$65$ couples go on a spring outing, there are a total of $65$ motorcycles, each motorcycle carries one boy and one girl. Suppose for any two motorcycles, the following two propositions are exactly one true:
(1) The boys on these two motorcycles know each other.
(2) The boyfriends of the girls on these two motorcycles know each other.
Prove that there must be a couple sitting together on the same motorcycle.
My attempt:
Using proof by contradiction, let's assume that there does not exist a pair of lovers on the same motorcycle.
The couples are numbered from $1$ to $65$, and we define the function $f(a) = b$ to represent the number of the boy on the motorcycle where the girlfriend in the couple $a\ (1 \le a \le 65)$ is seated.
Consider the cycle formed by iteratively applying $f$: $a\to f(a) \to f(f(a)) \to \cdots \to \ f^{(k)}(a) = a$
Thus, we partition the $65$ pairs of lovers into several cycles.
Since $65$ is an odd number, there must be a circle of odd length, let's assume it is: $a_1 \to a_2 \to \cdots \to a_k \to a_1 \enspace (k \ \text{is odd})$
If $k=1$, the conclusion is obviously true.
Therefore, we only need to discuss the case where $k>1$.
Your attempt gave me some ideas. I think this problem is handled by finding the proper labeling.
Denote each of the $65$ motorcycles by $(1,b_1),(2,b_2),(3,b_3),...,(65,b_{65})$, where the first components encode the girls, and second boys, so that any girl and boy of same label are a couple. (although this encoding will only be used much later), where $b_i\in\{1,2,...,65\}$ and $b_i\neq b_j$ whenever $i\neq j$.
Suppose that the conclusion is false. Hence, there cannot be a motorcycle of the form $(a,a)$.
Define a graph $G$ such that all motorcycles are the vertices, and two vertices $(a,b)$ and $(c,d)$ are adjacent if and only if $|\{a,b\}\cap\{c,d\}|=1$.
Hence by our motorcycle labeling, any $(a,b)$ is adjacent to at most two vertices $(b,c)$ and $(d,a)$ for some $c,d$.
We claim that each component of this graph is either an isolated edge or a cycle. To see this, start with any $(x_1,x_2)$, which is adjacent to $(x_2,x_3)$ for some $x_3$, which is adjacent to $(x_3,x_4)$ for some $x_4$, and so on. Continuing this observation and by finiteness, we end at a vertex $(x_i,x_j)$, where $1\leq j<i$. If $j>1$, then $(x_i,x_j)$ is adjacent to $(x_j,x_{j+1})$, which is also adjacent to $(x_{j-1},x_j)$ and $(x_{j+1},x_{j+2})$, which is impossible since one vertex is only adjacent to at most two others, unless $x_{j+2}=x_j$, which is again impossible. Hence, $j=1$. If there is an outside vertex to be adjacent to a vertex in this sequence, it will give a vertex of degree $3$ if the sequence is already a cycle, or if it is to an isolated edge, it will have one of its components equal to the corresponding component of one of the endpoints. Both are again impossible. Hence, the claim is shown.
Since there are odd number of vertices, there is an odd cycle. We can consider an odd cycle which must take the form $(x_1,x_2)(x_2,x_3)...(x_k,x_1)$ by our construction for the claim. Now, you can start with any edge. For simplicity and WLOG, you can start with the edge $(x_{k-1},x_k)(x_k,x_1)$. Assume that this edge satisfies rule (1), meaning that boys $x_k,x_1$ know each other. Hence, it must not satisfy rule (2), which implies boys $x_{k-1},x_k$ do not know each other. This however means the edge before, which is $(x_{k-2},x_{k-1})(x_{k-1},x_k)$, must satisfy rule (2), etc. If you assume the starting edge satisfies rule (2) instead, you can derive the similar conclusion.
Correction/improvement is much appreciated!