Putnam 2022, Problem B-2

Let $\times$ represent the cross product in $\mathbb{R}^3$. For what positive integers $n$ does there exist a set $S \subset \mathbb{R}^3$ with exactly $n$ elements such that $S = \{v \times w : v, w \in S\}$?

Solution:
The possible values of \( n \) are 1 and 7.  

Clearly the set \( S = \{0\} \) works. Suppose that \( S \neq \{0\} \) is a finite set satisfying the given condition; in particular, \( S \) does not consist of a collection of collinear vectors, since otherwise \( \{v \times w : v, w \in S\} = \{0\} \). We claim that \( S \) cannot contain any nonzero vector \( v \) with \( \|v\| \neq 1 \). Suppose otherwise, and let \( w \in S \) be a vector not collinear with \( v \). Then \( S \) must contain the nonzero vector \( u_1 = v \times w \), as well as the sequence of vectors \( u_n \) defined inductively by \( u_n = v \times u_{n-1} \). Since each \( u_n \) is orthogonal to \( v \) by construction, we have \( \|u_n\| = \|v\| \|u_{n-1}\| \) and so \( \|u_n\| = \|v\|^{n-1} \|u_1\| \). The sequence \( \|u_n\| \) consists of all distinct numbers and thus \( S \) is infinite, a contradiction. This proves the claim, and so every nonzero vector in \( S \) is a unit vector.  

Next note that any pair of vectors \( v, w \in S \) must either be collinear or orthogonal: by the claim, \( v, w \) are both unit vectors, and if \( v, w \) are not collinear then \( v \times w \in S \) must be a unit vector, whence \( v \perp w \). Now choose any pair of non-collinear vectors \( v_1, v_2 \in S \), and write \( v_3 = v_1 \times v_2 \). Then \( \{v_1, v_2, v_3\} \) is an orthonormal basis of \( \mathbb{R}^3 \), and it follows that all of these vectors are in \( S \): \( 0, v_1, v_2, v_3, -v_1 = v_3 \times v_2, -v_2 = v_1 \times v_3, -v_3 = v_2 \times v_1 \). On the other hand, \( S \) cannot contain any vector besides these seven, since any other vector \( w \) in \( S \) would have to be simultaneously orthogonal to all of \( v_1, v_2, v_3 \).  

Thus any set \( S \neq \{0\} \) satisfying the given condition must be of the form \( \{0, \pm v_1, \pm v_2, \pm v_3\} \) where \( \{v_1, v_2, v_3\} \) is an orthonormal basis of \( \mathbb{R}^3 \). It is clear that any set of this form does satisfy the given condition. We conclude that the answer is \( n = 1 \) or \( n = 7 \).