Atrisinājums

Tā kā \(0 \leq a^{2} \leq 1\) un \(0 \leq b^{2} \leq 1\), tad \(0 \leq \sqrt{1-a^{2}} \leq 1\) un \(0 \leq \sqrt{1-b^{2}} \leq 1\). Tātad \(-\frac{1}{2} \leq \frac{1}{2}-\sqrt{1-a^{2}} \leq \frac{1}{2} \quad\) un \(\quad-\frac{1}{2} \leq \frac{1}{2}-\sqrt{1-b^{2}} \leq \frac{1}{2} \quad\) jeb \(\quad\left|\frac{1}{2}-\sqrt{1-a^{2}}\right| \leq \frac{1}{2} \quad\) un \(\left|\frac{1}{2}-\sqrt{1-b^{2}}\right| \leq \frac{1}{2}\), tātad \(\left(\frac{1}{2}-\sqrt{1-a^{2}}\right)^{2} \leq \frac{1}{4}\) un \(\left(\frac{1}{2}-\sqrt{1-b^{2}}\right)^{2} \leq \frac{1}{4}\). Uzdevumā dotā vienādība izpildās tikai gadījumā, ja \(\left(\frac{1}{2}-\sqrt{1-a^{2}}\right)^{2}=\frac{1}{4}\) un \(\left(\frac{1}{2}-\sqrt{1-b^{2}}\right)^{2}=\frac{1}{4}\) jeb \(\sqrt{1-a^{2}}=0\) vai \(\sqrt{1-a^{2}}=1\) un \(\sqrt{1-b^{2}}=0\) vai \(\sqrt{1-b^{2}}=1\), tātad \(a^{2}=0\) vai \(a^{2}=1\) un \(b^{2}=0\) vai \(b^{2}=1\). Tāpēc iegūstam \(9\) atrisinājumus \((1; 1),(1;-1),(-1; 1),(-1;-1),(1; 0)\), \((-1; 0),(0; 1),(0;-1),(0; 0)\).