On perfect powers that are sums of cubes of a five term arithmetic progression

TL;DR

This paper proves that certain perfect powers expressed as sums of five cubes in an arithmetic progression only have trivial solutions for a large range of parameters, advancing understanding of sums of powers.

Contribution

It establishes a comprehensive result for the equation involving sums of five cubes in arithmetic progression equaling a perfect power, with solutions only when either x or y is zero.

Findings

Solutions only occur when x or y is zero for the specified range

No non-trivial solutions exist for p ≥ 5 prime and 1 ≤ r ≤ 10^6

The result extends the understanding of sums of powers in arithmetic progressions

Abstract

We prove that the equation $(x-2r)^3 + (x-r)^3 + x^3 + (x+r)^3 + (x+2r)^3= y^p$ only has solutions which satisfy $xy=0$ for $1\leq r\leq 10^6$ and $p\geq 5$ prime.

Equations

$$(x-2r{)}^3+(x-r{)}^3+x^3+(x+r{)}^3+(x+2r{)}^3=y^{p}x,r,y,p\in \mathbb{Z},\gcd (x,r)=1,$$

$$\mid a{x}^n-b{y}^n\mid \le c,\text{with }a,b,c\in {\mathbb{Z}}_{\ge 0}\text{ and }a\ne b$$

$$n\le \max \left\{3\log (1.5\mid c/b\mid ),\frac{7400\log A}{\log \left(1+\log A/\log (\mid a/b\mid )\right)}\right\}.$$

$$a{w}_2^p-b{w}_1^{2p}=c{r}^2$$

$$\mu (p,q)=\{{\eta }^{2p}:\eta \in {\mathbb{F}}_q\}=\{0\}\cup \{\zeta \in {\mathbb{F}}_q^{\ast }:{\zeta }^k=1\}$$

$$B(p,q)=\left\{\zeta \in \mu (p,q):((b\zeta +c)/a{)}^{2k}\in \{0,1\}\right\}.$$

$$a{w}_2^p-b{g}^{2p}{w_1^{\prime }}^{2p}=c.$$

$$a^{\prime }{w_2}^p-b^{\prime }{w_1^{\prime }}^{2p}=c^{\prime },$$

$$a^{\prime }{w_2^{\prime }}^p-b^{\prime }{w_1}^{2p}=c^{\prime },$$

$$A{\rho }^p-B{\sigma }^{2p}=C$$

$$B^{\prime }=\prod _{\text{ordq(B) is odd}}q.$$

$$(v{\sigma }^p+n\sqrt{-m})(v{\sigma }^p-n\sqrt{-m})=u{\rho }^p.$$

$$K(\mathfrak{S},p)=\{ϵ\in K^{\ast }/{K^{\ast }}^p:\text{ordP(ϵ)≡0modp for all P∈/S}\}.$$

$$\mathcal{E}=\{ϵ\in K(\mathfrak{S},p):Norm(ϵ)/u\in {{\mathbb{Q}}^{\ast }}^p\}.$$

$$v{\sigma }^p+n\sqrt{-m}=ϵ{\eta }^p,$$

$$\chi (p,q)=\{{\eta }^p:\eta \in {\mathbb{F}}_q\}.$$

$$C(p,q)=\{\zeta \in \chi (p,q):((v\zeta +n\sqrt{-m})/ϵ{)}^{2k}\equiv \text{0 or 1modqj for j=1, 2}\}.$$

$$a{\sigma }^p-b{\tau }^p=c$$

$${\tilde{u}}_n={\tilde{u}}_n(\alpha ,\beta )=\{\begin{array}{ll}\frac{{\alpha }^n-{\beta }^n}{\alpha -\beta }& n\text{is odd}\\ \frac{{\alpha }^n-{\beta }^n}{{\alpha }^2-{\beta }^2}& n\text{is even}\end{array}$$

$$x(x^2+6r^2)=5^{p-1}{w}^p.$$

$$43\le p\le 34365$$

$$Y^2=3X^5-3\cdot 2^7\cdot 5^8$$

$$37\le p\le 56565$$

$$p$$

$$Y^2=2X^5-2\cdot 3^7\cdot 5^8$$

$$41\le p\le 69551$$

$$p$$

$$Y^2=X^5-6^7\cdot 5^8$$

$$41\le 91751$$

$$6w_2^p$$

Full text

The author was supported by the Research Grant 221183 FOMIX, CONACYT - Government of Yucatán.

