Problem: Let $n$ be a positive integer.
Suppose we are given $2^n+1$ distinct sets,
each containing finitely many objects.
Place each set into one of two categories, the red sets and the blue sets,
so that there is at least one set in each category.
We define the \textit{symmetric difference} of two sets as
the set of objects belonging to exactly one of the two sets.
Prove that there are at least $2^n$ different sets which
can be obtained as the symmetric difference of a red set and a blue set.

\end{enumerate}

Solution: We can interpret the problem as working with
binary strings of length $\ell \ge n+1$,
with $\ell$ the number of elements across all sets.

Let $F$ be a field of cardinality $2^\ell$,
hence $F \cong \FF_2^{\oplus \ell}$.

Then, we can think of red/blue as elements of $F$,
so we have some $B \subseteq F$, and an $R \subseteq F$.
We wish to prove that $|B+R| \ge 2^n$.
Want $|B + R| \ge 2^n$.

Equivalently, any element of a set $X$ with $|X| = 2^n - 1$
should omit some element of $|B+R|$.
To prove this: we know $|B| + |R| = 2^n+1$, and define
\[ P(b,r) = \prod_{x \in X} (b+r-x). \]
Consider $b^{|B|-1} r^{|R|-1}$.
The coefficient of is $\binom{2^n-1}{|B|-1}$,
which is odd (say by Lucas theorem),
so the nullstellensatz applies.