Any answer? For 2 years?

I think answer is 1007.1008 Generally: If n=2x$\implies$ x.x, If n=2x+1$\implies$ x.(x+1) n=5 clearly 6 And if we will make induction on n it's not hard to obtain this conclusion.(I chose the student who has least handshakes but maybe there exist another ways)

@above what are you meaning about 1007.1008?

1007*1008

Math-wiz wrote: 1007*1008 Thx

Answer:$\lfloor \frac{n}{2} \rfloor \lfloor \frac{n+1}{2} \rfloor$ for all $n\in \mathbb Z$. Example: A bipartite graph which two sides have $\lfloor \frac{n}{2} \rfloor$ and $\lfloor \frac{n+1}{2} \rfloor$ members. Proof: We will use induction. It's true for $n=1,2,3,4,5,6$. Let it be true for $n=1,2,...,2k$. Let $A$ be the person who has at least friends. The graph without $A$ be $S$.$|S|=k$ so we claim that there are at most $k$ edges between $S$ and $A$. Let $\deg A\geq k+1$. So every vertices degree is larger than $k$.Take two people $X$ and $Y$ such that they are friends. They have at least one common friend. If $X$ and $Y$ have one common friend: Let the common friend be $Z$.Let $1,2,...,k-1,Y,Z$ be the friends of $X$ and $k+1,...,2k-2,X,Z$ be the friends of $Y$. $X,Y,Z$ is a triangle. $1$ has at least $k$ more friends. We have $k\geq 2$, then take two friends of $Z$ and $X,Y,Z$. There are at least $7$ edges between them which gives us a contradiction. If $X$ and $Y$ have two common friends: Let the common friends be $Z$ and $T$. $k\geq 3$. $Z$ has at least $k-1$ more friends. If $X$ and $Z$ or $Y$ and $Z$ have a common friend other than $X,Y,T$, then there are at least $7$ edges between those $5$ vertices which gives us a contradiction. Also, if $Z$ and $T$ are friends, then we get a contradiction again. So $k-2\leq 1$ but we know that $k\geq 4$. If $X$ and $Y$ have three or more common friends, then take three friends of them and $X,Y$. There should be at least $7$ edges which gives a contradiction. We can do the same things for the induction from $2k+1$ to $2k+2$ by writing $k+1$ instead of $k$.