TITLE: How prove can't exist a set $A$ of positive integers satisfying the two conditions
QUESTION [6 upvotes]: Question:

Show that there can't exist a set $A$ of positive integers such that the  following two condition hold:
(1) For every integer $m>1$, there exist $a,b\in A$ such $a+b=m$.
(2) If $a,b,c,d\in A$, $a,b,c,d>10$, and $a+b=c+d$, then we have $a=c$ or $a=d$.

My idea: Assume that there exists such a set $A$, and let $A(n)=|A\bigcap [1,n]|$.
Then by condition $(1)$, we have
$$\binom{A(n)}{2}\ge n-1 \Longrightarrow A(n)>\sqrt{2n}$$
but I can't use condition $(2)$.

REPLY [2 votes]: $\color{blue}{(2)}$ gives that $A_{\geq 10}$ is a Sidon set (aka Golomb ruler), and a theorem of Erdos and Turan  gives that
$$ S(n)\leq \sqrt{n}+O\left(n^{1/4}\right)\tag{A}$$
holds for any Sidon set $S$, so $A(n)\leq 10+\sqrt{n}+O(n^{1/4})$. However, the last inequality contradicts 
$$ A(n)\geq\sqrt{2n}\tag{B}$$
that is a consequence of $\color{blue}{(1)}$.