Mongolian Mathematical Olympiad

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

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:
$$[a(x)]^3 - 3a(x) \cdot a(x^2) + 2a(x^3) = [b(x)]^3 - 3b(x) \cdot b(x^2) + 2b(x^3) \quad (*)$$

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:
$$f(x) = (x-1)^k \cdot h(x), \quad h(1) \neq 0 \qquad (**)$$
Also, by (*) we have
$$(x-1)^k \cdot h(x) ([a(x)]^2 + a(x) \cdot b(x) + [b(x)]^2) - \\ -3 \left( a(x)(x^2 - 1)^k \cdot h(x^2) + b(x^2)(x-1)^k \cdot h(x) \right) + \\ +2 \cdot (x^3 - 1)^k \cdot h(x^3) = 0.$$
The last equation is the same as the following:
$$h(x) ([a(x)]^2 + a(x) \cdot b(x) + [b(x)]^2) - \\ -3 \left( a(x) \cdot (x+1)^k \cdot h(x^2) + b(x^2) \cdot h(x) \right) + \\ +2 \cdot (x^2 + x + 1)^k \cdot h(x^3) = 0. \qquad (***)$$
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 $(n, 6) = 1$.