Consecutive integers in the form $a^x+y^b$

1.   Introduction

Catalan's conjecture, one of the famous classical problems in number theory, was first enunciated by Catalan ^[1] in 1844. It states that the equation

has no solutions in positive integers $x$ and $y$, other than $3^2-2^3 = 1$, where $p$ and $q$ are different prime numbers. In 1976, by applying the Gelfond-Baker method Tijdeman ^[2] succeeded in solving Catalan's conjecture (see Mignotte ^[3] for an excellent survey of developments). Thirty years later, the conjecture was re-proved by Mihuailesc ^[4], who used completely different approaches with the theory of cyclotomic fields. Also, in a series of papers in the 1930s and 1940s, some scholars (see, for example, ^[5,6,7,8]) studied the solutions to the general equation

where $a, b$ and $c$ are fixed positive integers. Furthermore, in 1936 Pillai conjectured that the number of positive integer solutions $(a, b, x, y)$, with $x\ge 2, y\ge 2$, to (1.1) is finite, which is still open for all $c > 1$.

Let $n$ be an integer. We say $n$ is a power if $n = x^y$ for some nonnegative integers $x$ and $y$, and a perfect power if $n = u^v$ for some integers $u$ and $v$ with $u\ge 2$ and $v\ge 2$. Then, Pillai's conjecture amounts to saying that the distance between two consecutive terms in the sequence of all perfect powers tends to infinity. In particular, Catalan's conjecture is equivalent to the statement that no two consecutive integers are perfect powers, other than $2^3$ and $3^2$. Note that there are no four consecutive integers with all of them being perfect powers, since any set of four consecutive integers must contain one integer congruent to $2$ modulo $4$ which cannot be a perfect power. Are there three consecutive integers with all of them being perfect powers? In 1962, Ko ^[9], by supplying a sufficient and necessary condition for the equation $x^p-y^q = 1$ to be solvable with positive integers $x$ and $y$, showed that no three consecutive integers are powers of other positive integers.

Naturally, one shall ask, for a given positive integer $k$ with $k\ge 2$, if there exist $k$ consecutive integers such that each of them is a sum of two powers (or two perfect powers)? If exist, how many such $k$-tuples are there? In this paper, we concentrate the investigation on consecutive integers in a fixed form and would like to give an answer to this question. Let $a$ and $b$ be integers no less than $2$. Let $T_k = (t_1, t_2, \cdots, t_k)$ be a $k$-tuple of $k$ consecutive integers, where $t_{i+1}-t_i = 1$ for any $1\le i\le k-1$. We call $T_k$ a nice $k$-tuple with type $(a, b)$ if each integer of $T_k$ is a sum of two powers of the form $a^x+y^b$ and a perfect $k$-tuple with type $(a, b)$ if each integer of $T_k$ is a sum of two perfect powers of the form $a^x+y^b$. Let $N_k(a, b)$ be the number of nice $k$-tuples with type $(a, b)$ and $\widetilde{N}_k(a, b)$ the number of perfect $k$-tuples with type $(a, b)$. It is interesting to study the formulas for $N_k(a, b)$ and $\widetilde{N}_k(a, b)$. In 2020, Lin and Cheng ^[10] obtained the formula for $N_k(2, 2)$. In this paper, by using the method of modulo coverage together with some elementary techniques, we present the formulas for $\widetilde{N}_k(2, 2)$, $\widetilde{N}_k(3, 2)$ and $N_k(3, 2)$. To be more precise, we report the main results as follows.

Theorem 1.1. Let $k$ be a positive integer with $k\ge 2$. The following statements are true.

(a) Let $\widetilde{N}_k(2, 2)$ and $\widetilde{N}_k(3, 2)$ be the numbers of perfect $k$-tuples with type $(2, 2)$ and $(3, 2)$, respectively. Then, we have

(b) Let ${N}_k(3, 2)$ be the number of nice $k$-tuples with type $(3, 2)$, and we have

Moreover, the only $3$ nice $4$-tuples with type $(3, 2)$ are $(1, 2, 3, 4)$, $(2, 3, 4, 5)$ and $(25, 26, 27, 28)$, and the only nice $5$-tuple with type $(3, 2)$ is $(1, 2, 3, 4, 5)$.

