Let $A$ be a $0$-$1$ matrix of order $n$. Is the following assertion true?
Conjecture: $ AA^T\equiv 0\pmod{4}$ implies that $rank_2(A)\le \lfloor\frac{n}{4}\rfloor$, where $0$ is the zero matrix, $A^T$ is the transpose of $A$, and $rank_2(A)$ is the rank of $A$ over the binary field.
It is easy to construct an $0$-$1$ matrix $A$ with $rank_2(A)= \lfloor\frac{n}{4}\rfloor$. Indeed, the following constructions for small orders can be easily generalized to any order:
\begin{pmatrix} 1&1&1&1\\ 0&0&0&0\\ 0&0&0&0\\ 0&0&0&0\\ \end{pmatrix} \begin{pmatrix} 1&1&1&1&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ 0&0&0&0&0\\ \end{pmatrix} \begin{pmatrix} 1&1&1&1&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ 0&0&0&0&0&0\\ \end{pmatrix} \begin{pmatrix} 1&1&1&1&0&0&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0\\ \end{pmatrix} \begin{pmatrix} 1&1&1&1&0&0&0&0\\ 0&0&0&0&1&1&1&1\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0\\ \end{pmatrix} This means that the above conjecture is best possible (if true). Note that a similar assertion is a simple exercise in Linear Algebra: $ AA^T\equiv 0\pmod{2}$ implies that $rank_2(A)\le \lfloor\frac{n}{2}\rfloor$
How to prove the conjecture above?