1. The product of four numbers - the roots of the equations $x^{2}+2 b x+c=0$ and $x^{2}+2 c x+b=0$, is equal to 1. It is known that the numbers $b$ and $c$ are positive. Find them.
Solution. Let the roots of the first equation be $x_{1}$ and $x_{2}$, and the roots of the second equation be $x_{3}$ and $x_{4}$. By Vieta's Theorem, $1=x_{1} x_{2} x_{3} x_{4}=b c$. Let's write down the discriminants of both equations: $4 b^{2}-4 c \geqslant 0,4 c^{2}-4 b \geqslant 0$. Therefore, $b^{2} \geqslant c=1 / b$, but since $b>0$, we can multiply by it and get $b^{3} \geqslant 1$. From this, it follows that $b \geqslant 1$. Similarly, $c \geqslant 1$. But since their product is equal to one, then $b=c=1$.