I want to demonstrate the following statement.
Let $p,q\in \mathbb{GF}_2[x_1,\ldots,x_n]$ be of degree $n$ such that for all $v_1,\ldots,v_n\in\mathbb{GF}_2$, $p(v_1,\ldots,v_n)=q(v_1,\ldots,v_n)$. Prove that $p=q$.
I started by showing that if $v_1,\ldots,v_n$ are all equal to 0 then the know terms of $p$ and $q$ are equal. So $p(v_1,\ldots,v_n)=q(v_1,\ldots,v_n)$. Now I want to demonstrate the same thing for all the other coefficient.
How I can proceed?
I was thinking to proceed to do the same thing with the coefficient of the monomials of degree one and so on, just by showing with a proper combination of $v_1,\ldots,v_n$ that the coefficient are all the same. The only problem is that I don't know how to elegantly write this(or just in a "math" way).
It seems that I wasn't very cleary about my statment. I'll write and example so maybe it's more clear what I'm trying to do. $$p(x_1,x_2)= a_1+b_{11}x_1+b_{12}x_2+c_1x_1x_2$$ $$q(x_1,x_2)= a_2+b_{21}x_1+b_{22}x_2+c_2x_1x_2$$ Now I want to prove that if
$p(0,0)=q(0,0),p(1,0)=q(1,0),p(0,1)=q(0,1),p(1,1)=q(1,1)$
then
$a_1=a_2, b_{11}=b_{21},b_{12}=b_{22}, c_1=c_2$
because:
$p(0,0)=q(0,0) \Rightarrow a_1=a_2$
then $p(1,0)=q(1,0) \Rightarrow b_{11}=b_{21}$ because we already know that ($a_1=a_2$)
and so on
What the OP is apparently asking for is a proof that if $\ p\ $ and $\ q\ $ are polynomials over $\ \mathbb{GF}_2\ $ of the form \begin{align} p&=\sum_{S\subseteq \{1,2,\dots,n\}}\pi_S\prod_\limits{i\in S}x_i\ \text{ and }\\ q&= \sum_{S\subseteq \{1,2,\dots,n\}}\xi_S\prod_\limits{i\in S}x_i\ , \end{align} and $\ p\left(v\right)= q\left(v\right)\ $ for all $\ v\in \mathbb{GF}_2^n\ $, then $\ p=q\ $. This follows from the fact that if we define $\ w_S\in \mathbb{GF}_2^n\ $ by $$ w_{Si}=\cases{1&if $\ i\in S$\\ 0&if $\ i\not\in S\ $} $$ for any $\ S\subseteq \{1,2,\dots,n\} $, then $\ p\left(w_S\right)=\pi_S\ $ and $\ q\left(w_S\right)=\xi_S\ $, which gives $\ \pi_S=\xi_S\ $ for all $\ S\ $, and hence $\ p=q\ $.