The paper is organized as follows. First, in Section 2, we present some lemmas that will be used to prove Theorem 1.1. Particularly, by figuring out the $3$-adic representation for one of the variables, we show that a Diophantine equation $3^{2x}+2\cdot3^x+9-12\cdot9^y = (24z)^2$ in nonnegative integers $x, y, z$ has no solution, other than $(x, y, z) = (0, 0, 0)$ and $(x, y, z) = (2, 1, 0)$. In Section 3, the proof of Theorem 1.1 is given. Finally, in Section 4, two further research problems are raised for the interested.

2.   Lemmas

In this section, some useful lemmas are presented. In fact, we first give the results for $2^x+y^2\pmod{8}$, $3^x+y^2\pmod{8}$ and $3^x+y^2\pmod{9}$, and then we determine the solvability of a Diophantine equation.

Lemma 2.1. Let $m$ and $n$ be positive integers of the forms $m = 3^x+y^2$ and $n = 2^u+v^2$ with $x, y, u$ and $v$ being nonnegative integers. The following statements are true.

(a) If $u\ge 2$, then the set of the possible remainders of $n$ modulo $8$ is $\{0, 1, 4, 5\}$.

(b) The set of the possible remainders of $m$ modulo $8$ is $\{1, 2, 3, 4, 5, 7\}$. Precisely,

(i) $m\equiv 1\pmod{8}$ if and only if $x\equiv 0\pmod{2}$ and $y\equiv 0\pmod{4}$;

(ii) $m\equiv 2\pmod{8}$ if and only if $x\equiv 0\pmod{2}$ and $y\equiv \pm 1\pmod{4}$;

(iii) $m\equiv 3\pmod{8}$ if and only if $x\equiv 1\pmod{2}$ and $y\equiv 0\pmod{4}$;

(iv) $m\equiv 4\pmod{8}$ if and only if $x\equiv 1\pmod{2}$ and $y\equiv \pm 1\pmod{4}$;

(v) $m\equiv 5\pmod{8}$ if and only if $x\equiv 0\pmod{2}$ and $y\equiv 2\pmod{4}$; and

(vi) $m\equiv 7\pmod{8}$ if and only if $x\equiv 1\pmod{2}$ and $y\equiv 2\pmod{4}$.

(c) The set of the possible remainders of $m$ modulo $9$ equals $\{0, 1, 2, 3, 4, 5, 7, 9\}$. More concretely,

(i) $m\equiv 0\pmod{9}$ if and only if $x\ge 2$ and $y\equiv 0\pmod{3}$;

(ii) $m\equiv 1\pmod{9}$ if and only if $x = 0$ and $y\equiv 0\pmod{3}$, or $x = 1$ and $y\equiv \pm 4\pmod{9}$, or $x\ge 2$ and $y\equiv \pm 1\pmod{9}$;

(iii) $m\equiv 2\pmod{9}$ if and only if $x = 0$ and $y\equiv \pm 1\pmod{9}$;

(iv) $m\equiv 3\pmod{9}$ if and only if $x = 1$ and $y\equiv 0\pmod{3}$;

(v) $m\equiv 4\pmod{9}$ if and only if $x = 1$ and $y\equiv \pm 1\pmod{9}$, or $x\ge 2$ and $y\equiv \pm 2\pmod{9}$;

(vi) $m\equiv 5\pmod{9}$ if and only if $x = 0$ and $y\equiv \pm 2\pmod{9}$;

(vii) $m\equiv 7\pmod{9}$ if and only if $x = 1$ and $y\equiv \pm 2\pmod{9}$, or $x\ge 2$ and $y\equiv \pm 4\pmod{9}$; and

(viii) $m\equiv 8\pmod{9}$ if and only if $x = 0$ and $y\equiv \pm 4\pmod{9}$.

In particular, if $x\ge 2$, then the remainder of $m$ modulo $9$ runs over $\{0, 1, 4, 7\}$.

