41st Balkan Mathematical Olympiad

For $n=1$, the answer is negative. For $n>1$ we will show that answer is positive. In contrary, since $|S_n|=4^n$, we see that the set $S_n$ is complete system of remainders modulo $4^n$. Thus, $$\sum _{x\in S_n}{x}^3\equiv \sum _{i=1}^{4^n}{i}^3(\mathrm{mod}{4}^n)\equiv {\left(\frac{4^n(4^n+1)}{2}\right)}^2(\mathrm{mod}{4}^n)\equiv 0(\mathrm{mod}{4}^n).(♡)$$ Let us calculate ${\sum }_{x\in S_n}{x}^3$. Denote by $A_n={\sum }_{x\in S_n}x$, $B_n={\sum }_{x\in S_n}{x}^2$ and $C_n={\sum }_{x\in S_n}{x}^3$, for all $n\in \mathbb{N}$. Then: $$A_n=\sum _{k=1}^{n}10^{k-1}{4}^{n-1}(1+2+3+4)\equiv 2\cdot 4^{n-1}(\mathrm{mod}{4}^n),(1)$$ $$\begin{array}{rl}{B}_{n+1}& =\sum _{(c,x_n)\in \{1,2,3,4\}\times S_n}(10^n\cdot c+x_n{)}^2=\sum _{(c,x_n)\in \{1,2,3,4\}\times S_n}(10^{2n}\cdot c^2+2\cdot 10^n\cdot c\cdot x_n+x_n^2)\\ & =4^n\cdot 10^{2n}\cdot \sum _{c=1}^4{c}^2+2\cdot 10^n\cdot \sum _{c=1}^{4}c\cdot A_n+4\cdot B_n\equiv 4B_n(\mathrm{mod}{4}^{n+1}).\end{array}$$ Since $B_1=1^2+2^2+3^2+4^2=30\equiv 2(\mathrm{mod}4)$, by induction, it follows that $$B_n\equiv 2\cdot 4^{n-1}(\mathrm{mod}{4}^n)(2)$$ holds for all $n\in \mathbb{N}$.

So, by using (1) and (2), we have: $$\begin{array}{rl}{C}_{n+1}& =\sum _{(c,x_n)\in \{1,2,3,4\}\times S_n}(10^n\cdot c+x_n{)}^3\\ & =\sum _{(c,x_n)\in \{1,2,3,4\}\times S_n}(10^{3n}\cdot c^3+3\cdot 10^{2n}\cdot c^2\cdot x_n+3\cdot 10^n\cdot c\cdot x_n^2+x_n^3)\\ & =4^n\cdot 10^{3n}\cdot \sum _{c=1}^4{c}^3+3\cdot 10^{2n}\cdot \sum _{c=1}^4{c}^2\cdot A_n+3\cdot 10^n\cdot \sum _{c=1}^{4}c\cdot B_n+4\cdot C_n\\ & \equiv 0+0+3\cdot 10^{n+1}\cdot B_n+4C_n(\mathrm{mod}{4}^{n+1})\\ & \equiv 3\cdot 10^{n+1}\cdot (k\cdot 4^n+2\cdot 4^{n-1})+4C_n(\mathrm{mod}{4}^{n+1})\\ & \equiv 3\cdot 5^{n+1}\cdot 2^{n+2}\cdot 4^{n-1}+4C_n(\mathrm{mod}{4}^{n+1}).\end{array}\text{\hspace{1em}(3)(4)}$$ Since $C_1=100$, from (3), we obtain $C_2\equiv 1000(\mathrm{mod}16)\equiv 8(\mathrm{mod}16)$ and for $n\ge 2$, since $4^2\mid 2^{n+2}$, we get $$C_{n+1}\equiv 4C_n\equiv 4^{n+1}.(5)$$ Finally, from (5), by induction, we see that for all $n\in \mathbb{N}\setminus \{1\}$ $$C_n\equiv {}_{4n}2\cdot 4^{n-1}(\diamond )$$ holds and by (\cap) and (\diamond) we arrive at a contradiction.

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Alternative solution.

Alternative Solution. Assume such $x$ and $y$ do not exist and let $n>1$. Since $|S_n|=4^n$, we see that the set $S_n$ is complete system of remainders modulo $4^n$. Note that the numbers $$\overline{c_n{c}_{n-1}\dots c_{2}1},\overline{c_n{c}_{n-1}\dots c_{2}2},\overline{c_n{c}_{n-1}\dots c_{2}3},\overline{c_n{c}_{n-1}\dots c_{2}4}$$ give four consecutive remainders modulo $4^n$. Call such four consecutive remainders a block. Therefore all $4^n$ remainders modulo $4^n$ can be split into blocks. Thus, the difference of any two of the $4^{n-1}$ numbers $\overline{c_n{c}_{n-1}\dots c_{2}0}$ is congruent to $4t$ modulo $4^n$. But $$\overline{c_n{c}_{n-1}\dots c_{3}20}-\overline{c_n{c}_{n-1}\dots c_{3}10}=10,$$ implying $4t\equiv 10(\mathrm{mod}{4}^n)$, a contradiction.