1. Introduction

Finding perfect powers that are sums of terms in an arithmetic progression has received much interest; recent contributions can be found in [1], [2], [3], [4], [5] [9], [10], [14], [15], [16], [17], [19] [21], [23] and [24].

We are interested in solutions $(x,y,r)$ where $x$, $y$ and $r$ are coprime.

In this paper, we prove the following:

Theorem 1.1**.**

Let $p\geq 5$ be a prime. The equation

[TABLE]

with $0<r\leq 10^{6}$ only has solutions which satisfy $xy=0$.

The restriction $\gcd(x,r)=1$ is natural one, for otherwise it is easy to construct artificial solutions by scaling. We use a combination of techniques in the resolution of (1), the main ones being; a result of Mignotte based on linear form in logarithms ([8, Chapter 12, p. 423]); the method of Chabauty ([18, 20, 11]), the theorem due to Bilu, Hanrot and Voutier on primitive divisors ([6]), as well as some various elementary techniques.

2. Background

Here, we record some essential Theorems and Lemmas which are necessary for the computaions in Section 4.

Theorem 2.1**.**

(Mignotte) Assume that the exponential Diophantine inequality

[TABLE]

has a solution in strictly positive integers $x$ and $y$ with $\max\{x,y\}>1$. Let $A=\max\{a,b,3\}$. Then

[TABLE]

2.1. Criteria for eliminating equations of signature $(p,2p,2)$

We first apply a descent to equation (1) in Section 3. We are left with equations of the form:

[TABLE]

where $p$ is an odd prime and $a,b,c$ are positive integers satisfying $\gcd(a,b,c)=1$.

The criteria to follow in order to determine whether $(\ref{eqn:form})$ has solutions was previously presented in [13, 4, 1]. We start with the following lemma that give us a criteria for the nonexistence of solutions.

Lemma 2.2**.**

Let $p\geq 3$ be a prime. Let $a$, $b$ and $c$ be positive integers such that $\gcd(a,b,c)=1$. Let $q=2kp+1$ be a prime that does not divide $a$. Define

[TABLE]

and

[TABLE]

If $B(p,q)=\emptyset$, then equation (2) does not have integral solutions.

2.2. Local Solubility

After reducing the number of equations using Lemma 2.2, we give the next step and use classical local solubility methods to conclude nonexistence of solutions for many tuples $(a,b,c,p)$ in equation (2).

These methods work as follows: let $g=\operatorname{Rad}(\gcd(a,c))$ and suppose that $g>1$. Recall the condition $\gcd(a,b,c)=1$. Then $g\mid w_{1}$, and we can write $w_{1}=gw_{1}^{\prime}$. Thus

[TABLE]

Removing a factor of $g$ from the coefficients, we obtain

[TABLE]

where $a^{\prime}=a/c$ and $c^{\prime}=c/g<c$. Similarly, if $h=\gcd(b,c)>1$, we obtain

[TABLE]

where $c^{\prime}=c/h<c$. Applying these operations repeatedly, we arrive at an equation of the form

[TABLE]

where $A$, $B$, $C$ are now pairwise coprime. A necessary condition for the existence of solutions is that for any odd prime $q\mid A$, the residue $-BC$ modulo $q$ is a square. Besides this basic test, we also check for local solubility at the primes dividing $A$, $B$, $C$, and all primes $q\leq 19$.

2.3. A Descent

If local techniques previously presented failed to rule out solutions to equation (2) for particular coefficients and exponent $(a,b,c,p)$ then we may perform a further descent to rule out solutions. With $A$, $B$, $C$ as in (4) we let

[TABLE]

Thus $BB^{\prime}=v^{2}$. Write $AB^{\prime}=u$ and $CB^{\prime}=mn^{2}$ with $m$ squarefree. Rewrite (4) as

[TABLE]

Let $K=\mathbb{Q}(\sqrt{-m})$ and $\mathcal{O}$ be its ring of integers. Let $\mathfrak{S}$ contain the prime ideals of $\mathcal{O}$ that divide $u$ or $2n\sqrt{-m}$. Clearly $(v\sigma^{p}+n\sqrt{-m}){K^{*}}^{p}$ belongs to the “$p$-Selmer group”

[TABLE]

This is an $\mathbb{F}_{p}$-vector space of finite dimension can be computed by Magma using the command pSelmerGroup. Let

[TABLE]

It follows that

[TABLE]

where $\eta\in K^{*}$ and $\epsilon\in\mathcal{E}$.

