论文标题
关于二芬太丁方程的解决方案$ f_ {n_ {n_ {1}}+f_ {n_ {2}}}+f_ {n_ {3}}}+f_ {n_ {n_ {4}} = 2^a $
On solutions of the Diophantine equation $F_{n_{1}}+F_{n_{2}}+F_{n_{3}}+F_{n_{4}}=2^a$
论文作者
论文摘要
令$（f_n）_ {n \ geq 0} $为$ f_0 = 0给出的fibonacci序列，f_1 = 1 $和$ f_ {n+2} = f_ {n+1}+f_n $ for $ n \ geq 0 $。在本文中，我们解决了两个的所有功率，这些权力是四个斐波那契数的总和，我们表征了一些例外。
Let $(F_n)_{n\geq 0}$ be the Fibonacci sequence given by $F_0 = 0, F_1 = 1$ and $F_{n+2} = F_{n+1}+F_n$ for $n \geq 0$. In this paper, we solve all powers of two which are sums of four Fibonacci numbers with a few exceptions that we characterize.
