Arithmetic progressions consisting of unlike powers

In this paper we present some new results about unlike powers in arithmetic progression. We prove among other things that for given $k\geq 4$ and $L\geq 3$ there are only finitely many arithmetic progressions of the form $(x_0^{l_0},x_1^{l_1},...,x_{k-1}^{l_{k-1}})$ with $x_i\in{\Bbb Z},$ gcd$(x_0,x_1)=1$ and $2\leq l_i\leq L$ for $i=0,1,...,k-1.$ Furthermore, we show that, for L=3, the progression $(1,1,...,1)$ is the only such progression up to sign.Comment: 16 page

CUBES IN PRODUCTS OF TERMS IN ARITHMETIC PROGRESSION

Abstract. Euler proved that the product of four positive integers in arithmetic progression is not a square. Győry, using a result of Darmon and Merel showed that the product of three coprime positive integers in arithmetic progression cannot be an l-th power for l ≥ 3. There is an extensive literature on longer arithmetic progressions such that the product of the terms is an (almost) power. In this paper we prove that the product of k coprime integers in arithmetic progression cannot be a cube when 2 < k < 39. We prove a similar result for almost cubes. 1

Representations of reciprocals of Lucas sequences

In 1953 Stancliff noted an interesting property of the Fibonacci number $F_{11}=89.$ One has that $$\frac{1}{89}=\frac{F_0}{10}+\frac{F_1}{10^2}+\frac{F_2}{10^3}+\frac{F_3}{10^4}+\frac{F_4}{10^5}+\frac{F_5}{10^6}+\cdots.$$ De Weger determined a complete list of similar identities in case of the Fibonacci sequence, the solutions are as follows \begin{align*} \frac{1}{F_1}=\frac{1}{F_2}=\frac{1}{1}=\sum_{k=1}^{\infty}\frac{F_{k-1}}{2^k},\qquad& \frac{1}{F_5}=\frac{1}{5}=\sum_{k=1}^{\infty}\frac{F_{k-1}}{3^k},\\ \frac{1}{F_{10}}=\frac{1}{55}=\sum_{k=1}^{\infty}\frac{F_{k-1}}{8^k},\qquad& \frac{1}{F_{11}}=\frac{1}{89}=\sum_{k=1}^{\infty}\frac{F_{k-1}}{10^k}.\\ \end{align*} In this article we study similar problems in case of general Lucas sequences $U_n(P,Q)$. We deal with equations of the form \begin{equation*} \frac{1}{U_n(P_2,Q_2)}=\sum_{k=1}^{\infty}\frac{U_{k-1}(P_1,Q_1)}{x^k}, \end{equation*} for certain pairs $(P_1,Q_1)\neq(P_2,Q_2).$ We also consider equations of the form \begin{equation*} \sum_{k=1}^{\infty}\frac{U_{k-1}(P,Q)}{x^k}=\sum_{k=1}^{\infty}\frac{R_{k-1}}{y^k}, \end{equation*} where $R_n$ is a ternary linear recurrence sequence. The proofs are based on results related to Thue equations and elliptic curves