On the power values of power sums

AbstractUsing several currently available techniques, including Baker's method, Frey curves and modular forms, we prove that for odd values of k with 1⩽k<170, the equation1k+2k+⋯+xk=y2n in positive integers x,y,n with n>2 has only the trivial solution (x,y)=(1,1)

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, we first give an explicit formula for all positive integer solutions of the Diophantine equation $(x-d)^2+x^2+(x+d)^2=y^n$ (*) when $n$ is an odd prime and $d=p^r$, $p>3$ a prime. So this improves the results on the papers of A. Koutsianas and V. Patel cite{KP} and A. Koutsianas cite{Kou}. Secondly, under the assumption of our first result, we prove that (*) has at most one solution $(x,y)$. Next, for a general $d$, we prove the following two results: (i) if every odd prime divisor $q$ of $d$ satisfies $qnotequiv pm 1 pmod{2n},$ then (*) has only the solution $(x,y,d,n)=(21,11,2,3)$. (ii) if $n>228000$ and $d>8sqrt{2}$, then all solutions $(x,y)$ of (*) satisfy $y^n<2^{3/2}d^3$

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, we first give an explicit formula for all positive integer solutions of the Diophantine equation $(x-d)^2+x^2+(x+d)^2=y^n$ (*) when $n$ is an odd prime and $d=p^r$, $p>3$ a prime. So this improves the results on the papers of A. Koutsianas and V. Patel cite{KP} and A. Koutsianas cite{Kou}. Secondly, under the assumption of our first result, we prove that (*) has at most one solution $(x,y)$. Next, for a general $d$, we prove the following two results: (i) if every odd prime divisor $q$ of $d$ satisfies $qnotequiv pm 1 pmod{2n},$ then (*) has only the solution $(x,y,d,n)=(21,11,2,3)$. (ii) if $n>228000$ and $d>8sqrt{2}$, then all solutions $(x,y)$ of (*) satisfy $y^n<2^{3/2}d^3$

Perfect powers in an alternating sum of consecutive cubes

In this paper, we consider the problem about finding out perfect powers in an alternating sum of consecutive cubes. More precisely, we completely solve the Diophantine equation (x+1)3 - (x+2)3 + ∙∙∙ - (x + 2d)3 + (x + 2d + 1)3 = zp, where p is prime and x,d,z are integers with 1 ≤ d ≤ 50