USA Team Selection Test

Let $n \ge 4$ be an integer. We need to find all functions $W : \{1, \dots, n\}^2 \to \mathbb{R}$ such that for every partition $[n] = A \cup B \cup C$ into disjoint sets, the following condition holds:
$$\sum_{a \in A} \sum_{b \in B} \sum_{c \in C} W(a,b) W(b,c) = |A| |B| |C|.$$

To solve this, we denote the function $W$ by $f$ for simplicity. We start by considering specific partitions of $[n]$.

First, consider the partition $P(\{1\}, \{2\}, \{3, 4, 5, \ldots\})$:
$$f(1,2)f(2,4) + f(1,2)f(2,5) + \cdots + f(1,2)f(2,n) = (n-2) - f(1,2)f(2,3).$$

Next, consider the partition $P(\{1\}, \{3\}, \{2, 4, 5, \ldots\})$:
$$f(1,3)f(3,4) + f(1,3)f(3,5) + \cdots + f(1,3)f(3,n) = (n-2) - f(1,3)f(3,2).$$

Now, consider the partition $P(\{1\}, \{2, 3\}, \{4, 5, \ldots\})$:
$$f(1,2)f(2,4) + f(1,2)f(2,5) + \cdots + f(1,2)f(2,n) + f(1,3)f(3,4) + f(1,3)f(3,5) + \cdots + f(1,3)f(3,n) = 2(n-3).$$

This simplifies to:
$$(n-2) - f(1,2)f(2,3) + (n-2) - f(1,3)f(3,2) = 2(n-3) \implies f(1,2)f(2,3) + f(1,3)f(3,2) = 2.$$

Similarly, for any distinct $a, b, c$:
$$f(a,b)f(b,c) + f(a,c)f(c,b) = 2.$$

Considering $P(\{3, 4, 5, \ldots\}, \{2\}, \{1\})$, $P(\{2, 4, 5, \ldots\}, \{3\}, \{1\})$, and $P(\{4, 5, \ldots\}, \{2, 3\}, \{1\})$, we get:
$$f(3,2)f(2,1) + f(2,3)f(3,1) = 2,$$
which generalizes to:
$$f(a,b)f(b,c) + f(b,a)f(a,c) = 2.$$

Thus, we see that:
$$f(a,c)f(c,b) = f(b,a)f(a,c) \implies f(a,c) = 0 \quad \text{or} \quad f(c,b) = f(b,a).$$

Suppose $f(a,c) = 0$ for some $a \neq c$. Considering $P(\{a\}, \{c\}, [n] - \{a,c\})$, we get $0 = n-2$, a contradiction. Hence, $f(c,b) = f(b,a)$ for all distinct $a, b, c$.

This implies that $f(x,b)$ and $f(b,x)$ are constants for each $b$ and all $x \neq b$. Consequently, $f(x,y)$ is a constant $k$ if $x \neq y$. From the condition $2k^2 = 2$, we find $k = 1$ or $k = -1$.

Therefore, all solutions are of the form where $W(a,a)$ can be any value, and for all distinct $a, b \in [n]$, $W(a,b)$ equals a constant $k$, where $k = 1$ or $k = -1$.

The answer is: $\boxed{W(a,b)}$ = k $\text{}$ for all distinct } a, b $\text{}$ and } k = 1 $\text{}$ or } k = -1.}