Question Number 175218 by mnjuly1970 last updated on 23/Aug/22
$$ \\ $$$$\:\:\lfloor\:\:\left(\:\mathrm{1}\:+\:\sqrt{\mathrm{5}}\:\right)^{\:\mathrm{8}} \:\rfloor\:=\:? \\ $$$$\:\:\: \\ $$
Answered by behi834171 last updated on 23/Aug/22
$$\varphi^{\mathrm{2}} −\varphi−\mathrm{1}=\mathrm{0}\Rightarrow\varphi=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\left(\boldsymbol{{golden}}\:\boldsymbol{{ratio}}\right) \\ $$$$\mathrm{1}+\sqrt{\mathrm{5}}=\mathrm{2}\varphi \\ $$$$\varphi^{\mathrm{2}} =\varphi+\mathrm{1}\Rightarrow\varphi^{\mathrm{4}} =\varphi+\mathrm{1}+\mathrm{2}\varphi+\mathrm{1}=\mathrm{3}\varphi+\mathrm{2} \\ $$$$\Rightarrow\varphi^{\mathrm{8}} =\mathrm{9}\varphi^{\mathrm{2}} +\mathrm{12}\varphi+\mathrm{4}=\mathrm{21}\varphi+\mathrm{13} \\ $$$$\Rightarrow\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{\mathrm{8}} =\mathrm{2}^{\mathrm{8}} ×\varphi^{\mathrm{8}} =\mathrm{2}^{\mathrm{8}} \left(\mathrm{21}\boldsymbol{\varphi}+\mathrm{13}\right)= \\ $$$$=\mathrm{2}^{\mathrm{8}} \left(\mathrm{21}×\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}+\mathrm{13}\right)=\mathrm{2}^{\mathrm{7}} \left(\mathrm{21}\sqrt{\mathrm{5}}+\mathrm{47}\right)\:.\blacksquare \\ $$
Commented by Rasheed.Sindhi last updated on 24/Aug/22
$${Sir},\:{you}'{ve}\:{not}\:{considerd}\:\lfloor\:\:\:\:\:\:\rfloor? \\ $$
Commented by behi834171 last updated on 24/Aug/22
$${no},{sir}. \\ $$$${sender}\:{hasn}'{t}\:{talk}\:{about}\:{this}. \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 24/Aug/22
$${Sir}\:{doesn}'{t}\:{the}\:{question}\:{include}\:\lfloor\:\:\:\:\rfloor? \\ $$$$\:\:\lfloor\:\:\left(\:\mathrm{1}\:+\:\sqrt{\mathrm{5}}\:\right)^{\:\mathrm{8}} \:\rfloor\:=\:? \\ $$