Hi, Omid Hatami! Are you sure that statement is correct? Maybe I am wrong, but it seems I disprove it.

Yes I'm sure that problem is correct. It was our exam last week.

yes the main idea of this problem was given from the paper from erdos about random graph (just a liittle manipulation)

Of course the official solution is easy using Dirichlet theorem in number theory

If you say that problem is correct, it means that I still don't understand statement.

Well I mean that we ahve a relation and $A \subset \mathbb{N}$ we say $A$ is golden if and only if for evey $U,V \subset A$ that $U \cap V= \emptyset$ then thre exist $z \in A$ that z has relation with every $x \in U$ and does'nt have relation with every $y\in V$

Thank you, but the same is written above

Well I don't realize where is the problem. Explain more that what don't you understand.

Let $P_1=\{5,7\}$. Set $P_1$ is not golden (U={5}, V={7}). Let $P_2=P\setminus P_1$. Suppose $P_2$ is golden. Consider $U=\{13\}\cup \{a\in P_2\,|,13\circ a\}$, $V=\{11\}$. We see $13\bullet 11$ and $2\not\in U$. For all remaining prime numbers we have $p \bullet p$, since binary representation of $p$ has 0 on p-th place. So if $z\circ x\ \forall x\in U$ then $z\circ 13$ $\Rightarrow$ $z\in U$, but $z\bullet z$ -- contradiction. What is wrong?

Excuse me I didn't say that $U,V$ must be finite but I thought I had said.

I am very angry because both Omid and Sam-n said that statement is correct... rrrrrr... Now problem become very clear and really simple if we are allowed to use Dirichlet theorem.

oh myth i think it is part of default (being finite ) anyway sorry