Hensel's Lemma and solutions to polynomials. After reading comments on an earlier version I have decided to completely restate my question:
Let $f_1(x),\dots, f_n(x) \in \mathbb{Z}[X]$ be polynomials. These are fixed. Let $p$ be a prime. If $\alpha$ is a simple root of $f_1(x),\dots, f_n(x)$. Then for each $i$ with $1 \leq i \leq n$, $\alpha$ 'lifts'  to a unique $\alpha_i \in \mathbb{Z}_p$ such $f_i(\alpha_i)=0$ and each $\alpha_i$ is congruent to $\alpha \mod p$.
Now, in general there is no reason why the $\alpha_i$ should be equal. My question is, are they equal if $p$ is sufficiently large, given that we have fixed $f_1,\dots,f_n$?    
 A: Edit add the statement I want to prove:

Fix non-zero polynomials $f_1, \dots, f_n\in \mathbb Z[X]$. There exists $N>0$ such that for all $p>N$, if $\alpha_1, \dots, \alpha_n\in \mathbb Z_p$ satisfy $f_i(\alpha_i)=0$ for all $i\le n$, then for any pair $i, j\le n$, 
  the condition $\alpha_i\equiv \alpha_j \mod p$ implies $\alpha_i=\alpha_j$.

Each $f_i(x)$ has finite many zeros in $\overline{\mathbb Q}$. Let $L$ be a finite extension of $\mathbb Q$ containing all these zeros $\beta_j$ (when $i$ varies). Only finitely many prime ideals of $L$ appear in the differences $\beta_\ell-\beta_j$. So for $p$ big enough (coprime with the above prime ideals), if $\beta_\ell\ne \beta_j$, then $v_p(\beta_\ell -\beta_j)=0$ 
and $\beta_\ell\not\equiv \beta_j \mod p$. As $\alpha_1, \dots, \alpha_n$ are among the $\beta_j$'s (one can embed $\mathbb Q[\alpha_1, \dots, \alpha_n]$ in $L$), the statement is proved. 
Remark All these have little to do with $f_1,\dots, f_n$ and one don't really have to suppose $\alpha_i\in \mathbb Z_p$ (the condition $\alpha_i\equiv \alpha_j \mod p$ then means that $(\alpha_i-\alpha_j)/p$ is integral over $\mathbb Z_p$).
A: If $f_1$ and $f_2$ have no common factor over the rationals, and $\alpha$ is a root of both of them mod $p$, then their resultant is a non-zero multiple of $p$. So if $f_1$ and $f_2$ are fixed, then there are only finitely many primes $p$ for which they have a common zero. Thus, "$p$ sufficiently large" just doesn't happen, unless the polynomials have a common factor over the rationals, and in that case the question is trivial, isn't it? 