Proof. With some direct computations, the results are immediate. □

Lemma 2.2. The Diophantine equation

in nonnegative integers $x$, $y$ and $z$ does not have any solutions except for $(x, y, z) = (0, 0, 0)$ and $(x, y, z) = (2, 1, 0)$.

Proof. Suppose that $(x, y, z)$ is a solution to Eq (2.1) with $(x, y, z)\ne (0, 0, 0), (2, 1, 0)$. First of all, one easily checks that $(x, y, z)$ with $0\le x\le 2$ cannot be the solution to Eq (2.1). So, one lets $x\ge 3$, $y\ge 0$ and $z\ge 0$ in the following. One then claims that

In fact, if $y\ge x-1$, one then checks that

since $x\ge 3$, a contradiction. So, the first inequality of (2.2) holds. Next, from (2.1) one finds that

This implies $3^x+1-24z > 0$. Note that $3^x+1-24z\equiv 2\pmod{4}$, and $3^x+1-24z\equiv 1\pmod{3}$. This implies that $3^x+1-24z\ge 10$. Therefore,

which implies that $x < 2y+\log_3(\frac{6}{5}) = 2y+0.165\cdots$, that is, $x\le 2y$ as desired. Then, it is immediate from (2.2) that $x\ge 4$. So, we only need to prove that Eq (2.1) has no solutions $(x, y, z)$ for $x\ge 4$, which will be done in what follows. Let $x\ge 4$, and rewrite (2.1) as

By (2.2), one has $x-2 < 2y-1 < 2x-2$. Taking remainders of modulo $3^{x-2}$ both sides of (2.3), one derives that

It then follows that $8z+\lambda\equiv 0\pmod{3^{x-2}}$ for some $\lambda\in\{\pm 1\}$. Let

for a positive integer $t_1$, and then (2.3) becomes

It is also noted that $2y-x+1 < x-2 < x$. By taking remainders of modulo $3^{2y-x+1}$ both sides of (2.5), one then has that $2t_1\lambda\equiv -2\pmod{3^{2y-x+1}}$, i.e., $t_1\lambda\equiv -1\pmod{3^{2y-x+1}}$. So, one may write

where $t_2$ is an integer having the same sign as $\lambda$. Putting (2.6) into (2.5), one deduces that

which implies that $t_2\equiv 2\pmod{3^{2x-2y-3}}$. Now, let

where either $t_3 = 0$ or $t_3$ is an integer having the same sign as $t_2$. Substituting it into (2.7), we then obtain $t_1^2-2t_3 = 9$. Suppose $t_3 = 0$. One then finds $t_1 = 3$, $t_2 = 2$ and $\lambda = 1$. By (2.6), we arrive at $3 = 2\cdot 3^{2y-x+1}-1\ge 5$, a contradiction. It then follows that $\frac{x_3}{\lambda}$ is a positive integer. Now, putting (2.8) and (2.6) into (2.4), we have that

Note that $\frac{x_3}{\lambda}$ is a positive integer, and $\lambda\in\{\pm 1\}$. It follows from (2.9) that

Then, one deduces that

for any $x\ge 6$. This means that Eq (2.1) has no integral solution $(x, y, z)$ with $x\ge 6$. Let $(4, y, z)$ be a solution to (2.1), and then by (2.2) one has $y = 2$. This implies that $z^2 = 10$, a contradiction. Similarly, one can confirm that $(5, y, z)$ cannot be a solution to Eq (2.1). Therefore, Eq (2.1) has no integral solution $(x, y, z)$ with $x\ge 4$.

This finishes the proof of Lemma 2.2. □

3.   Proof of Theorem 1.1

In this section, we present the proof of Theorem 1.1.

