least number of equal fibers Here is the generalization of a nice problem.
Let $0\not=p,q \in \mathbb{C}[z_1,...,z_n], \; n\geq 1.$ Let $w_1,...,w_k$ be distinct complex numbers s.t. $p,q$ have the same fibers on them, that is, $p^{-1}(w_i)=q^{-1}(w_i)$ for every $1 \leq i \leq k.$ 

Find the least (if exists) $k \in \mathbb{N}$ s.t. the condition implies $p=q.$

As an example, for $n=1$ it can be verified that $k=2$ is the least number. 
 A: Assuming that you are right that $k=2$ for $n=1$ (I have not verified this), it seems to me that $k=2$ for all $n>1$ as well.
One argument goes as follows. Suppose $p^{-1}(a) = q^{-1}(a)$ and $p^{-1}(b) = q^{-1}(b)$, where $a$ and $b$ are distinct complex numbers. Fix a point $x\in p^{-1}(a)$. For each point $y\in p^{-1}(b)$, let $L_y$ be the unique complex line through $x$ and $y$. Then the restrictions $p|L_y$ and $q|L_y$ are one-variable polynomials $L_y\to \mathbb{C}$ whose fibers over $a$ and $b$ are the same. Thus by the one-variable argument, you know $p|L_y = q|L_y$. In particular, $p$ and $q$ agree on $\bigcup_{y\in p^{-1}(b)} L_y$, which has non-empty interior. By analytic continuation, $p = q$.
[Edit: This argument is needlessly complicated. The fact of the matter is that on any complex line $L$ through $\mathbb{C}^n$ the restrictions $p|L$ and $q|L$ are polynomials $L\to \mathbb{C}$ with the same fibers over $a$ and $b$. Thus $p$ and $q$ agree on any complex line, and hence on all of $\mathbb{C}^n$.] 
