On the Diophantine equation ∑j=1kjFjp=Fnq\sum_{j=1}^kjF_j^p=F_n^q

Let $F_n$ denote the $n^{th}$ term of the Fibonacci sequence. In this paper, we investigate the Diophantine equation $F_1^p+2F_2^p+\cdots+kF_{k}^p=F_{n}^q$ in the positive integers $k$ and $n$, where $p$ and $q$ are given positive integers. A complete solution is given if the exponents are included in the set $\{1,2\}$. Based on the specific cases we could solve, and a computer search with $p,q,k\le100$ we conjecture that beside the trivial solutions only $F_8=F_1+2F_2+3F_3+4F_4$, $F_4^2=F_1+2F_2+3F_3$, and $F_4^3=F_1^3+2F_2^3+3F_3^3$ satisfy the title equation.Comment: 12 page

A note on the exponential Diophantine equation ((A^2n)^x+(B^2n)^y=((A^2+B^2)n)^z)

Let (A) $B$ be positive integers such that $min{A,B}>1$, $gcd(A,B) = 1$ and $2|B.$ In this paper, using an upper bound for solutions of ternary purely exponential Diophantine equations due to R. Scott and R. Styer, we prove that, for any positive integer $n$, if $A >B^3/8$, then the equation $(A^2 n)^x + (B^2 n)^y = ((A^2 + B^2)n)^z$ has no positive integer solutions $(x,y,z)$ with $x > z > y$; if $B>A^3/6$, then it has no solutions $(x,y,z)$ with $y>z>x$. Thus, combining the above conclusion with some existing results, we can deduce that, for any positive integer $n$, if $Bequiv 2 pmod{4}$ and $A >B^3/8$, then this equation has only the positive integer solution $(x,y,z)=(1,1,1)$