# Problem 5.

Numbers $a_{1}, b_{1}, a_{2}, b_{2}, \ldots, a_{n}, b_{n}$ satisfy the conditions

$$
\begin{gathered}
a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}=2018^{2}, \quad b_{1}^{2}+b_{2}^{2}+\cdots+b_{n}^{2}=2017^{2} \\
a_{1} b_{1}+a_{2} b_{2}+\cdots+a_{n} b_{n}=2017 \cdot 2018
\end{gathered}
$$

Find the ratios $a_{1} / b_{1}, a_{2} / b_{2}, \ldots, a_{n} / b_{n}$.
# Solution.

Let's perform the transformations.

$$
\begin{gathered}
\left(2017 a_{1}-2018 b_{1}\right)^{2}+\left(2017 a_{2}-2018 b_{2}\right)^{2}+\ldots+\left(2017 a_{n}-2018 b_{n}\right)^{2}= \\
=2017^{2}\left(a_{1}^{2}+a_{2}^{2}+\ldots+a_{n}^{2}\right)+2018^{2}\left(b_{1}^{2}+b_{2}^{2}+\ldots+b_{n}^{2}\right)- \\
-2 \cdot 2017 \cdot 2018\left(a_{1} b_{1}+a_{2} b_{2}+\ldots+a_{n} b_{n}\right)= \\
=2017^{2} \cdot 2018^{2}+2018^{2} \cdot 2017^{2}-2 \cdot 2017^{2} \cdot 2018^{2}=0
\end{gathered}
$$

Since the sum of squares can be zero if and only if each term is zero, we have

$$
2017 a_{k}-2018 b_{k}=0, \quad k=1, \ldots, n
$$

From this, we obtain

Answer: $a_{1} / b_{1}=a_{1} / b_{1}=\ldots=a_{n} / b_{n}=2018 / 2017$.