Proof of Theorem 1.1. First of all, we prove Item (a). Let $k$ be an integer with $k\ge 2$. From Lemma 2.1 we know that there do not exist any perfect $k$-tuples with types $(2, 2)$ and $(3, 2)$ if $k\ge 3$, that is, $\widetilde{N}_k(2, 2) = \widetilde{N}_k(3, 2) = 0$ for any $k\ge 3$. Now, let us compute $\widetilde{N}_2(2, 2)$, that is, the number of solutions to the Diophantine equation

in integers $x, y, u, v\ge 2$. For any nonnegative integer $k$, one observes that

where

Let $x$ be any positive integer with $x\equiv 1\pmod{4}$ and $x\ge 9$, and take $u = x-6$. It then follows from (3.2) that

So, we can write

Clearly, $\Delta_{x} = \frac{2^x-2^{x-6}+1}{5}$ is an odd integer no less than $101$ depending on $x$. If one lets $v = \frac{\Delta_{x}+5}{2}$ and $y = \frac{\Delta_{x}-5}{2}$, then (3.1) is satisfied. It follows that

is indeed a perfect $2$-tuple with type $(2, 2)$. Note that

is a one-to-one map from $\mathbb{N}$ to itself. Thus, there are infinitely many perfect $2$-tuples with type $(2, 2)$ in the form (3.3). This implies that $\widetilde{N}_2(2, 2) = +\infty$. For the purpose of deriving $\widetilde{N}_2(3, 2)$, one needs to consider another Diophantine equation

in integers $m, n, k, s\ge 2$. For any positive integer $z$, it is easy to see that

is a solution to (3.4). That is to say,

is a perfect $2$-tuple with type $(3, 2)$. Also, one checks that

is one-to one from $\mathbb{N}$ to itself as well. So, there are infinitely many perfect $2$-tuples with type $(3, 2)$ as the form in (3.5). It follows that $\widetilde{N}_2(3, 2) = +\infty$.

Next, we turn our attention to the proof of item (b). Let $k$ be any positive integer with $k\ge 2$. First, it is immediate that ${N}_2(3, 2) = +\infty$ since ${N}_2(3, 2)\ge \widetilde{N}_2(3, 2) = +\infty$.

Second, let $k = 3$ and $y$ be a nonnegative integer. Then, $(y^2+3^0, y^2+2, y^2+3^1)$ is a nice $3$-tuple with type $(3, 2)$ if and only if the equation

in nonnegative integers $y, u$ and $v$ has at least one solution. Note that

Then, one may take $y+u = 3^v-2$ and $y-u = 1$ and infer that

is a solution to (3.6) for any positive integer $v$. It then follows that

is a nice $3$-tuple with type $(3, 2)$ for any positive integer $v$. Together with $v\mapsto \left(\frac{3^v-1}{2}\right)^2+3^0$ being injective from $\mathbb{Z}^+$ to itself, this implies that there exist infinitely many nice $3$-tuples with type $(3, 2)$, i.e., $N_3(3, 2) = +\infty$.

Third, let $k = 4$. Let $(A, A+1, A+2, A+3)$ be a nice $4$-tuple with type $(3, 2)$, and $A = 3^{x_0}+y_0^2$ with $x_0, y_0$ being nonnegative integers. From Item (a) of Lemma 2.1, one knows that

CASE 1. $(A, A+1, A+2, A+3)\equiv (1, 2, 3, 4)\pmod{8}$. By Item (b-ⅰ) of Lemma 2.1, we have that $x_0$ is even, and $y_0\equiv 0\pmod{4}$. Now, we split all possible values of $x_0$ into the following subcases.

SUBCASE 1.1. $x_0 = 0$. Item (c) of Lemma 2.1 tells us that only three kinds of results for $(A, A+1, A+2, A+3)$ modulo $9$ would happen, that is,

which will be handled one by one in what follows.

SUBCASE 1.1.1. $(A, A+1, A+2, A+3)\equiv (1, 2, 3, 4)\pmod{9}$. Let

which is congruent to $2$ modulo $9$. It follows from Item (c-ⅲ) of Lemma 2.1 that $x_1 = 0$. Then, we have

This gives us that $y_1 = 1$ and $y_0 = 0$. So, $(A, A+1, A+2, A+3) = (1, 2, 3, 4)$. It is indeed a nice $4$-tuple with type $(3, 2)$ since $(1, 2, 3, 4) = (0^2+3^0, 1^2+3^0, 0^2+3^1, 1^2+3^1)$.

