Mongolian Mathematical Olympiad

Setting $|A|=n$, $|B|=m$ we get $|A^{\ast }|=C_n^2$, $|B^{\ast }|=C_m^2$ and since $A^{\ast }=B^{\ast }$, $C_n^2=C_m^2\Rightarrow n=m$. Let $A=\{a_1,a_2,...,a_n\}$; $B=\{b_1,b_2,...,b_n\}$. Consider polynomials $f(x)=x^{a_1}+x^{a_2}+...+x^{a_n}$; $g(x)=x^{b_1}+x^{b_2}+...+x^{b_n}$. From $A^{\ast }=B^{\ast }$ follows that $f(x{)}^2-f(x^2)=g(x{)}^2-g(x^2)$. Obviously $f(1)=g(1)=n$. Since $f(1)-g(1)=n-n=0$, there exists polynomial $h(x)$ such that $f(x)-g(x)=(x-1{)}^{k}h(x)$, $k\ge 1$, $h(1)\ne 0$. Furthermore we get $f(x)+g(x)=$ $$\frac{f(x{)}^2-g(x{)}^2}{f(x)-g(x)}=\frac{f(x^2)-g(x^2)}{f(x)-g(x)}=\frac{(x^2-1{)}^{k}h(x^2)}{(x-1{)}^{k}h(x)}\text{ and }f(x)+g(x)=(x+1{)}^k\frac{h(x^2)}{h(x)}$$ Setting $x=1$ in the latter follows that $n=2^{k-1}$ and thus we are done.