that (3) has a rational point $(x,y)$ with $y\neq 0$ and $\ell \geqslant 7.$ Let $3\leqslant p\leqslant k$ be a prime either dividing $d$ or dividing at least one term and at most two terms of $\unicode[STIX]{x1D6E5}^{(i)}$ . Then

$$\begin{eqnarray}\log \ell \leqslant \frac{16(\mathop{\prod }_{q<k,q\neq p}q)+1}{6}\log (\sqrt{p}+1)\leqslant 3^{k}.\end{eqnarray}$$

Lemma 3.4. Let $\ell \geqslant 3,\unicode[STIX]{x1D6FC}\geqslant 0$ be integers. Then the equation

$$\begin{eqnarray}x^{\ell }+y^{\ell }=2^{\unicode[STIX]{x1D6FC}}z^{\ell }\end{eqnarray}$$

in relatively prime integers $x,y,z\geqslant 1$ has no solution for $\unicode[STIX]{x1D6FC}\neq 1$ and, for $\unicode[STIX]{x1D6FC}=1$ , the equation has only the trivial solution $x=y=z=1.$

This result was established by Wiles [Reference Wiles16] for $\unicode[STIX]{x1D6FC}=0,$ by Darmon and Merel [Reference Darmon and Merel3] for $\unicode[STIX]{x1D6FC}=1$ and by Ribet [Reference Ribet8] for $\unicode[STIX]{x1D6FC}>1.$

Lemma 3.5 [Reference Saradha and Shorey10, Lemma 13].

Let $\ell \geqslant 5$ . Let $a,b,c$ be non-zero integers such that $P(abc)\leqslant 3$ . Then the equation

$$\begin{eqnarray}ax^{\ell }-by^{\ell }=cz^{\ell }\end{eqnarray}$$

in non-zero integers $x,y,z$ with

$$\begin{eqnarray}\gcd (ax^{\ell },by^{\ell },cz^{\ell })=1,\qquad \text{ord}_{2}(by^{\ell })\geqslant 4\end{eqnarray}$$

has no solution.

4 Proof of Theorem 1.1

We may assume that $\ell \geqslant 7$ . Let $k\in \{3,4,5\}$ . For $1\leqslant j_{1}<j_{2}\leqslant k$ with $i\notin \{j_{1},j_{2}\}$ , form the identity

$$\begin{eqnarray}(j_{2}-j_{1})d^{\ell }=n+j_{2}d^{\ell }-(n+j_{1}d^{\ell })=a_{j_{2}}x_{j_{2}}^{\ell }-a_{j_{1}}x_{j_{1}}^{\ell }.\end{eqnarray}$$

After cancelling the common factors, we obtain an equation of the form

$$\begin{eqnarray}ax^{\ell }-by^{\ell }=cz^{\ell }\quad \text{with }by^{\ell }\text{ even and }\gcd (ax^{\ell },by^{\ell },cz^{\ell })=1.\end{eqnarray}$$

Observe that $P(a_{j_{1}}a_{j_{2}}(j_{2}-j_{1}))\leqslant 3$ , implying that $P(abc)\leqslant 3$ . Further, since $k\leqslant 5$ , $\gcd (a_{j_{1}}x_{j_{1}}^{\ell },a_{j_{2}}x_{j_{2}}^{\ell },(j_{2}-j_{1})d^{\ell })$ divides $2^{2}\cdot 3$ . Hence, if $2|dx_{j_{1}}x_{j_{2}}$ , we use $\ell \geqslant 7$ and Lemma 3.5 to get a contradiction. We now assume that $2\nmid dx_{j_{1}}x_{j_{2}}$ .

Let $k=3$ . Then $i=2$ , $j_{1}=1,j_{2}=3$ and $P(abc)\leqslant 2$ and we get an equation of the form $x^{\ell }+y^{\ell }=2^{\unicode[STIX]{x1D6FC}}z^{\ell }$ with either $\unicode[STIX]{x1D6FC}=1$ or $\unicode[STIX]{x1D6FC}=\ell -2$ . This is a contradiction by Lemma 3.4, since $d\neq 1$ . Let $k\in \{4,5\}$ . Then it is possible to find $1\leqslant j,j^{\prime }\leqslant k$ such that $i\notin \{j,j^{\prime }\}$ , $j-j^{\prime }$ odd and ord $_{2}(n+jd^{\ell })$ is maximal and hence at least $\ell -3\geqslant 4$ . Taking $\{j_{1},j_{2}\}=\{j,j^{\prime }\}$ , the assertion follows by Lemma 3.5.

Let $k\geqslant 6$ . If $p|d$ with $3\leqslant p\leqslant k$ , then the inequality (2) is valid by Lemma 3.2. Hence, we assume that $p\nmid d$ for each $3\leqslant p\leqslant k$ . Let $k\in \{6,7,8\}$ . If $p\in \{5,7\},p\leqslant k$ and $p|\unicode[STIX]{x1D6E5}^{(i)}$ , then it divides at most two terms in $\unicode[STIX]{x1D6E5}^{(i)}$ and the assertion follows from Lemmas 3.1 and 3.2. Thus, we assume that $p\nmid \unicode[STIX]{x1D6E5}^{(i)}$ for $p\in \{5,7\},p\leqslant k$ . Then $P(a_{j})\leqslant 3$ and $3$ divides at most three terms in $\unicode[STIX]{x1D6E5}^{(i)}$ . So, there exist $1\leqslant j<j^{\prime }\leqslant 6$ with $i\notin \{j,j^{\prime }\}$ and $j^{\prime }-j\leqslant 2$ such that $P(a_{j}a_{j^{\prime }})\leqslant 2$ . From

$$\begin{eqnarray}(j^{\prime }-j)d^{\ell }=(n+j^{\prime }d^{\ell })-(n+jd^{\ell })=a_{j^{\prime }}x_{j^{\prime }}^{\ell }-a_{j}x_{j}^{\ell },\end{eqnarray}$$

we obtain an equation of the form $x^{\ell }+y^{\ell }=2^{\unicode[STIX]{x1D6FC}}z^{\ell }$ . This is a contradiction by Lemma 3.4.

We now take $k\geqslant 9$ . For $1\leqslant i\leqslant k$ , the product $(n+d^{\ell })\cdots (n+(i-1)d^{\ell })$ is divisible by all primes $3\leqslant p\leqslant i-1$ and the product $(n+(i+1)d^{\ell })\cdots (n+kd^{\ell })$ is divisible by all primes $3\leqslant p\leqslant k-i$ . Here the empty product is taken as equal to $1$ . Hence, $\unicode[STIX]{x1D6E5}^{(i)}$ is divisible by all primes $3\leqslant p\leqslant \max (i-1,k-i)$ . We have $\max (i-1,k-i)=k-i\geqslant p_{\unicode[STIX]{x1D703}}$ when $2\leqslant i\leqslant k-p_{\unicode[STIX]{x1D703}}$ and $\max (i-1,k-i)=i-1\geqslant p_{\unicode[STIX]{x1D703}}$ when $p_{\unicode[STIX]{x1D703}}<i<k$ . Thus, $p_{\unicode[STIX]{x1D703}}$ divides $\unicode[STIX]{x1D6E5}^{(i)}$ . Further, $p_{\unicode[STIX]{x1D703}}\geqslant k/2$ implies that $p_{\unicode[STIX]{x1D703}}\geqslant 3$ and $p_{\unicode[STIX]{x1D703}}$ divides at most two terms in $\unicode[STIX]{x1D6E5}^{(i)}$ . Hence, by Lemma 3.2, inequality (2) is valid.◻