45th Mongolian Mathematical Olympiad

Consider the following polynomials: $$a(x)=x^{a_1}+x^{a_2}+\cdots +x^{a_n}$$ $$b(x)=x^{b_1}+x^{b_2}+\cdots +x^{b_n}$$ Now assume the contrary, in other words $a(x)\ne b(x)$. $$\begin{array}{rl}[a(x){]}^3& =\left(\sum _{i=1}^n{x}^{a_i}\right)\cdot \left(\sum _{i\ne j}{x}^{a_i+a_j}+\sum _{i=1}^n{x}^{2a_i}\right)=\\ & =\left(\sum _{i=1}^n{x}^{a_i}\right)\left(\sum _{i\ne j}{x}^{a_i+a_j}\right)+\left(\sum _{i\ne j}{x}^{a_i}\right)\left(\sum _{i=1}^n{x}^{2a_i}\right)=\\ & =\left(\sum _{i\ne j\ne k}{x}^{a_i+a_j+a_k}+2\sum _{i\ne j}{x}^{a_i+2a_j}\right)+\left(\sum _{i\ne j}{x}^{a_i+2a_j}+\sum _{i=1}^n{x}^{3a_i}\right)\\ & \left(\sum _{i\ne j\ne k}{x}^{a_i+a_j+a_k}\right)=[a(x){]}^3-3a(x)\cdot a(x^2)+2a(x^3)\end{array}$$

Analogously, we have $[b(x){]}^3=[b(x){]}^3-3b(x)\cdot b(x^2)+2b(x^3)$. From the given condition, we get the following equality: $$[a(x){]}^3-3a(x)\cdot a(x^2)+2a(x^3)=[b(x){]}^3-3b(x)\cdot b(x^2)+2b(x^3)(\ast )$$

Now consider $f(x)=a(x)-b(x)$ polynomials. Therefore $f(1)=a(1)-b(1)=n-n=0$. Thus we can write $f$ the following way: $$f(x)=(x-1{)}^k\cdot h(x),h(1)\ne 0(\ast \ast )$$ Also, by () we have $$\begin{gathered} (x-1{)}^k\cdot h(x)([a(x){]}^2+a(x)\cdot b(x)+[b(x){]}^2)- \\ -3\left(a(x)(x^2-1{)}^k\cdot h(x^2)+b(x^2)(x-1{)}^k\cdot h(x)\right)+ \\ +2\cdot (x^3-1{)}^k\cdot h(x^3)=0. \end{gathered}$$ The last equation is the same as the following: $$\begin{gathered} h(x)([a(x){]}^2+a(x)\cdot b(x)+[b(x){]}^2)- \\ -3\left(a(x)\cdot (x+1{)}^k\cdot h(x^2)+b(x^2)\cdot h(x)\right)+ \\ +2\cdot (x^2+x+1{)}^k\cdot h(x^3)=0.(\ast \ast \ast ) \end{gathered}$$ In the equation (**) substituting $x=1$, we get $3n^2-3(n\cdot 2^k+n)+2\cdot 3^k=0$ which is the same as $n^2-n\cdot (2^k+1)+2\cdot 3^{k-1}=0$. Therefore, this implies that $n|2\cdot 3^{k-1}$. This is a contradiction with $(n,6)=1$.