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Question Number 175218 by mnjuly1970 last updated on 23/Aug/22

⌊ ( 1 + (√5) )^( 8) ⌋ = ?

$$ \\ $$$$\:\:\lfloor\:\:\left(\:\mathrm{1}\:+\:\sqrt{\mathrm{5}}\:\right)^{\:\mathrm{8}} \:\rfloor\:=\:? \\ $$$$\:\:\: \\ $$

Answered by behi834171 last updated on 23/Aug/22

ϕ^2 −ϕ−1=0⇒ϕ=((1+(√5))/2) (golden ratio) 1+(√5)=2ϕ ϕ^2 =ϕ+1⇒ϕ^4 =ϕ+1+2ϕ+1=3ϕ+2 ⇒ϕ^8 =9ϕ^2 +12ϕ+4=21ϕ+13 ⇒(1+(√5))^8 =2^8 ×ϕ^8 =2^8 (21𝛟+13)= =2^8 (21×((1+(√5))/2)+13)=2^7 (21(√5)+47) .■

$$\varphi^{\mathrm{2}} −\varphi−\mathrm{1}=\mathrm{0}\Rightarrow\varphi=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\left(\boldsymbol{{golden}}\:\boldsymbol{{ratio}}\right) \\ $$$$\mathrm{1}+\sqrt{\mathrm{5}}=\mathrm{2}\varphi \\ $$$$\varphi^{\mathrm{2}} =\varphi+\mathrm{1}\Rightarrow\varphi^{\mathrm{4}} =\varphi+\mathrm{1}+\mathrm{2}\varphi+\mathrm{1}=\mathrm{3}\varphi+\mathrm{2} \\ $$$$\Rightarrow\varphi^{\mathrm{8}} =\mathrm{9}\varphi^{\mathrm{2}} +\mathrm{12}\varphi+\mathrm{4}=\mathrm{21}\varphi+\mathrm{13} \\ $$$$\Rightarrow\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)^{\mathrm{8}} =\mathrm{2}^{\mathrm{8}} ×\varphi^{\mathrm{8}} =\mathrm{2}^{\mathrm{8}} \left(\mathrm{21}\boldsymbol{\varphi}+\mathrm{13}\right)= \\ $$$$=\mathrm{2}^{\mathrm{8}} \left(\mathrm{21}×\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}+\mathrm{13}\right)=\mathrm{2}^{\mathrm{7}} \left(\mathrm{21}\sqrt{\mathrm{5}}+\mathrm{47}\right)\:.\blacksquare \\ $$

Commented by Rasheed.Sindhi last updated on 24/Aug/22

Sir, you′ve not considerd ⌊ ⌋?

$${Sir},\:{you}'{ve}\:{not}\:{considerd}\:\lfloor\:\:\:\:\:\:\rfloor? \\ $$

Commented by behi834171 last updated on 24/Aug/22

no,sir. sender hasn′t talk about this.

$${no},{sir}. \\ $$$${sender}\:{hasn}'{t}\:{talk}\:{about}\:{this}. \\ $$$$ \\ $$

Commented by Rasheed.Sindhi last updated on 24/Aug/22

Sir doesn′t the question include ⌊ ⌋? ⌊ ( 1 + (√5) )^( 8) ⌋ = ?

$${Sir}\:{doesn}'{t}\:{the}\:{question}\:{include}\:\lfloor\:\:\:\:\rfloor? \\ $$$$\:\:\lfloor\:\:\left(\:\mathrm{1}\:+\:\sqrt{\mathrm{5}}\:\right)^{\:\mathrm{8}} \:\rfloor\:=\:? \\ $$

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