Atrisinājums

No Vjeta teorēmas izriet, ka

\[\left\{\begin{array}{l} a+b=\sqrt[4]{7} \cdot(1+\sqrt{5}) \\ ab=\frac{\sqrt[4]{7}}{1+\sqrt{5}} \end{array}\right.\]

Pārveidojam doto izteiksmi: \(a^{4}b+ab^{4}+3a^{3}b^{2}+3a^{2}b^{3}+16a^{4}b^{3}+16a^{3}b^{4}=ab\left(a^{3}+b^{3}+3a^{2}b+3ab^{2}+16a^{3}b^{2}+16a^{2}b^{3}\right)=\) \(=ab\left((a+b)^{3}+16a^{2}b^{2}(a+b)\right)=\frac{\sqrt[4]{7}}{1+\sqrt{5}} \cdot\left(\sqrt[4]{7^{3}} \cdot(1+\sqrt{5})^{3}+16 \cdot \frac{\sqrt[4]{7^{2}} \cdot \sqrt[4]{7} \cdot(1+\sqrt{5})}{(1+\sqrt{5})^{2}}\right)=\) \(=7 \cdot(1+\sqrt{5})^{2}+\frac{16 \cdot 7}{(1+\sqrt{5})^{2}}=7 \cdot\left((6+2 \sqrt{5})+\frac{16}{6+2 \sqrt{5}}\right)=\) \(=7 \cdot \frac{36+24 \sqrt{5}+20+16}{6+2 \sqrt{5}}=7 \cdot 12 \cdot \frac{6+2 \sqrt{5}}{6+2 \sqrt{5}}=84.\) Tā kā skaitlis \(84\) ir vesels skaitlis, tad prasītais ir pierādīts.