SUBCASE 1.1.2. $(A, A+1, A+2, A+3)\equiv (2, 3, 4, 5)\pmod{9}$. Let $A+1 = y_1^2+3^{x_1}$, that is,

Clearly, $y_1^2+3^{x_1}\equiv 3\pmod{9}$. By Item (c-iv) of Lemma 2.1, one has $x_1 = 1$. Then, (3.7) becomes $(y_0+y_1)(y_0-y_1) = 1$, i.e., $y_0 = 1$ and $y_1 = 0$, contradicting $y_0\equiv 0\pmod{4}$. Therefore, in this subcase there are no nice $4$-tuples with type $(3, 2)$.

SUBCASE 1.1.3. $(A, A+1, A+2, A+3)\equiv (8, 0, 1, 2)\pmod{9}$. Let $A+3 = y_3^2+3^{x_3}$. On the one hand, $y_3^2+3^{x_3}\equiv 4\pmod{8}$. It follows from Item (b-iv) of Lemma 2.1 that

On the other hand, one notes that $y_3^2+3^{x_3}\equiv 2\pmod{9}$. By Item (c-ⅲ) of Lemma 2.1, one then derives that $x_3 = 0$, a contradiction with (3.8). Thus, there does not exist any nice $4$-tuple with type $(3, 2)$ in the subcase.

SUBCASE 1.2. $x_0\ge 2$ is an even number. By Item (c) of Lemma 2.1, one checks that

Then, we have the following discussions.

SUBCASE 1.2.1. $(A, A+1, A+2, A+3)\equiv (0, 1, 2, 3)\pmod{9}$. Let $A+2 = 3^{x_2}+y_2^2$. Note that

It then follows from Items (b-ⅲ) and (c-ⅲ) of Lemma 2.1 that $x_2\equiv 1\pmod{2}$ and $x_2 = 0$, a contradiction. So, in this subcase we have no nice $4$-tuples with type $(3, 2)$.

SUBCASE 1.2.2. $(A, A+1, A+2, A+3)\equiv (1, 2, 3, 4)\pmod{9}$. First, applying Item (c-ⅱ) of Lemma 2.1 to the fact $3^{x_0}+y_0^2\equiv 1\pmod{9}$, we know that

Next, we let

which is congruent to $2$ modulo $9$. By Item (c-ⅲ) of Lemma 2.1, one then has $x_1 = 0$ and $y_1\equiv \pm 1\pmod{9}$. Putting $x_1 = 0$ into (3.10), one has $y_1^2-y_0^2 = 3^{x_0}$, i.e.,

If $x_0 = 2$, then (3.11) implies that

which contradicts (3.9). In the following, let $x_0\ge 4$. By (3.11), one may let $y_1+y_0 = 3^t$ and $y_1-y_0 = 3^{x_0-t}$ with $t\le x_0$ being a nonnegative integer. Note that $x_0\ge 4$ and $y_0\ne 0$. Then, $y_0 = \frac{3^t-3^{x_0-t}}{2}$ with $2\le \frac{x_0}{2} < t\le x_0$. Write $y_0 = \frac{3^t-1}{2}-\frac{3^{x_0-t}-1}{2}$, and one computes that

Obviously, all the results of (3.12)–(3.14) contradict with (3.9). Hence, we have no nice $4$-tuples with type $(3, 2)$ in this subcase.

SUBCASE 1.2.3. $(A, A+1, A+2, A+3)\equiv (7, 8, 0, 1)\pmod{9}$. In this subcase, $3^{x_0}+y_0^2\equiv 7\pmod{9}$, which gives us $y_0\equiv \pm 4\pmod{9}$. Let $A+1 = 3^{x_1}+y_1^2$, i.e.,

