<s>[INST] $2n \times 2n$ matrices with entries in $\{1, 0, -1\}$ with exactly $n$ zeroes in each row and each column with orthogonal rows and orthogonal columns I am interested in answering the following question:
Question
For a given $n$, does there exist a $2n \times 2n$ matrix with entries in $\{1, 0, -1\}$ having orthogonal rows and columns with exactly $n$ zeroes in each row and column?
Conjectures

*

*For $n=2^k$, $k\ge0$ such a matrix always exists.


*For $n=3$ such a matrix does not exist.


*For $n=5$ such a matrix exists, for example:
$$
\left(
\begin{array}{cccccccccc}
 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 \\
 1 & 0 & 0 & 0 & -1 & 0 & -1 & -1 & -1 & 0 \\
 1 & 0 & 0 & -1 & 0 & 0 & -1 & 1 & 1 & 0 \\
 0 & -1 & -1 & 1 & 1 & 0 & -1 & 0 & 0 & 0 \\
 1 & 0 & -1 & 0 & 0 & -1 & 1 & 0 & 0 & -1 \\
 1 & -1 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 1 \\
 0 & 0 & 0 & -1 & 1 & -1 & 0 & -1 & 0 & 1 \\
 0 & -1 & 1 & 0 & 0 & 0 & 0 & -1 & 1 & -1 \\
 0 & -1 & 1 & 0 & 0 & -1 & 0 & 1 & -1 & 0 \\
 0 & 0 & 0 & -1 & 1 & 1 & 0 & 0 & -1 & -1 \\
\end{array}
\right)
$$


*For $n=7$ such a matrix does not exist.
 [/INST] There is no such matrix if $n\equiv 3\pmod 4$.
Suppose otherwise. Each column represents a vector of length $\sqrt n$. Since those vectors are pairwise orthogonal, their sum is a vector whose scalar square is $2n^2$.
On the other hand, the sum of all columns has odd entries, so their squares are all  congruent to $1$ modulo $8$. Hence the sum of those squares is congruent to $2n$ modulo $8$. Hence we should have $2n^2\equiv 2n\pmod 8$, or $8\mid 2n(n-1)$ which does not hold.
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