Middle European Mathematical Olympiad, 2020

Solution:
- $n=1$ :
Here, $x_{1}:=1$ provides a solution, since
$$\frac{1}{1^{2}}=1$$

- $n=2$ :
Here, no solution exists. Indeed, $x_{1}=1$ or $x_{2}=1$ yields $\frac{1}{x_{1}^{2}}+\frac{2}{x_{2}^{2}}>1$, while $x_{1}, x_{2} \geq 2$ leads to
$$\frac{1}{x_{1}^{2}}+\frac{2}{x_{2}^{2}} \leq \frac{1}{4}+\frac{2}{4}=\frac{3}{4}<1$$

- $n=4$ :
Here, $(x_{1}, x_{2}, x_{3}, x_{4}):=(3,3,3,6)$ provides a solution, since
$$\frac{1}{3^{2}}+\frac{2}{3^{2}}+\frac{4}{3^{2}}+\frac{8}{6^{2}}=\frac{7}{9}+\frac{8}{36}=\frac{7}{9}+\frac{2}{9}=1$$

- Induction step from $n$ to $(n+2)$ :
Let $(y_{1}, y_{2}, \ldots, y_{n})$ be a solution for $n$, i.e.,
$$\frac{1}{y_{1}^{2}}+\frac{2}{y_{2}^{2}}+\frac{4}{y_{3}^{2}}+\cdots+\frac{2^{n-1}}{y_{n}^{2}}=1$$
Then
$$(x_{1}, x_{2}, \ldots, x_{n+2}):=(2,2,4 y_{1}, 4 y_{2}, \ldots, 4 y_{n})$$
is a solution for $(n+2)$, since
$$\begin{aligned} \frac{1}{x_{1}^{2}}+\frac{2}{x_{2}^{2}}+\frac{4}{x_{3}^{2}}+\cdots+\frac{2^{n+1}}{x_{n+2}^{2}} & =\frac{1}{2^{2}}+\frac{2}{2^{2}}+\frac{4}{(4 y_{1})^{2}}+\cdots+\frac{2^{n+1}}{(4 y_{n})^{2}} \\ & =\frac{1}{4}+\frac{2}{4}+\frac{4}{16}\left[\frac{1}{(y_{1})^{2}}+\cdots+\frac{2^{n-1}}{(y_{n})^{2}}\right] \\ & =\frac{3}{4}+\frac{1}{4} \cdot 1 \\ & =1 \end{aligned}$$
Using this induction step and the solutions for $n=1$ and $n=4$, we can construct solutions for all $n \geq 3$.

Solution 1a. There are other induction steps possible. For example from $n$ to $(n+3)$ :
Let $(y_{1}, y_{2}, \ldots, y_{n})$ be a solution for $n$, i.e.,
$$\frac{1}{y_{1}^{2}}+\frac{2}{y_{2}^{2}}+\frac{4}{y_{3}^{2}}+\cdots+\frac{2^{n-1}}{y_{n}^{2}}=1$$
Then
$$(x_{1}, x_{2}, \ldots, x_{n+3}):=(3,3,3,6 y_{1}, 6 y_{2}, \ldots, 6 y_{n})$$
is a solution for $(n+3)$, since
$$\begin{aligned} \frac{1}{x_{1}^{2}}+\frac{2}{x_{2}^{2}}+\frac{4}{x_{3}^{2}}+\cdots+\frac{2^{n+2}}{x_{n+3}^{2}} & =\frac{1}{3^{2}}+\frac{2}{3^{2}}+\frac{4}{3^{2}}+\frac{8}{(6 y_{1})^{2}}+\cdots+\frac{2^{n+2}}{(6 y_{n})^{2}} \\ & =\frac{1}{9}+\frac{2}{9}+\frac{4}{9}+\frac{8}{36}\left[\frac{1}{(y_{1})^{2}}+\cdots+\frac{2^{n-1}}{(y_{n})^{2}}\right] \\ & =\frac{7}{9}+\frac{2}{9} \cdot 1 \\ & =1 \end{aligned}$$

