Dots, ka \(a\) ir tāds reāls skaitlis, ka kvadrātvienādojumam \(x^{2}-x+a=0\) ir divas dažādas reālas saknes \(x_{1}\) un \(x_{2}\). Pierādīt: \(\left|x_{1}^{2}-x_{2}^{2}\right|=1\) tad un tikai tad, ja \(\left|x_{1}^{3}-x_{2}^{3}\right|=1\).
Saskaņā ar Vjeta teorēmu \(\left|x_{1}^{2}-x_{2}^{2}\right|=1 \Leftrightarrow\left|\left(x_{1}-x_{2}\right)\left(x_{1}+x_{2}\right)\right|=1 \Leftrightarrow\left|x_{1}-x_{2}\right|=1 \Leftrightarrow 2 \sqrt{\frac{1}{4}-a}=1 \Leftrightarrow a=0\), kā arī \(\left|x_{1}^{3}-x_{2}^{3}\right|=1 \Leftrightarrow\left|\left(x_{1}-x_{2}\right)\left(x_{1}^{2}+x_{1} x_{2}+x_{2}^{2}\right)\right|=1 \Leftrightarrow \mid \sqrt{1-4 a}\left(\left(x_{1}+x_{2}\right)^{2}-x_{1} x_{2}\right)=1 \Leftrightarrow\) \(\Leftrightarrow \sqrt{1-4 a}|1-a|=1 \Leftrightarrow 4 a^{3}-9 a^{2}+6 a=0 \Leftrightarrow a=0\).