TITLE: Graph theory for friendship in three cities
QUESTION [5 upvotes]: There are three cities each of which has exactly the same number of citizens, say $n$. Every citizen in each city has exactly a total of $n+1$ friends in the other two cities. Show that there exists three people, one from each city, such that they are friends. We assume that friendship is mutual
ISI B MATH UGB 7/18/2021

I thought of doing this by induction , let $P(k)$ be the statement that the required is true for a city of size $k$
$P(1)$ is trivially true
assume $P(k)$ to be true , then for $P(k+1)$ we can imagine putting one person in each city. Now, when we do this, the previous friendship triplet still exists and hence $P(k+1)$ is true as well.
I discussed this with a friend and he criticized that the question since mathematical, there maybe a possibility that everyone gets amnesia when the new tourists are introduced and hence a new friendship pairs occur and the previous triplet vanishes. I thought it was a fair point, but then I thought, maybe if I could prove that the friendships for three cities of size $n$ is unique, then my proof would hold... but I don't think this is the case.
Can I salvage the proof somehow? If not, how would I do it correctly?

REPLY [7 votes]: Given a person $p$ let $f(p)$ be the minimum of the number of friends that $p$ has on each of the other two cities. Select $p$ such that $f(p)$ is minimized and let $k=f(p)$ and without loss of generality $p$ is in city $1$ and has $k$ friends in city $2$. We deduce $p$ has exactly $n+1−k$ friends in city $3$.
Pick a friend $q$ of $p$ in city $2$. Notice that $q$ has at least $k$ friends in city $3$ and since $p$ has exactly $n+1-k$ friends in city $3$ there must be a friend in city $3$ that is common to $p$ and $q$.