Nice exercise with nice tech

 Exercise:

Determine the values of $a\And b$ such that ${{\left( \frac{1+\sqrt{5}}{2} \right)}^{12}}=a+b\sqrt{5}$

Solution:

Let $u=\frac{1+\sqrt{5}}{2}\Leftrightarrow 2u=1+\sqrt{5}\Leftrightarrow 2u-1=\sqrt{5}$

Squaring both sides to get $4{{u}^{2}}-4u+1=5\Leftrightarrow 4\left( {{u}^{2}}-u-1 \right)=0\Leftrightarrow $

Notice that, ${{u}^{3}}={{u}^{2}}+u=2u+1\And {{u}^{6}}={{\left( {{u}^{3}} \right)}^{2}}={{\left( 2u+1 \right)}^{2}}=4{{u}^{2}}+4u+1$

                                                                $=4\left( u+1 \right)+4u+1=8u+5$

It follows that , ${{u}^{12}}={{\left( {{u}^{6}} \right)}^{2}}={{\left( 8u+5 \right)}^{2}}=64{{u}^{2}}+25+80u=64\left( u+1 \right)+25+80u$

                          $=64u+80u+64+25=144u+89$

So that, ${{\left( \frac{1+\sqrt{5}}{2} \right)}^{12}}=\frac{144\left( 1+\sqrt{5} \right)}{2}+89=72\left( 1+\sqrt{5} \right)+89=161+72\sqrt{5}=a+b\sqrt{5}$

Therefore, after comparing we obtain $a=161\And b=72$