Nice exercise asked by japan teacher Kunihiko Chikaya in Imad zak group and solved by Bassam Barakat

Exercise:

Let ${{\left\{ {{u}_{n}} \right\}}_{n\ge 1}}$ be a sequence defined to be ${{u}_{0}}=0\,\,\,,\,\,\,{{u}_{1}}=1\,\,\,\,\And \,\,\,\,{{u}_{n+1}}={{u}_{n}}+{{u}_{n-1}}$ and

Let $b={{u}_{n+1}}+{{u}_{n-1}}\,\,\,,\,\,\,\,c={{u}_{n+1}}{{u}_{n-1}}-{{u}_{n}}^{2}$ determine the zeros for ${{x}^{2}}-bx+c$

Solution: the sequence ${{\left\{ {{u}_{n}} \right\}}_{n\ge 1}}$ is Fibonacci sequence hence there is a strong relation between

 Fibonacci and the golden ratio $\varphi =\frac{1\pm \sqrt{5}}{2}$ which comes from solving ${{x}^{2}}-x-1=0$

Put $\varphi =\frac{1+\sqrt{5}}{2}\,\,\,\,\,\,\And \,\,\,\,\psi =\frac{1-\sqrt{5}}{2}$  by Cassini and Catalan identity $c={{\left( -1 \right)}^{n}}$

Hence ${{u}_{n}}=\frac{{{\varphi }^{n}}-{{\psi }^{n}}}{\sqrt{5}}$  So $b={{u}_{n+1}}+{{u}_{n-1}}={{u}_{n}}+2{{u}_{n-1}}=\frac{{{\varphi }^{n}}-{{\psi }^{n}}+2\left( {{\varphi }^{n-1}}-{{\psi }^{n-1}} \right)}{\sqrt{5}}$

$=\frac{1}{\sqrt{5}}\left[ {{\varphi }^{n}}+2{{\varphi }^{n-1}} \right]-\frac{1}{\sqrt{5}}\left[ {{\psi }^{n}}+2{{\psi }^{n-1}} \right]$  but $\psi =-\frac{1}{\varphi }$

Observe that ${{\varphi }^{n}}+2{{\varphi }^{n-1}}={{\varphi }^{n}}\left( 1+\frac{2}{\varphi } \right)={{\varphi }^{n}}\left( 1-2\psi  \right)={{\varphi }^{n}}\left( 1-1+\sqrt{5} \right)=\sqrt{5}\,{{\varphi }^{n}}$

And ${{\psi }^{n}}+2{{\psi }^{n-1}}={{\psi }^{n}}\left( 1+\frac{2}{\psi } \right)={{\psi }^{n}}\left( 1-2\varphi  \right)={{\psi }^{n}}\left( 1-1-\sqrt{5} \right)=-\sqrt{5}{{\psi }^{n}}$

Hence ${{u}_{n+1}}+{{u}_{n-1}}=\frac{\sqrt{5}{{\varphi }^{n}}+\sqrt{5}{{\psi }^{n}}}{\sqrt{5}}={{\varphi }^{n}}+{{\psi }^{n}}$ so $b={{\varphi }^{n}}+{{\psi }^{n}}$

So ${{x}^{2}}-\left( {{\varphi }^{n}}+{{\psi }^{n}} \right)x+{{\left( -1 \right)}^{n}}=0$

But $\varphi \psi =\frac{1+\sqrt{5}}{2}\times \frac{1-\sqrt{5}}{2}=\frac{1-5}{4}=\frac{-4}{4}=-1$

so${{\left( -1 \right)}^{n}}={{\left( \varphi \psi  \right)}^{n}}={{\varphi }^{n}}{{\psi }^{n}}$

thus the obtain equation is ${{x}^{2}}-\left( {{\varphi }^{n}}+{{\psi }^{n}} \right)x+{{\varphi }^{n}}{{\psi }^{n}}=0$

${{x}^{2}}-Sx+P=0\Leftrightarrow {{\left( x-\frac{S}{2} \right)}^{2}}-\frac{{{S}^{2}}}{4}+P=0\Leftrightarrow {{\left( x-\frac{S}{2} \right)}^{2}}=-P+\frac{{{S}^{2}}}{4}=\frac{-4P+{{S}^{2}}}{4}$

$\Rightarrow x=\frac{S\pm \sqrt{{{S}^{2}}-4P}}{2}$ but ${{S}^{2}}-4P={{\varphi }^{2n}}+{{\psi }^{2n}}+2{{\varphi }^{n}}{{\psi }^{n}}-4{{\varphi }^{n}}{{\psi }^{n}}={{\left( {{\varphi }^{n}}-{{\psi }^{n}} \right)}^{2}}$

$\Rightarrow S+\sqrt{{{S}^{2}}-4P}={{\varphi }^{n}}+{{\psi }^{n}}+{{\varphi }^{n}}-{{\psi }^{n}}=2{{\varphi }^{n}}$ and $S-\sqrt{{{S}^{2}}-4P}=2{{\psi }^{n}}$

Hence ${{x}_{1}}={{\varphi }^{n}}\,\,or\,\,{{x}_{2}}={{\psi }^{n}}$