Note that $3^{x_1}+y_1^2\equiv 8\pmod{9}$. By Item (c-ⅷ) of Lemma 2.1, one has $x_1 = 0$. Then, (3.15) can be simplified to $y_1^2-y_0^2 = 3^{x_0}$. If $x_0 = 2$, one then deduces that $y_1 = 5$ and $y_0 = 4$. This gives a nice $4$-tuple $(A, A+1, A+2, A+3) = (25, 26, 27, 28)$ with type $(3, 2)$, since $(25, 26, 27, 28) = (4^2+3^2, 5^2+3^0, 0^2+3^3, 5^2+3^1)$. Now, let $x_0\ge 4$. As in Subcase 1.2.2, (3.12)–(3.14) can also be derived. Note that $y_0\equiv \pm 4\pmod{9}$. It then follows that only (3.14) would occur among (3.12)–(3.14). So, we have

implying that

Let $A+2 = 3^{x_2}+y_2^2$. We know that $3^{x_2}+y_2^2\equiv 0\pmod{9}$ and $3^{x_2}+y_2^2\equiv 3\pmod{8}$. It follows from Items (b-ⅲ) and (c-ⅰ) of Lemma 2.1 that $y_2\equiv 0\pmod{12}$ and $x_2\equiv 1\pmod{2}$. Write $y_2 = 12t$ and $x_2 = 2k+1$ with $t\ge 0$ and $k\ge 1$ being integers. Hence, we have that $(A, A+1, A+2, A+3)$ is a nice $4$-tuple with type $(3, 2)$ if and only if the equation

in nonnegative integers $x_0, t, k$ with $x_0\ge 4$ being even and $k\ge 1$ has at least one solution. However, Lemma 2.2 tells that (3.16) has no solutions. So, $(A, A+1, A+2, A+3)$ is not a nice $4$-tuple with type $(3, 2)$ in this subcase.

Next, we discuss the second case.

CASE 2. $(A, A+1, A+2, A+3)\equiv (2, 3, 4, 5)\pmod{8}$. In this case, $A = 3^{x_0}+y_0^2\equiv 2\pmod{8}$. From Item (b-ⅱ) of Lemma 2.1, one has $x_0\equiv 0\pmod{2}$ and $y_0\equiv \pm 4\pmod{4}$. Then, the following subcases are considered.

SUBCASE 2.1. $x_0 = 0$. Item (c) of Lemma 2.1 tells us that

First, we assume that $(A, A+1, A+2, A+3)\equiv (1, 2, 3, 4)\pmod{9}$. If let $A+1 = 3^{x_1}+y_1^2$, then $3^{x_1}+y_1^2\equiv 2\pmod{9}$. It follows from Item (c-ⅲ) of Lemma 2.1 that $x_1 = 0$. So, one derives that $y_1^2-y_0^2 = 1$, which implies that $y_0 = 0$. This is impossible since $y_0\equiv \pm 1\pmod{4}$.

Second, assume that $(A, A+1, A+2, A+3)\equiv (2, 3, 4, 5)\pmod{9}$. Let $A+1 = 3^{x_1}+y_1^2$, which is $3$ modulo $9$. Then, by Item (c-iv) of Lemma 2.1, one has $x_1 = 1$. Now, putting $x_1 = 1$ and $x_0 = 0$ into $3^{x_0}+y_0^2+1 = 3^{x_1}+y_1^2$, one derives $y_1 = 0$ and $y_0 = 1$. This is to say $(A, A+1, A+2, A+3) = (2, 3, 4, 5)$, which is really a nice $4$-tuple with type $(3, 2)$ since $(2, 3, 4, 5) = (3^0+1^2, 3^1+0^2, 3^1+1^2, 3^0+2^2)$.

Finally, assume that $(A, A+1, A+2, A+3)\equiv (8, 0, 1, 2)\pmod{9}$. In this case, $3^{x_0}+y_0^2\equiv 8\pmod{9}$. It is implied from Item (c-ⅷ) of Lemma 2.1 that

Let $A+3 = 3^{x_3}+y_3^2$, which clearly is $2$ modulo $9$. Then, by Item (c-ⅲ) of Lemma 2.1, one gets $x_3 = 0$. Substituting $x_0 = x_3 = 0$ into $3^{x_0}+y_0^2+3 = 3^{x_3}+y_3^2$, one deduces that $y_3^2-y_0^2 = 3$. Consequently, $y_0 = 1$, and $y_3 = 2$, which contradicts (3.17).

