Dots, ka \(a\) un \(b\) - tādi reāli skaitļi, ka \(a+b\) ir vesels skaitlis un \(a^{2}+b^{2}=2\). Atrast visus šādus \(a\) un \(b\) pārus un pierādīt, ka citu bez Jūsu atrastajiem nav.
Apzīmēsim \(a+b=n\). Tad \(2\left(a^{2}+b^{2}\right)=(a+b)^{2}+(a-b)^{2} \geq^{2}\), tāpēc \(n^{2} \leq 2 \cdot 2=4\) un \(n=0;\ \pm 1;\ \pm 2\). Risinot vienādojumu sistēmas \(\left\{\begin{array}{c}a+b= \pm 2 \\ a^{2}+b^{2}=2\end{array},\ \left\{\begin{array}{c}a+b= \pm 1 \\ a^{2}+b^{2}=2\end{array},\ \left\{\begin{array}{c}a+b=0 \\ a^{2}+b^{2}=2\end{array}\right.\right.\right.\), iegūstam meklētos pārus \((1;\ 1),\ (-1;\ -1),\ (1;\ -1),\ (-1;\ 1)\), \(\left(\frac{1+\sqrt{3}}{2}; \frac{1-\sqrt{3}}{2}\right),\ \left(\frac{1-\sqrt{3}}{2}; \frac{1+\sqrt{3}}{2}\right),\ \left(\frac{-1+\sqrt{3}}{2}; \frac{-1-\sqrt{3}}{2}\right)\), \(\left(\frac{-1-\sqrt{3}}{2}; \frac{-1+\sqrt{3}}{2}\right)\)