Powerful numbers in $(1^{\ell}+q^{\ell})(2^{\ell}+q^{\ell})\cdots (n^{\ell}+q^{\ell})$

TL;DR

This paper extends previous results by proving that certain products involving powers and a positive integer q are not powerful numbers for large enough n, covering odd prime powers and all positive odd integers.

Contribution

It generalizes earlier work by establishing non-powerfulness of products for broader classes of exponents and sufficiently large n, including odd prime powers and all positive odd integers.

Findings

For odd prime powers , the product is not powerful if n ;

For all positive odd , there exists N_{q,} such that for n  N_{q,}, the product is not powerful.

The results extend the non-powerfulness property to more general exponents and larger ranges of n.

Abstract

Let $q$ be a positive integer. Recently, Niu and Liu proved that if $n\ge \max\{q,1198-q\}$, then the product $(1^3+q^3)(2^3+q^3)\cdots (n^3+q^3)$ is not a powerful number. In this note, we prove that (i) for any odd prime power $\ell$ and $n\ge \max\{q,11-q\}$, the product $(1^{\ell}+q^{\ell})(2^{\ell}+q^{\ell})\cdots (n^{\ell}+q^{\ell})$ is not a powerful number; (2) for any positive odd integer $\ell$, there exists an integer $N_{q,\ell}$ such that for any positive integer $n\ge N_{q,\ell}$, the product $(1^{\ell}+q^{\ell})(2^{\ell}+q^{\ell})\cdots (n^{\ell}+q^{\ell})$ is not a powerful number.

Equations

$${\nu }_p((1^{\ell }+q^{\ell })(2^{\ell }+q^{\ell })\cdots (n^{\ell }+q^{\ell }))={\nu }_p((p-q{)}^{\ell }+q^{\ell })=1,$$

$$P(m)=\{p:p\text{is a prime},\frac{m+1}{2}<p\le m+1\},$$

$$P(m;k,1)=\{p:p\text{is a prime},\frac{m+1}{2}<p\le m+1,p\equiv 1(\text{mod}k)\}.$$

$${\nu }_p(x^{\ell }+y^{\ell })={\nu }_p(x+y)+{\nu }_p(\ell ).$$

$$|P(n+q-1)|\ge 2.$$

$$\prod _{a=1}^n(a^{\ell }+q^{\ell })=\prod _{a=1}^n(a+q)\cdot \prod _{a=1}^n\frac{a^{\ell }+q^{\ell }}{a+q},$$

$${\nu }_p(a^{\ell }+q^{\ell })={\nu }_p(a+q)+{\nu }_p(\ell )={\nu }_p(a+q).$$

Full text

2010 Mathematics Subject Classification: Primary 11A25.

Keywords and phrases: shifted power, powerful number, $p$-adic valuation, Dirichlet’s theorem

1. School of Mathematics and Statistics, Nanjing University of Information

Science and Technology, Nanjing 210044, China

2. Wentian College, Hohai University, Maanshan 243031, China

1 Introduction

A positive integer $t$ is called a powerful number, if $t>1$ and $p^{2}\mid t$ for every prime divisor $p$ of $t$ (see [8]). In 2008, Cilleruelo [4] proved that, for any integer $n>3$, the product $(1^{2}+1)(2^{2}+1)\cdots(n^{2}+1)$ is not a square. Amdeberhan, Medina and Moll [1] claimed that if $n>12$ and $\ell$ is an odd prime, then $(1^{\ell}+1)(2^{\ell}+1)\cdots(n^{\ell}+1)$ is not a square. Gürel and Kisisel [11] confirmed the claim for $\ell=3$, while Zhang and Wang [17] confirmed the claim for any prime $\ell\geq 5$. In fact, they proved that $(1^{\ell}+1)(2^{\ell}+1)\cdots(n^{\ell}+1)$ is not a powerful number. Later, Chen et al. [2, 3] proved that if $\ell$ is an odd integer with at most two distinct prime factors, then $(1^{\ell}+1)(2^{\ell}+1)\cdots(n^{\ell}+1)$ is not a powerful number. There are many related results on this topic, one can refer to [5, 7, 9, 10, 12, 14, 15, 18].

Recently, Niu and Liu [13] extended the work of Gürel and Kisisel and proved that the following theorem.

Theorem A. For any positive integers $q$ and $n\geq\max\{q,1198-q\}$, the product $(1^{3}+q^{3})(2^{3}+q^{3})\cdots(n^{3}+q^{3})$ is not a powerful number.

In this paper, we generalize the results of Niu and Liu in the following theorem.

Theorem 1**.**

Let $q$ be a positive integer and $\ell$ be an odd prime power. For any integer $n\geq\max\{q,11-q\}$, the product $(1^{\ell}+q^{\ell})(2^{\ell}+q^{\ell})\cdots(n^{\ell}+q^{\ell})$ is not a powerful number.

The next theorem is a generalization of Theorem 2 in [3].

