Is the eventown problem solution unique up to isomorphism Suppose we have a town with a set of residents $X$, where $|X| = n$ (suppose $n$ even for simplicity here). The residents like forming clubs, and we have clubs $C_1,C_2,\ldots,C_m \subseteq X$. The eventown problem is interested in the maximum number of clubs $m$ that can be formed with these specific rules:

*

*All clubs are distinct.

*A club must have an even number of members. $\forall i, |C_i|$ is even.

*Any pair of two clubs shares an even number of members: $\forall i \neq j, |C_i \cap C_j|$ is even.

It is easy to show that the largest number of clubs possible is $m=2^{n/2}$. An easy construction to attain this bound goes as follow : Take any partition of $X$ into sets of size 2, $B_1,\ldots,B_{n/2}$, and consider the family $\mathcal{F}=\{\bigcup_{i\in S}B_i \ :\ S\subseteq [n/2] \}$
For any partition, this matches the required bound, and all these possibilities are pairwise isomorphic.

Is the solution unique up to isomorphism, or are there some other constructions?

 A: The proof of eventown tells us that at equality, the set of clubs can be represented in $ \mathbb{F}_2^n$, as a linear subspace $F$ of dimension $n/2$ (and $ F = F^\perp$). We will use this context.
Here is a counterexample when $n = 8$.
Let the clubs be spanned by
$$\begin{pmatrix} 
1 & 1 & 0 & 0 & 1 & 0 & 0 & 1 \\ 
1 & 1 & 1 & 0 & 0 & 0 & 1 & 0 \\
1 & 1 & 0 & 1 & 0 & 1 & 0 & 0 \\
1 & 0 & 1 & 1 & 1 & 0 & 0 & 0 \\
\end{pmatrix}$$
Check that

*

*every vector has 4 1's,

*the intersection of any 2 vectors has 2 1's

*Linear independence: Vectors 1, 2, 3 are the only ones with a 1 in the last / second last / third last column, so can't be involved in the equation. Hence these vectors are linearly independent.

However, there isn't a pairing that we can do with the first coordinate.

Here is a counterexample when $n = 7$ (so not quite what is asked).
Let the clubs be spanned by
$$\begin{pmatrix} 
1 & 1 & 0 & 0 & 1 & 0 & 1 \\
1 & 0 & 1 & 0 & 1 & 1 & 0 \\
1 & 1 & 1 & 1 & 0 & 0 & 0 \\ \end{pmatrix}$$
Check that

*

*every vector has 4 1's,

*the intersection of any 2 vectors has 2 1's

*Linear independence: Vectors 1 and 2 are the only ones with a 1 in the last / second last column, so can't be involved in the equation. Hence these vectors are linearly independent.

However, there isn't a pairing that we can do with the first coordinate.
A: Just writing down some details from the counter-example found by Calvin Lin.
Note that in a maximum solution derived from a partition of $X$, for any 3 clubs $C_1,C_2,C_3$, we have $\vert C_1\cap C_2 \cap C_3 \vert$ is also even.
Let $C_1,C_2,C_3$ be the following clubs in $\mathbb{F_2}^n$ (where the $\dots$ are all $0$ )
$$\begin{cases} 
C_1 &= &( & 1 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & \dots &)\\
C_2 &= &( & 1 & 0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & \dots &)\\
C_3 &= &( & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & \dots &)\\ 
\end{cases}$$
Note that they form an Eventown, but their global intersection is odd, therefore the will never be from a partition of the ground set $X$.
Now, we use the fact that in Eventown, any maximal solution (not extendable) is maximum (of size $2^{n/2}$).
Therefore we will have a maximal solution not isomorphic to the partition solution.
