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

@article{ArgaezGarcia2019OnPP,
  title={On perfect powers that are sums of cubes of a five term arithmetic progression},
  author={Alejandro Arg'aez-Garc'ia},
  journal={Journal of Number Theory},
  year={2019}
}
Using only elementary arguments, Cassels and Uchiyama (independently) determined all squares that are sums of three consecutive cubes. Zhongfeng Zhang extended this result and determined all perfect powers that are sums of three consecutive cubes. Recently, the equation $(x-r)^k + x^k + (x+r)^k$ has been studied for $k=4$ by Zhongfeng Zhang and for $k=2$ by Koutsianas. In this paper, we complement the work of Cassels, Koutsianas and Zhang by considering the case when $k=3$ and showing that the… 

Tables from this paper

Perfect powers in sum of three fifth powers
In this paper we determine the perfect powers that are sums of three fifth powers in an arithmetic progression. More precisely, we completely solve the Diophantine equation $$ (x-d)^5 + x^5 + (x +
On the sum of fourth powers of numbers in arithmetic progression
We prove that the equation ${ (x - y)^4 + x^4 + (x + y)^4 = z^n }$ has no integer solutions ${ x, y, z}$ with ${ \gcd(x, y) = 1 }$ for all integers ${ n > 1 }$. We mainly use a modular approach with
Perfect powers that are sums of squares of an arithmetic progression
In this paper, we determine all primitive solutions to the equation $(x+r)^2 +(x+2r)^2 +\cdots +(x+dr)^2 = y^n$ for $2\leq d\leq 10$ and for $1\leq r\leq 10^4$. We make use of a factorization
On perfect powers that are sums of cubes of a seven term arithmetic progression
Abstract We prove that the equation ( x − 3 r ) 3 + ( x − 2 r ) 3 + ( x − r ) 3 + x 3 + ( x + r ) 3 + ( x + 2 r ) 3 + ( x + 3 r ) 3 = y p only has solutions which satisfy x y = 0 for 1 ≤ r ≤ 10 6 and
On the solutions of the Diophantine equation $(x-d)^2+x^2+(x+d)^2=y^n$ for $d$ a prime power
In this paper, we determine the primitive solutions of the Diophantine equation $(x-d)^2+x^2+(x+d)^2=y^n$ when $n\geq 2$ and $d=p^b$, $p$ a prime and $p\leq 10^4$. The main ingredients are the
The Diophantine equation $(x+1)^k+(x+2)^k+\cdots+(\ell x)^k=y^n$ revisited
Let $k,\ell\geq2$ be fixed integers and $C$ be an effectively computable constant depending only on $k$ and $\ell$. In this paper, we prove that all solutions of the equation
The equation $(x-d)^5+x^5+(x+d)^5=y^n$
In this paper, we solve the equation of the title under the assumption that $\gcd(x,d)=1$ and $n\geq 2$. This generalizes earlier work of the first author, Patel and Siksek [BPS16]. Our main tools
Perfect powers that are sums of squares in a three term arithmetic progression
We determine primitive solutions to the equation [Formula: see text] for [Formula: see text], making use of a factorization argument and the Primitive Divisors Theorem due to Bilu, Hanrot and Voutier.
On the power values of the sum of three squares in arithmetic progression
In this paper, using a deep result on the existence of primitive divisors of Lehmer numbers due to Y. Bilu, G. Hanrot and P. M. Voutier, firstly, we give an explicit formula for all positive integer
On the sum of fourth powers in arithmetic progression
We prove that the equation ${ (x - y)^4 + x^4 + (x + y)^4 = z^n }$ has no integer solutions ${ x, y, z}$ with ${ \gcd(x, y) = 1 }$ for all integers ${ n > 1 }$. We mainly use a modular approach with

References

SHOWING 1-10 OF 61 REFERENCES
Superelliptic equations arising from sums of consecutive powers
Using only elementary arguments, Cassels solved the Diophantine equation $(x-1)^3+x^3+(x+1)^3=z^2$ in integers $x$, $z$. The generalization $(x-1)^k+x^k+(x+1)^k=z^n$ (with $x$, $z$, $n$ integers and
On powers that are sums of consecutive like powers
TLDR
It is shown that for almost all d d≥2 (in the sense of natural density), the equation xk+(x+r) k+⋯+(x+(d-1)r)k=yn,x,y,n∈Z,n≥ 2, has no solutions.
Perfect powers that are sums of squares of an arithmetic progression
In this paper, we determine all primitive solutions to the equation $(x+r)^2 +(x+2r)^2 +\cdots +(x+dr)^2 = y^n$ for $2\leq d\leq 10$ and for $1\leq r\leq 10^4$. We make use of a factorization
On perfect powers that are sums of cubes of a seven term arithmetic progression
Abstract We prove that the equation ( x − 3 r ) 3 + ( x − 2 r ) 3 + ( x − r ) 3 + x 3 + ( x + r ) 3 + ( x + 2 r ) 3 + ( x + 3 r ) 3 = y p only has solutions which satisfy x y = 0 for 1 ≤ r ≤ 10 6 and
On the solutions of the Diophantine equation $(x-d)^2+x^2+(x+d)^2=y^n$ for $d$ a prime power
In this paper, we determine the primitive solutions of the Diophantine equation $(x-d)^2+x^2+(x+d)^2=y^n$ when $n\geq 2$ and $d=p^b$, $p$ a prime and $p\leq 10^4$. The main ingredients are the
Perfect powers that are sums of consecutive cubes
Euler noted the relation 63 = 33 + 43 + 53 and asked for other instances of cubes that are sums of consecutive cubes. Similar problems have been studied by Cunningham, Catalan, Gennochi, Lucas,
Rational isogenies of prime degree
In this table, g is the genus of Xo(N), and v the number of noncuspidal rational points of Xo(N) (which is, in effect, the number of rational N-isogenies classified up to "twist"). For an excellent
On modular representations of $$(\bar Q/Q)$$ arising from modular forms
where G is the Galois group GaI ( I ) /Q) and F is a finite field of characteristic I > 3. Suppose that p is modular of level N, i.e., that it arises from a weight-2 newform of level dividing N and
On the Diophantine equation $1^k+2^k+\dotsb+x^k=y^n$
In this paper, we resolve a conjecture of Schäffer on the solvability of Diophantine equations of the shape $1^k + 2^k + \dotsb + x^k = y^n$, for $1 \leq k \leq 11$. Our method, which may, with a
The Modular Approach to Diophantine Equations
One of the most powerful tools in the study of Diophantine equations, extensively developed in the past few years, has been the use of special types of elliptic curves associated with possible
...
1
2
3
4
5
...