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Question Number 17454 by Tinkutara last updated on 06/Jul/17

$$\mathrm{Find}\mathrm{all}\mathrm{integers}n\mathrm{such}\mathrm{that}$$$${(n^2-n-1)}^{n+2}=1$$

Answered by mrW1 last updated on 06/Jul/17

$${\mathrm{n}}^2-\mathrm{n}-1=1\mathrm{if}\mathrm{n}\mathrm{is}\mathrm{odd}\mathrm{or}\mathrm{even}\left(\mathrm{i}\right)$$$${\mathrm{n}}^2-\mathrm{n}-1=-1\mathrm{if}\mathrm{n}\mathrm{is}\mathrm{even}\left(\mathrm{ii}\right)$$$$\mathrm{n}+2=0\mathrm{and}{\mathrm{n}}^2-\mathrm{n}-1\ne 0\left(\mathrm{iii}\right)$$$$$$$${\mathrm{n}}^2-\mathrm{n}-1=1$$$${\mathrm{n}}^2-\mathrm{n}-2=0$$$$(\mathrm{n}-2)(\mathrm{n}+1)=0$$$$\Rightarrow \mathrm{n}=2\mathrm{even}\left(\mathrm{ok}\right)$$$$\Rightarrow \mathrm{n}=-1\mathrm{odd}\left(\mathrm{ok}\right)$$$$$$$${\mathrm{n}}^2-\mathrm{n}-1=-1$$$$\mathrm{n}(\mathrm{n}-1)=0$$$$\Rightarrow \mathrm{n}=0\mathrm{even}\left(\mathrm{ok}\right)$$$$\Rightarrow \mathrm{n}=1\mathrm{odd}\left(\mathrm{not}\mathrm{ok}\right)$$$$$$$$\mathrm{n}+2=0$$$$\mathrm{n}=-2\left(\mathrm{ok}\right)$$$${\mathrm{n}}^2-\mathrm{n}-1=5\ne 0$$$$$$$$\Rightarrow \mathrm{all}\mathrm{solutions}:-2,-1,0,2$$

Commented by Tinkutara last updated on 06/Jul/17

$$\mathrm{Thanks}\mathrm{Sir}!$$

Commented by RasheedSoomro last updated on 06/Jul/17

$$\mathrm{Also}\mathrm{n}=-2$$

Commented by mrW1 last updated on 06/Jul/17

$$\mathrm{you}\mathrm{are}\mathrm{right}\mathrm{sir}!$$

Commented by RasheedSoomro last updated on 09/Jul/17

$$\mathrm{G}\underset{⌣}{\stackrel{⌢}{⊚}}\stackrel{⌢}{⊚}\mathrm{D}$$$$\mathrm{approach}!$$

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