I'd like to know if this is a valid proof. It is the friendship Friendship Puzzle, and i understand that while this question exists,i still want to know if my method is valid
The theorem is that:
At any party of two or more people, show that there are always two people with exactly the same number of friends at the party.
My attempt:
Assume the party has n people. There two possible cases, a person, x, has at least one friend, or a person has no friends.
case 1: at least one friend. Then x can have up to n-1 friends. Since everyone at the party can have up to n-1 friends, and the party has n people, by the pigeonhole principle, at least two people have n-1 friends.
case 2:
then for n-1 people at the party (excluding x), everyone can have up to n-2 friends (excluding themselves and x). Similary by the Pigeonhole principle, at least two people have the same number of friends.
We are done.
As stated in the comments, I got OP to realize that in their case 2, they have $n-1$ pigeons and $n-1$ holes, so can't quite apply Pigeonhole principle and get the stated result.
I suggested that to complete the proof, they could proceed by induction, or proceed in a similar manner (which would essentially be induction).