We end up with the last criteria.

Lemma 2.3**.**

Let $\mathfrak{q}$ be a prime ideal of $K$. Suppose one of the following holds:

1. $(i)$

$\operatorname{ord}_{\mathfrak{q}}(v)$, $\operatorname{ord}_{\mathfrak{q}}(n\sqrt{-m})$, $\operatorname{ord}_{\mathfrak{q}}(\epsilon)$ are pairwise distinct modulo $p$; 2. $(ii)$

$\operatorname{ord}_{\mathfrak{q}}(2v)$, $\operatorname{ord}_{\mathfrak{q}}(\epsilon)$, $\operatorname{ord}_{\mathfrak{q}}(\overline{\epsilon})$ are pairwise distinct modulo $p$; 3. $(iii)$

$\operatorname{ord}_{\mathfrak{q}}(2n\sqrt{-m})$, $\operatorname{ord}_{\mathfrak{q}}(\epsilon)$, $\operatorname{ord}_{\mathfrak{q}}(\overline{\epsilon})$ are pairwise distinct modulo $p$.

Then there is no $\sigma\in\mathbb{Z}$ and $\eta\in K$ satisfying (5).

Lemma 2.4**.**

Let $q=2kp+1$ be a prime. Suppose $q\mathcal{O}=\mathfrak{q}_{1}\mathfrak{q}_{2}$ where $\mathfrak{q}_{1}$, $\mathfrak{q}_{2}$ are distinct, and such that $\operatorname{ord}_{\mathfrak{q}_{j}}(\epsilon)=0$ for $j=1$, $2$. Let

[TABLE]

Let

[TABLE]

Suppose $C(p,q)=\emptyset$. Then there is no $\sigma\in\mathbb{Z}$ and $\eta\in K$ satisfying (5).

2.4. Thue equations

Finally for the remaining equations that couldn’t be ruled out, they can be considered Thue equations by letting $\sigma=w_{2}$ and $\tau=w_{1}^{2}$

[TABLE]

where the exponent is a prime $p$. We use Magma’s Thue solver [7] and PARI/GP’s thueinit, thue commands as the final test to determine whether the equations has solutions.

2.5. Prime divisors of Lehmer sequences

A Lehmer pair is a pair $\alpha$, $\beta$ of algebraic integers such that $(\alpha+\beta)^{2}$ and $\alpha\beta$ are nonzero coprime rational integers and $\alpha/\beta$ is not a root of unity. The Lehmer sequence associated to the Lehmer pair $(\alpha,\beta)$ is

[TABLE]

A prime $p$ is called a primitive divisor of $\tilde{u}_{n}$ if it divides $\tilde{u}_{n}$ but does not divide $(\alpha^{2}-\beta^{2})\cdot\tilde{u}_{1}\cdots\tilde{u}_{n-1}$. We now state the following celebrated theorem due to Bilu, Hanrot and Voutier [6].

Theorem 2.5**.**

Let $\alpha$, $\beta$ be a Lehmer pair. Then $\tilde{u}_{n}(\alpha,\beta)$ has a primitive root for all $n>30$ and for all prime $n>13$.

It will be necessary to use Lehmer pairs to prove for certain values of $r$ that $(\ref{eq:main})$ has no solutions.

3. Descent to eight cases

Equation (1) can be rewritten as: $5x(x^{2}+6r^{2})=y^{p}.$ Letting $y=5w$, we can rewrite as:

[TABLE]

Note that $\gcd(x,x^{2}+6r^{2})\in\{1,2,3,6\}$ depending on whether $2,3$ divides $x$ or not. Thus, we are now able to divide into eight cases.

