I am reading an article by F beukers of recurrence sequence, and I have not been able to justify this statement that says it is trivial. If someone can help me or give me a clue or a book to consult on the subject, I would greatly appreciate it. Consider the equation $$ u_n=\alpha_{1}\theta_{1}^{n}+\alpha_{2}\theta_{2}^{n}+\alpha_{3}\theta_{3}^{n}=0, n\in \mathbb{Z} $$ where $\theta_1, \theta_2, \theta_3 $ are roots of $X^3-PX^2-QX-R$. And suppose that not all are real, and no one of ratios $\frac{\theta_i}{\theta_j} (i \neq j)$ are distinct of zero.
Consider the finite valuations of the field $\mathbb{Q}(\theta_1,\theta_2,\theta_3)$. If there exist a valuation $\boldsymbol{v}$ such that $|\theta_1|_{\boldsymbol{v}},|\theta_2|_{\boldsymbol{v}},|\theta_3|_{\boldsymbol{v}}$ are all distinct, then it is easy to see that this equation has at most three solutions. Thanks...
If the three $|\alpha_j\theta_j^n|_v$ are all distinct, then $\sum_j\alpha_j\theta_j^n$ must be nonzero. This is a basic property of non-Archimedean valuations.
But $|\alpha_1\theta_1^n|_v=|\alpha_2\theta_2^n|_v$ implies $|\alpha_1|_v|\theta_1|^n_v=|\alpha_2|_v|\theta_2|_v^n$, and as $|\theta_1|_v\ne|\theta_2|_v$ then there is at most one $n$ for which $|\alpha_1\theta_1^n|_v=|\alpha_2\theta_2^n|_v$. Likewise there is at most one $n$ for which $|\alpha_1\theta_1^n|_v=|\alpha_3\theta_3^n|_v$ and one $n$ for which $|\alpha_2\theta_2^n|_v=|\alpha_3\theta_3^n|_v$.