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

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

Answered by behi834171 last updated on 23/Aug/22

$${\phi }^2-\phi -1=0\Rightarrow \phi =\frac{1+\sqrt{5}}{2}\left(\mathit{g}\mathit{o}\mathit{l}\mathit{d}\mathit{e}\mathit{n}\mathit{r}\mathit{a}\mathit{t}\mathit{i}\mathit{o}\right)$$$$1+\sqrt{5}=2\phi$$$${\phi }^2=\phi +1\Rightarrow {\phi }^4=\phi +1+2\phi +1=3\phi +2$$$$\Rightarrow {\phi }^8=9{\phi }^2+12\phi +4=21\phi +13$$$$\Rightarrow {(1+\sqrt{5})}^8=2^8\times {\phi }^8=2^8(21\mathit{\phi }+13)=$$$$=2^8(21\times \frac{1+\sqrt{5}}{2}+13)=2^7(21\sqrt{5}+47).◼$$

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

$$Sir,{you}^{\prime }venotconsiderd\lfloor \rfloor ?$$

Commented by behi834171 last updated on 24/Aug/22

$$no,sir.$$$$sender{hasn}^{\prime }ttalkaboutthis.$$$$$$

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

$$Sir{doesn}^{\prime }tthequestioninclude\lfloor \rfloor ?$$$$\lfloor {(1+\sqrt{5})}^8\rfloor =?$$

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