Case Conditions on $x$ Descent equations Equation of signature $(p,p,2)$

$1$ $5\mid x$ and $6\nmid x$

$x=5^{p-1}w_{1}^{p}$

$x^{2}+6r^{2}=w_{2}^{p}$

$w_{2}^{p}-5^{2p-2}w_{1}^{2p}=6r^{2}$

$2$ $5\mid x$ and $2\mid x$ and $3\nmid x$

$x=2^{p-1}5^{p-1}w_{1}^{p}$

$x^{2}+6r^{2}=2w_{2}^{p}$

$w_{2}^{p}-2^{2p-3}5^{2p-2}w_{1}^{2p}=3r^{2}$

$3$ $5\mid x$ and $3\mid x$ and $2\nmid x$

$x=3^{p-1}5^{p-1}w_{1}^{p}$

$x^{2}+6r^{2}=3w_{2}^{p}$

$w_{2}^{p}-3^{2p-3}5^{2p-2}w_{1}^{2p}=2r^{2}$

$4$ $5\mid x$ and $6\mid x$

$x=6^{p-1}5^{p-1}w_{1}^{p}$

$x^{2}+6r^{2}=6w_{2}^{p}$

$w_{2}^{p}-6^{2p-3}5^{2p-2}w_{1}^{2p}=r^{2}$

$5$ $5\nmid x$ and $6\nmid x$

$x=w_{1}^{p}$

$x^{2}+6r^{2}=5^{p-1}w_{2}^{p}$

$5^{p-1}w_{2}^{p}-w_{1}^{2p}=6r^{2}$

$6$ $5\nmid x$ and $2\mid x$ and $3\nmid x$

$x=2^{p-1}w_{1}^{p}$

$x^{2}+6r^{2}=2\cdot 5^{p-1}w_{2}^{p}$

$5^{p-1}w_{2}^{p}-2^{2p-3}w_{1}^{2p}=3r^{2}$

$7$ $5\nmid x$ and $3\mid x$ and $2\nmid x$

$x=3^{p-1}w_{1}^{p}$

$x^{2}+6r^{2}=3\cdot 5^{p-1}w_{2}^{p}$

$5^{p-1}w_{2}^{p}-3^{2p-3}w_{1}^{2p}=2r^{2}$

$8$ $5\nmid x$ and $6\mid x$

$x=6^{p-1}w_{1}^{p}$

$x^{2}+6r^{2}=6\cdot 5^{p-1}w_{2}^{p}$

$5^{p-1}w_{2}^{p}-6^{2p-3}w_{1}^{2p}=r^{2}$

4. Tables of computations

In this section, we apply the Theorems and Lemmas of Section 2 in order to fully resolve equation (1) and prove Theorem 1.1.

4.1. Case 1

We first apply Theorem 2.1 to obtain the bound $p\leq 34365$ when $|r|\leq 1.7\times 10^{2486}$. Thus, when we focus on $5\leq p\leq 34365$ and $1\leq r\leq 10^{6}$ we have the following table containing the obtained information after computational calculations.

[TABLE]

4.2. Case 2

For $p=5$ we have $w_{2}^{5}-2^{7}\cdot 5^{8}w_{1}^{10}=3r^{2}$. Letting $X=w_{2}/w_{1}^{2}$ and $Y=3r/w_{1}^{5}$ we obtain the hyperelliptic curve

[TABLE]

whose Jacobian has rank 1. We can show, using Magma, that $C(\mathbb{Q})=\{\infty\}$ which implies the curve only has trivial solutions.

Using Theorem 2.1 we bound $p\leq 56565$ for $|r|\leq 6.8\times 10^{4092}$. Thus, when we focus on $7\leq p\leq 56565$ and $1\leq r\leq 10^{6}$ we have the following table.

[TABLE]

The Thue equations that could not be solved by Magma, were solved by PARI/GP

[TABLE]

using the commands thueinit and thue. Observe that the solutions found contradict the condition $w_{1}\cdot w_{2}\neq 0$. Moreover, observe that all the equations that were solved are of signature $(p,2p,p)$.

