Let $\mathbf{1}$ be the all-ones vector, and suppose $\mathbf{1}, \mathbf{v_1}, \mathbf{v_2}, \ldots, \mathbf{v_{n-1}} \in \{-1,0,1\}^n$ are mutually orthogonal non-zero vectors. Does it follow that $n \in \{1,2\} \cup \{4k : k \in \mathbb{N}\}$?

A few notes are in order:

1. If we do not allow "$0$" entries in the vectors then this is a well-known necessary condition for the existence of an $n \times n$ Hadamard matrix, so the answer to my question is "yes" if we replace $\{-1,0,1\}$ with$\{-1,1\}$.

2. With some computer help, I have shown that there do not exist collections of vectors with these properties when $n = 3$, $5$, or $6$, so the answer to my question is "yes" when $n \leq 6$.

Let $\mathbf{1}$ be the all-ones vector, and suppose $\mathbf{1}, \mathbf{v_1}, \mathbf{v_2}, \ldots, \mathbf{v_{n-1}} \in \{-1,0,1\}^n$ are mutually orthogonal non-zero vectors. Does it follow that $n \in \{1,2\} \cup \{4k : k \in \mathbb{N}\}$?

A few notes are in order:

1. If we do not allow "$0$" entries in the vectors then this is a well-known necessary condition for the existence of an $n \times n$ Hadamard matrix, so the answer to my question is "yes" if we replace $\{-1,0,1\}$ with$\{-1,1\}$.

2. With some computer help, I have shown that there do not exist collections of vectors with these properties when $n = 3$, $5$, or $6$, so the answer to my question is "yes" when $n \leq 6$.

Edit (May 19, 2023): In the comments, Max Alekseyev has shown that the answer is "yes" when $n \leq 12$, and Ilya Bogdanov has shown that the answer is "yes" when $n$ is an odd prime.

Let $\mathbf{1}$ be the all-ones vector, and suppose $\mathbf{1}, \mathbf{v_1}, \mathbf{v_2}, \ldots, \mathbf{v_{n-1}} \in \{-1,0,1\}^n$ are mutually orthogonal vectors. Does it follow that $n \in \{1,2\} \cup \{4k : k \in \mathbb{N}\}$?

A few notes are in order:

1. If we do not allow "$0$" entries in the vectors then this is a well-known necessary condition for the existence of an $n \times n$ Hadamard matrix, so the answer to my question is "yes" if we replace $\{-1,0,1\}$ with$\{-1,1\}$.

2. With some computer help, I have shown that there do not exist collections of vectors with these properties when $n = 3$, $5$, or $6$, so the answer to my question is "yes" when $n \leq 6$.

Let $\mathbf{1}$ be the all-ones vector, and suppose $\mathbf{1}, \mathbf{v_1}, \mathbf{v_2}, \ldots, \mathbf{v_{n-1}} \in \{-1,0,1\}^n$ are mutually orthogonal non-zero vectors. Does it follow that $n \in \{1,2\} \cup \{4k : k \in \mathbb{N}\}$?

A few notes are in order:

1. If we do not allow "$0$" entries in the vectors then this is a well-known necessary condition for the existence of an $n \times n$ Hadamard matrix, so the answer to my question is "yes" if we replace $\{-1,0,1\}$ with$\{-1,1\}$.

2. With some computer help, I have shown that there do not exist collections of vectors with these properties when $n = 3$, $5$, or $6$, so the answer to my question is "yes" when $n \leq 6$.