论文标题
在Diophantine方程式上$ u_n-b^m = c $
On the Diophantine equation $U_n-b^m = c$
论文作者
论文摘要
令$（u_n）_ {n \ in \ mathbb {n}} $为在整数上定义的固定线性复发序列（具有一些技术限制）。我们证明存在有效计算常数$ b $和$ n_0 $，因此，对于任何$ b，c \ in \ mathbb {z} $，带有$ b> b $带有方程$ u_n -b^m = c $在\ mathbb {n} n}^2 $ n_ $ n_ $ n_ $ n_ $ n_ $ n_ $ n_ $ n_ $ n_ $ n_ $ n_ $和$ n_的最多两个不同的解决方案$（n，m）\ que q $ n_此外，我们将结果应用于由$ t_1 = t_1 = 1 $，$ t_3 = 2 $和$ t_ {n} = t_ {n-1}+t_ {n-2}+t_ {n-3} $ for $ n \ geq 4 $。通过LLL-Algorithm和持续的分数减少，我们能够证明$ n_0 = 1.1 \ cdot 10^{37} $和$ b = e^{438} $。相应的还原算法在SAGE中实现。
Let $(U_n)_{n\in \mathbb{N}}$ be a fixed linear recurrence sequence defined over the integers (with some technical restrictions). We prove that there exist effectively computable constants $B$ and $N_0$ such that for any $b,c\in \mathbb{Z}$ with $b> B$ the equation $U_n - b^m = c$ has at most two distinct solutions $(n,m)\in \mathbb{N}^2$ with $n\geq N_0$ and $m\geq 1$. Moreover, we apply our result to the special case of Tribonacci numbers given by $T_1= T_2=1$, $T_3=2$ and $T_{n}=T_{n-1}+T_{n-2}+T_{n-3}$ for $n\geq 4$. By means of the LLL-algorithm and continued fraction reduction we are able to prove $N_0=1.1\cdot 10^{37}$ and $B=e^{438}$. The corresponding reduction algorithm is implemented in Sage.