4.3. Case 3

For $p=5$ we have $w_{2}^{5}-3^{7}\cdot 5^{8}w_{1}^{10}=2r^{2}$. Choosing the change of variable $X=w_{2}/w_{1}^{2}$ and $Y=2r/w_{1}^{5}$ we obtain the hyperelliptic curve

[TABLE]

whose Jacobian has rank 1. We can show, using Magma, that $C(\mathbb{Q})=\{\infty\}$ which implies the curve only has trivial solutions.

Using Theorem 2.1 we bound $p\leq 69551$ for $|r|\leq 3.8\times 10^{5881}$. Thus, when we focus on $7\leq p\leq 69551$ and $1\leq r\leq 10^{6}$ we have the following table.

[TABLE]

The Thue equations that could not be solved by Magma, were solved by PARI/GP

[TABLE]

using the commands thueinit and thue. Observe that the solutions found contradict the condition $w_{1}\cdot w_{2}\neq 0$. Moreover, observe that all the equations that were solved are of signature $(p,2p,p)$.

4.4. Case 4

For $p=5$ we have $w_{2}^{5}-6^{7}\cdot 5^{8}w_{1}^{10}=r^{2}$. Choosing the change of variable $X=w_{2}/w_{1}^{2}$ and $Y=r/w_{1}^{5}$ we obtain the hyperelliptic curve

[TABLE]

whose Jacobian has rank 1. We can show, using Magma, that $C(\mathbb{Q})=\{\infty\}$ which implies the curve only has trivial solutions.

Using Theorem 2.1 we bound $p\leq 91751$ for $|r|\leq 1.9\times 10^{6639}$. Thus, when we focus on $7\leq p\leq 91751$ and $2\leq r\leq 10^{6}$ we have the following table.

[TABLE]

We noticed a remarkable behaviour when $r=1$ and $7\leq p\leq 91751$.

We observed that application of Lemma 2.2 and the local solubility tests were always failing for every $p$. Due to this situation, we decided to use Lehmer pairs to solve it.

Let $K=\mathbb{Q}(\sqrt{-6})$ and write $\mathcal{O}_{K}$ for its ring of integers. This has class group isomorphic to $\mathbb{Z}/2\mathbb{Z}$. We consider the equation $x^{2}+6=6w_{2}^{p}$ where $x=30^{p-1}w_{1}^{p}$ and $w_{1}$, $w_{2}$ as before. Take

[TABLE]

It follows that

[TABLE]

where $\mathfrak{p}_{2}=(2,\sqrt{-6})$, $\mathfrak{p}_{3}=(3,\sqrt{-6})$ are the primes above $2$, $3$ and $\zeta$ is an ideal in $\mathcal{O}_{K}$. We write

[TABLE]

and $\mathfrak{q}=\mathfrak{p}_{2}\mathfrak{p}_{3}$. It follows that $\mathfrak{q}\zeta=(\gamma)\in\mathcal{O}_{K}$ is principal and $\gamma=u+v\sqrt{-6}\in\mathcal{O}_{K}$ with $u,v\in\mathbb{Z}$. After a possible change of sign we obtain

[TABLE]

Subtracting and conjugating we obtain

[TABLE]

or equivalently

[TABLE]

Let $L=\mathbb{Q}(\sqrt{-1},\sqrt{6})$. Taking $\alpha=\gamma/\sqrt{6}$ and $\beta=\overline{\gamma}/\sqrt{6}$.

Lemma 4.1**.**

Let $\alpha$ and $\beta$ as above. Then $\alpha$ and $\beta$ are algebraic integers. Moreover $(\alpha+\beta)^{2}$ and $\alpha\beta$ are nonzero coprime rational integers and $\alpha/\beta$ is not a unit.

Proof.

Let $\mathfrak{q}\mathcal{O}_{L}=\sqrt{6}\mathcal{O}_{L}$. By definition $\mathfrak{q}|\gamma,\overline{\gamma}$ which implies that $\alpha$ and $\beta$ are algebraic integers. Now we compute $(\alpha+\beta)^{2}=4u^{2}/3$.

