Problem 4. Let $n \in \mathbb{N}^{*}, n \geq 2$. For real numbers $a_{1}, a_{2}, \ldots, a_{n}$, we denote $S_{0}=1$ and

$$
S_{k}=\sum_{1 \leq i_{1}<i_{2}<\ldots<i_{k} \leq n} a_{i_{1}} a_{i_{2}} \ldots a_{i_{k}}
$$

the sum of all products of $k$ numbers chosen from $a_{1}, a_{2}, \ldots, a_{n}, k \in\{1,2, \ldots, n\}$.

Determine the number of $n$-tuples $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ for which the following relation holds:

$$
\left(S_{n}-S_{n-2}+S_{n-4}-\ldots\right)^{2}+\left(S_{n-1}-S_{n-3}+S_{n-5}-\ldots\right)^{2}=2^{n} S_{n}
$$
Solution and grading: The identity holds

$$
\prod_{k=1}^{n}\left(a_{k}+i\right)=\left(S_{n}-S_{n-2}+S_{n-4}-\cdots\right)+i\left(S_{n-1}-S_{n-3}+S_{n-4}-\cdots\right)
$$

It follows that

$$
\begin{aligned}
& \left(S_{n}-S_{n-2}+S_{n-4}-\cdots\right)^{2}+\left(S_{n-1}-S_{n-3}+S_{n-4}-\cdots\right)^{2} \\
= & \left|\prod_{k=1}^{n}\left(a_{k}+i\right)\right|^{2}=\prod_{k=1}^{n}\left|a_{k}+i\right|^{2}=\prod_{k=1}^{n}\left(a_{k}^{2}+1\right)
\end{aligned}
$$

The relation from the statement is equivalent to

$$
\prod_{k=1}^{n}\left(a_{k}^{2}+1\right)=2^{n} a_{1} a_{2} \cdots a_{n}
$$

From the inequalities $a_{k}^{2}+1 \geq 2\left|a_{k}\right|$, for all $k \in\{1,2, \ldots, n\}$, it follows that equality in the relation from the statement holds if and only if $\left|a_{1}\right|=\left|a_{2}\right|=\cdots=\left|a_{n}\right|=1$, and the number of values equal to -1 is even.

Therefore, the number of $n$-tuples $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ for which the relation from the statement holds is

$$
C_{n}^{0}+C_{n}^{2}+C_{n}^{4}+\ldots=2^{n-1}
$$

## National Mathematics Olympiad  Final Stage, April 4, 2018

GRADE 11

## Solutions and approximate grading