Solution 2. As in the previous solution, we can show that there does not exist a solution for $n=2$ and find explicit solutions for $n=1$ and $n=3$.
We construct further solutions by induction. Let $(y_{1}, y_{2}, \ldots, y_{n})$ be a solution for $n \geq 3$.
Then setting $x_{n-2}=x_{n+1}:=3 y_{n-2}$ and $x_{i}:=y_{i}$ for all other $i$ is a solution for $n+1$, since for the sum of the terms corresponding to $x_{n-2}$ and $x_{n+1}$ (which are the only ones that differ between the solutions for $n$ and $n+1$ ) we get
$$\frac{2^{n-3}}{x_{n-2}^{2}}+\frac{2^{n}}{x_{n+1}^{2}}=\frac{2^{n-3}}{9 y_{n-2}^{2}}+\frac{2^{n}}{9 y_{n-2}^{2}}=\frac{(1+8) \cdot 2^{n-3}}{9 y_{n-2}^{2}}=\frac{2^{n-3}}{y_{n-2}^{2}}$$
thereby keeping the sum of all terms equal.

Solution 3. As in the other solutions, we can show that there does not exist a solution for $n=2$. Also, we can find explicit solutions for $n=4, n=6$ and $n=8$. Now we prove that we can find suitable integers for all other $n$ :
For $n=2 k+1$ (where $k$ is a suitable non-negative integer), we can choose
$$x_{1}=\cdots=x_{n-1}=2^{k}, \quad x_{n}=2^{2 k}$$
to obtain
$$\sum_{i=1}^{n} \frac{2^{i-1}}{x_{i}^{2}}=\sum_{i=1}^{2 k} \frac{2^{i-1}}{2^{2 k}}+\frac{2^{2 k}}{2^{4 k}}=\frac{\frac{2^{2 k}-1}{2-1}}{2^{2 k}}+\frac{1}{2^{2 k}}=1$$
For $n=2 k$ with an integer $k \geq 5$, observe that, by setting $x_{i}=2^{\frac{i-1}{2}}$ for odd $i$ and $x_{i}=2^{\frac{i}{2}-1}$ for even $i$, we have
$$\sum_{i=1}^{2 k} \frac{2^{i-1}}{x_{i}^{2}}=\sum_{j=1}^{k}\left(\frac{2^{2 j-2}}{2^{2 j-2}}+\frac{2^{2 j-1}}{2^{2 j-2}}\right)=\sum_{j=1}^{k}(1+2)=3 k$$
If we can modify the $x_{i}$ in order to obtain some square number $m^{2}$ on the right hand side, we can in a second step multiply each $x_{i}$ by $m$ to obtain a sum of 1 .
Observe that all $x_{i}$ for $i>2$ are even. We show that there is a square divisible by 3 that can be obtained by replacing some of the $x_{i}$ (with $i>2$ ) by $x_{i}^{\prime}:=x_{i} / 2$ :
Note that for odd $i$, this increases the sum by $4-1=3$ and for even $i$, this increases the sum by $8-2=6$. Since there are $k-1$ odd and $k-1$ even indexes to choose from, we can increase the sum by every number $l$ that is divisible by three and satisfies $0 \leq l \leq 3(k-1)+6(k-1)=9 k-9$.
Because of $k \geq 5$ and therefore $\sqrt{3 k}<k$, the smallest square $m^{2} \geq 3 k$ which is divisible by 3 certainly satisfies
$$(\sqrt{3 k})^{2}=3 k \leq m^{2} \leq(\sqrt{3 k}+3)^{2}=3 k+6 \sqrt{3 k}+9<9 k+9$$
In particular, $m^{2} \leq 9 k+6$ because $m$ is divisible by 3 . This means that in order to increase $3 k$ to $m^{2}$, we have to add a number between 0 and $6 k+6$ to the sum above. However, $6 k+6 \leq 9 k-9$ because $k \geq 5$ and by the above argument, we can always do that.
To summarize, we first set the $x_{i}$ as above, then select up to $2 k-2$ of them to be divided by 2 , then multiply all of them by $m$, yielding a right hand side of 1 .