AI 中文总结
研究方程\\(\sum\limits_{j = 1}^{k}jU_{j}(x,y)^{p}=U_{n}(x,y)^{q}\\)的解，通过特定方法求解该方程，推广了相关方程结果，还找出\\(k = 2\\)且\\(y = \pm1\\)时的所有解。
AI 中文摘要
我们求解方程\\(\sum\limits_{j = 1}^{k}jU_{j}(x,y)^{p}=U_{n}(x,y)^{q}\\)，其中\\(x,p,q,k,n\\)为正整数，\\(y = \pm1\\)且\\(\max\{p,q\}\leq11\\)，\\(U_{m}(x,y)=\frac{\alpha^{m}-\beta^{m}}{\alpha - \beta}\\)，\\(\alpha\\)和\\(\beta\\)是多项式\\(t^2 - xt + y\\)的根。这推广了类似方程的现有结果，还找出了\\(k = 2\\)且\\(y = \pm1\\)时的所有解。
英文摘要
We solve the equation $\sum\limits_{j=1}^{k}jU_{j}(x,y)^{p}=U_{n}(x,y)^{q}$ positive integers $x,p,q,k,n$, with $y=\pm1$ and $\max\{p,q\}\leq11$, where $U_{m}(x,y)=\frac{α^{m}-β^{m}}{α-β}$ for $α$ and $β$ roots of the polynomial $t^2-xt+y$. This generalizes existing results on similar equations, wherein the sequence was fixed as either the Fibonacci or Pell numbers. In addition, we find all solutions with $k=2$ and $y=\pm1$.
Comments17 Pages. Author's accepted manuscript. Final version published in the Fibonacci Quarterly
Journal refFibonacci Quart. 62 (2024), no. 3, 241-257
DOI：10.1080/00150517.2024.12459540