Theorem 2**.**

For any positive integer $q$ and odd positive integer $\ell$, there exists an integer $N_{q,\ell}$ such that for any positive integer $n\geq N_{q,\ell}$, the product $(1^{\ell}+q^{\ell})(2^{\ell}+q^{\ell})\cdots(n^{\ell}+q^{\ell})$ is not a powerful number.

2 Preliminary lemmas

Lemma 1**.**

Let $p$ be a prime and $q,\ell$ be positive integers with $2\nmid\ell$ and $\gcd(\ell,p-1)=1$. Then the congruence equation $x^{\ell}+q^{\ell}\equiv 0~{}(\text{mod}~{}p)$ has only one solution $x\equiv-q~{}(\text{mod}~{}p)$.

Proof.

If $p\mid q$, then the congruence equation has only one solution $x\equiv 0\equiv-q~{}(\text{mod}~{}p)$, the result is true. Now we assume $p\nmid q$. Let $g$ be a primitive root modulo $p$. Then $g^{\frac{p-1}{2}}\equiv-1~{}(\text{mod}~{}p)$. Let $x\equiv g^{t}~{}(\text{mod}~{}p)$, $q\equiv g^{m}~{}(\text{mod}~{}p)$, where $0\leq t,m\leq p-2$. Then the congruence equation $x^{\ell}+q^{\ell}\equiv 0~{}(\text{mod}~{}p)$ is equivalent to $g^{t\ell}\equiv g^{\frac{p-1}{2}+m\ell}~{}(\text{mod}~{}p)$, that is, $\ell t\equiv\frac{p-1}{2}+m\ell~{}(\text{mod}~{}p-1)$. Since $(\ell,p-1)=1$, it follows that $t$ has only one solution. Hence $x$ also has only one solution. By $2\nmid\ell$, it is easy to see that $x\equiv-q~{}(\text{mod}~{}p)$ is the only solution. ∎

Corollary 1**.**

Let $q$ be a positive integer and $\ell=k^{s}$, where $k$ is an odd prime and $s$ is a positive integer. If $p$ is a prime with $k\nmid p-1$, then the congruence equation $x^{\ell}+q^{\ell}\equiv 0~{}(\text{mod}~{}p)$ has only one solution $x\equiv-q~{}(\text{mod}~{}p)$.

For a nonzero integer $m$ and a prime $p$, let $\nu_{p}(m)$ denote the smallest nonnegative integer $k$ such that $p^{k}\mid m$ and $p^{k+1}\nmid m$.

Lemma 2**.**

Let $\ell=k^{s}$ be an odd prime power, $p$ be a prime and $q$ be a positive integer such that $p>q$, $p\not=k$ and $k\nmid p-1$. If $p-q\leq n\leq 2p-q-1$, then the product $(1^{\ell}+q^{\ell})(2^{\ell}+q^{\ell})\cdots(n^{\ell}+q^{\ell})$ is not a powerful number.

Proof.

By Corollary 1, the smallest two positive integers $x$ satisfying $x^{\ell}+q^{\ell}\equiv 0~{}(\text{mod}~{}p)$ are $p-q$ and $2p-q$. Noting that $p>q$ and $p\not=k$, we have $p^{2}\nmid((p-q)^{\ell}+q^{\ell})$. Hence, if $p-q\leq n\leq 2p-q-1$, then

[TABLE]

and so the product $(1^{\ell}+q^{\ell})(2^{\ell}+q^{\ell})\cdots(n^{\ell}+q^{\ell})$ is not a powerful number. ∎

For any positive integers $m$ and $k$, let

[TABLE]

[TABLE]

Lemma 3**.**

(See [17, Lemma 2.3].) If $m\not=1,3,5$ or $9$, then $|P(m)|\geq 2.$

Lemma 4**.**

(See [17, Lemma 2.4].) If $m\geq 4k$, where $k$ is an odd prime with $k\geq 5$, then $|P(m)|>|P(m;k,1)|.$

Lemma 5**.**

(See [3, Lemma 2].) Let $m$ be an integer with $m\geq 4$ and $m\not=9$. Then there is always an odd prime $p\in P(m)$ with $p\equiv 2~{}(\text{mod}~{}3)$.

The following lemma is a powerful lemma for solving exponential Diophantine equations. It is pretty well-known in the Olympiad folklore (see, e.g., [6]) though its origins are hard to trace.

Lemma 6**.**

(Lifting The Exponent Lemma.) Let $x,y$ be two integers, $\ell$ be an odd positive integer, and $p$ be an odd prime such that $p\mid x+y$ and none of $x$ and $y$ is divisible by $p$. We have

[TABLE]

3 Proofs of Theorems 1 and 2

Proof of Theorem 1..

