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Register a new userPerfect powers in alternating sum of consecutive cubes
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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 $(x+1)^3 - (x+2)^3 + cdots - (x + 2d)^3 + (x + 2d + 1)^3 = z^p$, where $p$ is prime and $x,d,z$ are integers with $1 leq d leq 50$.
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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 $$ (x-d)^5 + x^5 + (x + d)^5 = z^n,~ngeq 2, $$ where $d,x,z in mathbb{Z}$ and $d = 2^a5^b$ with $a,bgeq 0$.
We give an explicit form of Inghams Theorem on primes in the short intervals, and show that there is at least one prime between every two consecutive cubes $xsp{3}$ and $(x+1)sp{3}$ if $loglog xge 15$.
We solve Diophantine equations of the type $ , a , (x^3 + y^3 + z^3 ) = (x + y + z)^3$, where $x,y,z$ are integer variables, and the coefficient $a eq 0$ is rational. We show that there are infinite families of such equations, including those where $a$ is any ratio of cubes or certain rational fractions, that have nontrivial solutions. There are also infinite families of equations that do not have any nontrivial solution, including those where $1/a = 1 - 24/m$ with certain restrictions on the integer $m$. The equations can be represented by elliptic curves unless $a = 9$ or 1. If $a$ is an integer and two variables are equal and nonzero, there exist nontrivial solutions only for $a=4$ or 9; there are no solutions for $a = 4$ when $xyz eq 0$. Without imposing constraints on the variables, we find the general solution for $a = 9$, which depends on two integer parameters. These cubic equations are important in particle physics, because they determine the fermion charges under the $U(1)$ gauge group.
We show that the diophantine equation $n^ell+(n+1)^ell + ...+ (n+k)^ell=(n+k+1)^ell+ ...+ (n+2k)^ell$ has no solutions in positive integers $k,n ge 1$ for all $ell ge 3$.
Let $n$ be a positive integer and $f(x) := x^{2^n}+1$. In this paper, we study orders of primes dividing products of the form $P_{m,n}:=f(1)f(2)cdots f(m)$. We prove that if $m > max{10^{12},4^{n+1}}$, then there exists a prime divisor $p$ of $P_{m,n}$ such that ord$_{p}(P_{m,n} )leq ncdot 2^{n-1}$. For $n=2$, we establish that for every positive integer $m$, there exists a prime divisor $p$ of $P_{m,2}$ such that ord$_{p} (P_{m,2}) leq 4$. Consequently, $P_{m,2}$ is never a fifth or higher power. This extends work of Cilleruelo who studied the case $n=1$.
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