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Let $Q_{4n-1}$ be a unit hypercube of dimension $4n-1$. Has the following statement been proven?

There are $4n$ vertices in $Q_{4n-1}$ such that the distance between each pair of them is $2\sqrt{2n}$.

In other words, such vertices form a complete graph of equal sides (=$2\sqrt{2n}$).

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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 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.

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  • $\begingroup$ Thank you. I know that this conjecture has been proved for $4n = 2^km$ with $m^2 \leq 2^k$. where can I find this special case? $\endgroup$ – C.F.G Oct 1 '17 at 11:20

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