By Lemma 2, it is enough to prove that there exists a prime $p>q$ with $p\not=k$ and $k\nmid p-1$ such that $p-q\leq n\leq 2p-q-1$. It is easy to see that $p-q\leq n\leq 2p-q-1$ is equivalent to $\frac{n+q}{2}<p\leq n+q$. Since $n\geq q$, it follows that $p>\frac{n+q}{2}\geq q$. Hence we need to prove that there exists a prime $p\not=k$ with $p\not\equiv 1~{}(\text{mod}~{}k)$ such that $\frac{n+q}{2}<p\leq n+q$.

By $n\geq 11-q$, we have $n+q-1\geq 10$. Hence, by Lemma 3, we obtain

[TABLE]

Suppose that $k=3$. Since $n+q-1\geq 10$, by Lemma 5, there exists an odd prime $p$ with $p\equiv 2~{}(\text{mod}~{}3)$ such that $\frac{n+q}{2}<p\leq n+q$. It is clear that $p\not=3$.

Now we assume $k\geq 5$.

Case 1. $n<2k-q+1$. If $p\in P(n+q-1;k,1)$, then $p\equiv 1~{}(\text{mod}~{}k)$ and $p\geq 2k+1>n+q$, a contradiction. Hence $|P(n+q-1;k,1)|=0$ in this case. Therefore, by (1), there exists at least one prime $p\not=k$ with $p\not\equiv 1~{}(\text{mod}~{}k)$ such that $\frac{n+q}{2}<p\leq n+q$.

Case 2. $2k-q+1\leq n<4k-q+1$. Suppose that $|P(n+q-1;k,1)|=|P(n+q-1)|$. Then $|P(n+q-1;k,1)|\geq 2$. Hence, there exist two primes $p_{1}$ and $p_{2}$ satisfying $p_{1}<p_{2}\leq n+q<4k+1$ and $p_{1}\equiv p_{2}\equiv 1~{}(\text{mod}~{}k)$. It follows that $p_{1}\geq 2k+1$ and $p_{2}\geq 4k+1$, a contradiction. Hence $|P(n+q-1)|>|P(n+q-1;k,1)|$. Therefore, there exists a prime $p$ with $p\not\equiv 1~{}(\text{mod}~{}k)$ such that $\frac{n+q}{2}<p\leq n+q$. Clearly, $p>\frac{n+q}{2}\geq\frac{2k+1}{2}>k$.

Case 3. $n\geq 4k-q+1$. It follows that $n+q-1\geq 4k$. By Lemma 4, there exists a prime $p$ with $p\not\equiv 1~{}(\text{mod}~{}k)$ such that $\frac{n+q}{2}<p\leq n+q$. Clearly $p>\frac{n+q}{2}\geq\frac{4k+1}{2}>k$.

By three cases above, there exists a prime $p\not=k$ with $p\not\equiv 1~{}(\text{mod}~{}k)$ such that $\frac{n+q}{2}<p\leq n+q$.

Therefore, the product $(1^{\ell}+q^{\ell})(2^{\ell}+q^{\ell})\cdots(n^{\ell}+q^{\ell})$ is not a powerful number. ∎

Proof of Theorem 2..

By Dirichlet’s theorem on arithmetic progressions (see [16, p. 285]), there exists an integer $N_{q,\ell}>q$ such that for any integer $n\geq N_{q,\ell}$, there is an odd prime $p\in P(n+q-1)$ with $p\equiv 2~{}(\text{mod}~{}\ell)$. Clearly, $\frac{n+q+1}{2}\leq p\leq n+q$ and $\gcd(p-1,\ell)=1$. Suppose that the product $(1^{\ell}+q^{\ell})(2^{\ell}+q^{\ell})\cdots(n^{\ell}+q^{\ell})$ is a powerful number. Noting that $\frac{n+q+1}{2}\leq p\leq n+q$ and $n\geq N_{q,\ell}>q$, we have $p\geq\frac{n+q+1}{2}\geq q+1$, and so $\nu_{p}(\prod_{a=1}^{n}(a+q))=1$. Hence, by

[TABLE]

it follows that $p\mid\frac{a^{\ell}+q^{\ell}}{a+q}$ for some $1\leq a\leq n$. Since $p\mid a^{\ell}+q^{\ell}$, $2\nmid\ell$ and $\gcd(p-1,\ell)=1$, by Lemma 1, we have $p\mid a+q$. On the other hand, by $p\equiv 2~{}(\text{mod}~{}\ell)$ and $p\geq q+1$, we have $p\nmid\ell$ and $p\nmid q$, and so $p\nmid a$. Hence, by Lemma 6, we have

[TABLE]

That is, $p\nmid\frac{a^{\ell}+q^{\ell}}{a+q}$, a contradiction.

This completes the proof of Theorem 2.

∎

4 Acknowledgement

This work was supported by the National Natural Science Foundation for Youth of China, Grant No. 11501299, the Natural Science Foundation of Jiangsu Province, Grant Nos. BK20150889, 15KJB110014 and the Startup Foundation for Introducing Talent of NUIST, Grant No. 2014r029.