Since $\mathfrak{p}_{3}|\sqrt{-6}$ we conclude that $\mathfrak{p}_{3}|u$ and so $3|u$. So $(\alpha+\beta)^{2}$ is a rational integer, i.e., $(\alpha+\beta)^{2}\in\mathbb{Z}$. If $(\alpha+\beta)^{2}=0$ then $u=0$, however this will imply that $30^{p}w_{1}^{p}=0$ contradicting $w_{1}\cdot w_{2}\neq 0$. Thus $(\alpha+\beta)^{2}$ is a nonzero ration integer. Moreover $\alpha\beta=\gamma\overline{\gamma}/6$ is a nonzero rational integer since $3|u$ and $\mathfrak{p}_{2}|\gamma,\overline{\gamma}$.

We now check that $(\alpha+\beta)^{2}$ and $\alpha\beta$ are coprime. Suppose they are not coprime. Then there exists a prime $\mathfrak{p}$ of $\mathcal{O}_{L}$ which divides both. Then $\mathfrak{p}$ divides $\alpha$ and $\beta$ and from the equation above $\mathfrak{p}$ divides $(w_{2})\mathcal{O}_{L}$ and $2\sqrt{-6}$. We contradicted our assumption of $(w_{1},w_{2})$ being a nontrivial coprime solution.

Finally, $\alpha/\beta=\gamma/\overline{\gamma}\in\mathcal{O}_{K}$ is not a root of unity because the only roots of unity in $K$ are $\pm 1$ which implie $\gamma=\pm\overline{\gamma}$ implying either $u=0$ or $v=0$ which is a contradiction. ∎

From Lemma 4.1 we have that the pair $(\alpha,\beta)$ is Lehmer pair and we denote by $\tilde{u}_{k}$ the associate Lehmer sequence. Substituting we see that

[TABLE]

Hence, we get

[TABLE]

thus $v=\pm 1$. By theorem 2.5, we immediately deduce that $p\in\{5,7,11,13\}$. For a given prime $p$, using equation (8), we see that $u$ is a root of the polynomial:

[TABLE]

Computing the roots of these polynomials, we find that there are no solutions.

4.5. Case 5

For $p=5$ we have $5^{4}w_{2}^{5}-w_{1}^{10}=6r^{2}$. Choosing the change of variable $X=w_{2}/w_{1}^{2}$ and $Y=6r/w_{1}^{5}$ we obtain the hyperelliptic curve

[TABLE]

whose Jacobian has rank 0. We can show, using Magma, that $C(\mathbb{Q})=\{\infty\}$ which implies the curve only has trivial solutions.

Using Theorem 2.1 we bound $p\leq 17183$ for $|r|\leq 8.3\times 10^{1242}$. Thus, when we focus on $7\leq p\leq 17183$ and $1\leq r\leq 10^{6}$ we have the following table.

[TABLE]

4.6. Case 6

Using Theorem 2.1 we bound $p\leq 9101$ for $|r|\leq 9.4\times 10^{657}$. Thus, when we focus on $5\leq p\leq 9101$ and $1\leq r\leq 10^{6}$ we have the following table.

[TABLE]

4.7. Case 7

Using Theorem 2.1 we bound $p\leq 22515$ for $|r|\leq 5.4\times 10^{1628}$. Thus, when we focus on $5\leq p\leq 22515$ and $1\leq r\leq 10^{6}$ we have the following table.

[TABLE]

4.8. Case 8

For $p=5$ we have $5^{4}w_{2}^{5}-6^{7}w_{1}^{10}=r^{2}$. Choosing the change of variable $X=w_{2}/w_{1}^{2}$ and $Y=r/w_{1}^{5}$ we obtain the hyperelliptic curve

[TABLE]

whose Jacobian has rank 0. We can show, using Magma, that $C(\mathbb{Q})=\{\infty\}$ which implies the curve only has trivial solutions.

Using Theorem 2.1 we bound $p\leq 44855$ for $|r|\leq 2.8\times 10^{3245}$. Thus, when we focus on $7\leq p\leq 44855$ and $1\leq r\leq 10^{6}$ we have the following table.

[TABLE]

This completes the proof of Theorem 1.1.

Acknowledgements

I would like to thank CIMAT Unidad Mérida, where I worked under a postdoctoral fellowship provided by CONACYT grant FOMIX-YUC 221183. I would also like to thank Professor John E. Cremona and The University of Warwick for allowing me access to the Number Theory Group computers where the computations were done.