Graph theory problem , acquaintances Prove that in a group of 60 people one can always find two people with even number of common acquaintances. 
I just want a small hint to this problem , not a full solution .
 A: I've tried to give a simple hint, but things got complicated, so there are several hints, and the solution is divided into two parts (so you can just check one of them if you are stuck again).
Hint 1:

 Prove that in a group of $2n$ there exist two people with even number of common acquaintances.

Hint 2:

 Assume that any pair has odd number of common acquaintances and proof that everybody has an even number of friends.

Solution part I:

 Denote by $F(x)$ the set of friends of $x$.
 Assume that $$\forall x \neq y.\ \big|F(x) \cap F(y)\big| \text{ is odd}. \tag{$\spadesuit$}$$
 Observe that $\sum_{y \in F(x)} |F(x) \cap F(y)|$ is even, because $z \in F(y)$ implies $y \in F(z)$ (in the sum both are acquaintances of $x$). However, by $(\spadesuit)$ we know that each intersection is of odd size and so $|F(x)|$ is even.

Hint 3:

 Prove that in a group of $2n$ for any person there exist another such that they share an even number of acquaintances.

Hint 4:

 Consider $$\sum_{y \in F(x)} \big|F(y) \setminus \{x\}\big|$$ and change the summation order (now you sum people other than $x$ by friends of $x$, try to sum friends of $x$ per each person other than $x$).

Solution part II:

 Consider $$\sum_{y \in F(x)} \big|F(y) \setminus \{x\}\big|.$$
 It's even, because $|F(x)|$ is even and $|F(y)\setminus \{x\}|$ is odd. 
 This stays true when we change the summation order 
 $$ \sum_{z \neq x} \Big|\big\{y \in F(x)\ \big|\ z \in F(y) \big\}\Big| = \sum_{z \neq x} \Big|F(x) \cap F(z)\Big|, $$ 
 but by $(\spadesuit)$ the sum of $2n-1$ elements of odd size has to be odd,
 contradiction. $\blacksquare$

I hope this helps $\ddot\smile$
