Saudi Arabia Mathematical Competitions 2012

If $a_1<a_2<\cdots <a_n$ are the elements of $A$, then $a_1^2+a_2^2,a_1^2+a_3^2,\dots ,a_1^2+a_n^2,a_2^2+a_n^2,\dots ,a_{n-1}^2+a_n^2$ belong to $A$, and we have $a_1^2+a_2^2<a_1^2+a_3^2<\cdots <a_1^2+a_n^2<a_2^2+a_n^2<\cdots <a_{n-1}^2+a_n^2$, which implies $2n-3\le n$. It follows that $A$ has at most 3 elements. If $n=3$ and $A=\{a,b,c\}$ with $a<b<c$, then $a^2<b^2<c^2$, hence $a^2+b^2<a^2+c^2<b^2+c^2$. Since $A=\{a,b,c\}=\{a^2+b^2,a^2+c^2,b^2+c^2\}$, it follows $$a^2+b^2=a,a^2+c^2=b,b^2+c^2=c.(1)$$ From the first and the last relation in (1) we get $a^2-c^2=a-c$, hence $a+c=1$. Substituting in the second relation of (1) it follows $$b=a^2+(1-a{)}^2=2a^2-2a+1=2(a^2-a)+1=-2b^2+1,$$ therefore $(2b-1)(b+1)=0$, hence $b=\frac{1}{2}$. Then $a^2-a+\frac{1}{4}=0$ leads to $a=\frac{1}{2}$, which contradicts $a<b$.