Condition for dim of the Euclidean space with orthogonal basis

Question

I would like to show that if the orthogonal basis of the $\Bbb R^n$ Euclidean space with the standard dot product has the vectors whose elements are exclusively $1$ or $-1$, then $n \le 2$ or $n$ is multiple of $4$.

First, the dimension has be $ > 1$. Otherwise, the dot product would not work, since it is defined for such vectors $v_i$, $v_j$ that $i,j$ are not equal. So we immediately have $\dim = 2$.

Secondly, take two vectors $v_1$ and $v_2$ which in total have $w$ elements, then for the dot product to work there should at least $w/4$ and at most $w/2$ elements of $(-1)$ or $1$.

All I produced so far are such intuitions. Can anyone help with the formal proof?

Edit: some last minute thoughts. Actually, we cannot have $w/2$ elements of -1 or 1. This is because that could produce a vector whose entries are all 1's or (-1)'s. Then the only way to effectively dotproduct such a vector for odd dimensions would be to have another vector, which would be a -1 multiple of the first one. That would not make a basis however. Hence it collapses to even dimensions, and as we cannot have $w/2$ elements of -1 or 1, we can have $w/4.$.

Answer 1

So, as I understand it, you want to show that $(\mathbb{R}^n, \cdot)$ has an orthogonal basis consisting of vectors whose entries are all $\pm 1$ iff $n = 1$ or $n$ is even.

This is trivial for $n = 1$: since there's only one element in a basis for $\mathbb{R}^1$, every basis is orthogonal, so take either $\{1\}$ or $\{-1\}$.

For $n=2$, we may take the basis $\{(1, 1), (1, -1)\}$.

For $n \geq 3$ odd, notice that if the entries of $v$ and $w$ are all $\pm 1$, then $$v \cdot w = v_1 w_1 + \cdots + v_n w_n$$ is the sum of an odd number of terms, all of which are $\pm 1$, so the sum can't possibly be zero.

So now you just need to eliminate the cases $n \equiv 2 \pmod{4}$ and construct bases for the cases $n \equiv 0 \pmod{4}$.