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How to prove that the maximum number of mutually equidistant points in an n-dimensional Euclidean space is (n+1)?

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closed as too localized by Robin Chapman, Gjergji Zaimi, Yemon Choi, S. Carnahan Jul 2 '10 at 13:17

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Induction. The points equidistant to two points in R^n form a hyperplane which is isomorphic to R^{n-1}. – Qiaochu Yuan Jul 2 '10 at 5:00
how to prove the induction step then? thanks – Nick Jul 2 '10 at 5:18
Start with a point $P_1$, all the others must lie in some (n-1)-dimensional sphere. The same applies to $P_2$ and so $P_3,\dots$ will lie in an (n-2)-dimensional sphere. Iterate. This gives you a bound of $\le n+2$ but it is easy to see that when you reach the 1-dimensional sphere (two points) in the end, you can only pick at most one of them, and so you have the bound $\le n+1$ – Gjergji Zaimi Jul 2 '10 at 5:29
Put one point at the origin. The Gram matrix of $n$ further points equidistant from the origin and each other is nonsingular. – Robin Chapman Jul 2 '10 at 6:19
Prove by induction that any two sets of $n+1$ equidistant points in $\mathbb{R}^n$ are isometric. In the induction step, observe that the distance between the two possible choices of the last point is greater than $1$ (this requires a small computation). This proof is close to Gjergij's argument. – Victor Protsak Jul 2 '10 at 6:31

Assume one of the points is the origin, the others are the vectors a1,...,an+1. If the pairwise distance is 1, then $a^2_i=1$ and $(a_i-a_j)^2=1$ (i < j)(scalar multiplication). This gives $(a_i,a_j)=\frac{1}{2}$. Now show that $a_1,\dots,a_{n+1}$ are linearly independent: if $\lambda_1,\dots,\lambda_{n+1}$ are scalars and $\sum \lambda_ia_i=0$, then scalar multiplication by $a_i$ gives $\lambda_i+\frac{1}{2}\sum_{j\neq i}\lambda_j=0$, then we have $\lambda_i=-\Lambda$ for each $i$ where $\Lambda=\lambda_1+\cdots+\lambda_{n+1}$, summing gives $\Lambda=0$ so each $\lambda_i=0$.

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