Orthogonal basis of $\mathbb R^n$ consisting of $\pm 1$ vectors

Question

I have to prove that if orthogonal basis of $\mathbb R^n$ containing only vectors which coordinates are $1$ or $-1$ exists then $n \leqslant 2$ or $n$ is divisible by 4.

It's obvious that $n$ has to be even but I have no idea what to do next. I'm not allowed to use determinants.

Answer 1

Since exchanging the columns, and flipping the sign of an entire column, preserve orthogonality, we may assume that

1. The top row of the matrix is all-ones.
2. The second row is $n/2$ ones followed by $-1$s.

So far so good, but what should the third row be? Let $k$ be the number of $1$s in the first half of the row. There must be $k$ negative entries in the second half. But then the scalar product of the second and third rows is $$ k -(n/2-k) +k-(n/2-k) = 4k-n $$ which is not zero.