Prove that $n+2$ points in $\Bbb R^n$ cannot all be at a unit distance from each other. I saw this on the FB group "Actually good math problems" where one solution was an illegible photo and the other involved intersecting spheres. I have a solution. I'd like to see how many good answers I get before I post it.
If $S$ is a set of $n+2$ points in $\Bbb R^n$ (with $n\in \Bbb N$) with the usual (Cartesian) norm, show there exist $2$ distinct $p,q\in S$ with $\|p-q\|\ne 1.$
 A: Let us assume that $\{P_1,\ldots,P_{n+2}\}$ is a set of points in $\mathbb{R}^n$ such that $\|P_i-P_j\|=1$ for any $i\neq j$.
We may assume without loss of generality that $P_{n+2}=O$. From the polarization formula
$$ 2\cos\theta_{ij}=2\langle P_i, P_j \rangle = \|P_i-P_j\|^2 + \|P_i+P_j\|^2 = 1+\|P_i+P_j\|^2\geq 1$$
it follows that $\widehat{P_i O P_j}\geq \frac{\pi}{3}$ and $\langle P_i,P_j\rangle \geq \frac{1}{2}$ for any $i\neq j$ such that $i,j\in{1,\ldots,n+1}$. We have
$$ n=\sum_{k=1}^{n}\|P_{n+1}-P_k\|^2=2n-2\sum_{k=1}^{n}\langle P_k,P_{n+1}\rangle \leq n$$
holding as a strict inequality unless $\langle P_{n+1},P_k\rangle=\frac{1}{2}$ for any $k\in\{1,\ldots,n\}$. This implies that $P_1,\ldots,P_n$ all lie 
on a affine hyperplane $\pi$, while $P_{n+1}$ and $P_{n+2}$ lie on opposite sides of it, since $\langle P_{n+1},P_{n+1}\rangle = 1, \langle P_{n+1},P_{n+2}\rangle=0$. These equalities give that $O=P_{n+2}$ and $P_{n+1}$ are  actually symmetric with respect to $\pi$. Let $M\in \pi$ be the midpoint of $P_{n+1} P_{n+2}$. By the Pythagorean theorem $P_k M=\frac{1}{2}\sqrt{3}$ for any $k\in\{1,\ldots,n\}$, so we have $n$ points, $1$-apart from each other, on a $S^{n-2}$ with radius $\frac{\sqrt{3}}{2}$ in $\mathbb{R}^{n-1}\simeq \pi$. Let us consider $P_1$ and $P_2$: all the points $P_3,\ldots,P_n$ have to lie on the previous sphere and on the spheres centered at $P_1,P_2$ with unit radius. These spheres do not have a common intersection, so we cannot have $n+2$ points in $\mathbb{R}^n$ which are $1$-apart from each other.