SUBCASE 2.2. $x_0\ge 2$ is an even number. By Item (c) of Lemma 2.1 we have that

Now, one claims that all the congruences in (3.18) cannot happen, so there is not any nice $4$-tuple with type $(3, 2)$ in this subcase. First, suppose $(A, A+1, A+2, A+3)\equiv (0, 1, 2, 3)\pmod{9}$. Let $A+2 = 3^{x_2}+y_2^2$, and we then know that $3^{x_2}+y_2^2\equiv 4\pmod{8}$ and $3^{x_2}+y_2^2\equiv 2\pmod{9}$. It follows from Items (b-iv) and (c-ⅲ) of Lemma 2.1 that $x_2\equiv 1\pmod{2}$ and $x_2 = 0$, a contradiction. Second, suppose $(A, A+1, A+2, A+3)\equiv (1, 2, 3, 4)\pmod{9}$. Let $A+1 = 3^{x_1}+y_1^2$. Note that $3^{x_1}+y_1^2\equiv 3\pmod{8}$ and $3^{x_1}+y_1^2\equiv 2\pmod{9}$. From Items (b-ⅲ) and (c-ⅲ) of Lemma 2.1, one derives that $x_1\equiv 1\pmod{2}$ and $x_1 = 0$, a contradiction as well. Third, suppose $(A, A+1, A+2, A+3)\equiv (7, 8, 0, 1)\pmod{9}$. If $A+1 = 3^{x_1}+y_1^2$, one then finds that $3^{x_1}+y_1^2\equiv 3\pmod{8}$ and $3^{x_1}+y_1^2\equiv 8\pmod{9}$. From Items (b-ⅲ) and (c-ⅷ) of Lemma 2.1, it follows that $x_1\equiv 1\pmod{2}$ and $x_1 = 0$, still a contradiction.

Combining all cases above, we have that there are only three nice $4$-tuples with type $(3, 2)$. More precisely, all of the nice $4$-tuples with type $(3, 2)$ are $(1, 2, 3, 4)$, $(2, 3, 4, 5)$ and $(25, 26, 27, 28)$. From this, we see that $(1, 2, 3, 4, 5)$ is the only nice $5$-tuple with type $(3, 2)$, and there is no nice $k$-tuple with type $(3, 2)$ for any $k\ge 6$.

The proof of Theorem 1.1 is complete. □

4.   Conclusions

The gaps in integer sequences are wide problems in number theory. The gap of primes $|p_n-p_{n+1}|$ is one of the most important topics in analytic Number Theory. In the field of Diophantine analysis, there are many open questions on the gap of the powers $|x^m-y^n|$. In this paper, we considered the gap $|(a^{x_1}+y_1^b)-(a^{x_2}+y_2^b)|$. In fact, we studied $k$-tuples of consecutive integers $(a_1, a_2, \cdots, a_k)$ such that each of them is the sum of powers. We used the method of modulo coverage together with some elementary techniques to present the formulas for $\widetilde{N}_k(2, 2)$, $\widetilde{N}_k(3, 2)$ and $N_k(3, 2)$. Note that in this paper we obtain that $\widetilde{N}_2(2, 2) = +\infty$ and $N_2(3, 2) = +\infty$. However, we neither give all perfect $2$-tuples with types $(2, 2)$ and $(3, 2)$ nor present all nice $k$-tuples with type $(3, 2)$ when $k\ge 6$. It seems a little difficult to do this. Here, we post a problem as future research.

Problem 4.1. Let $a, b$ and $k$ be integers with $k, a, b\ge 2$. Let $N_k(a, b)$ be the number of nice $k$-tuples with type $(a, b)$.

(A) Find $(a, b)$ such that $N_k(a, b)$ can be completely determined.

(B) For fixed $a, b$ and $k$, figure out the set of nice $k$-tuples with type $(a, b)$ completely if $N_k(a, b) = +\infty$.

Acknowledgments

The authors would like to thank the anonymous referees for careful readings of the manuscript and helpful comments. The corresponding author Cheng thanks Professor Fangwei Fu for his warm help during the author's visit to the Shiing-Shen Chern Institute of Mathematics of Nankai University. This work was supported partially by China's Central Government Funds for Guiding Local Scientific and Technological Development (No. 2021ZYD0013) and the National Natural Science Foundation of China (No. 12226335).

Conflict of interest

There is no conflict of interest.