Let $Q\_{4n-1}$ be a unit [hypercube](https://en.wikipedia.org/wiki/Hypercube) of dimension $4n-1$. Has the following statement been proven?



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> There are $4n$ vertices in $Q\_{4n-1}$ such that the distance between each pair of them is $2\sqrt{2n}$.
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In other words, such vertices form a complete graph of equal sides (=$2\sqrt{2n}$).



Distance relation among points in high-dimensional hypercubes
You probably mean that the hypercube is $Q\_{4n-1}=\{-1,+1\}^{4n-1}$. If $u,v\in Q\_{4n-1}$ and $\|u-v\|=2\sqrt{2n}$, then $8n-2-2(u,v)=(u-v)^2=8n$,
$(u,v)=-1$. Add $(4n)$-th coordinate 1 to $u$, $v$. We get two vectors $U,V\in Q\_{4n}$ such that $(U,V)=(u,v)+1=0$. So your question reduces to the famous [Hadamard conjecture](https://en.wikipedia.org/wiki/Hadamard_matrix) about Hadamard matrices of order $4k$. Actually two questions are equivalent: if Hadamard's matrix exists, we may suppose without loss of generality that the last column consists only on 1's, and removing the last columns we get $4n$ rows of length $4n-1$ satisfying your property.


