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

Find all integers n such that (n^2 − n − 1)^(n + 2) = 1

$$\mathrm{Find}\:\mathrm{all}\:\mathrm{integers}\:{n}\:\mathrm{such}\:\mathrm{that} \\ $$$$\left({n}^{\mathrm{2}} \:−\:{n}\:−\:\mathrm{1}\right)^{{n}\:+\:\mathrm{2}} \:=\:\mathrm{1} \\ $$

Answered by mrW1 last updated on 06/Jul/17

n^2 −n−1=1 if n is odd or even (i) n^2 −n−1=−1 if n is even (ii) n+2=0 and n^2 −n−1≠0 (iii) n^2 −n−1=1 n^2 −n−2=0 (n−2)(n+1)=0 ⇒n=2 even (ok) ⇒n=−1 odd (ok) n^2 −n−1=−1 n(n−1)=0 ⇒n=0 even (ok) ⇒n=1 odd (not ok) n+2=0 n=−2 (ok) n^2 −n−1=5≠0 ⇒all solutions: −2,−1,0,2

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

Commented by Tinkutara last updated on 06/Jul/17

Thanks Sir!

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

Commented by RasheedSoomro last updated on 06/Jul/17

Also n=−2

$$\mathrm{Also}\:\mathrm{n}=−\mathrm{2} \\ $$

Commented by mrW1 last updated on 06/Jul/17

you are right sir!

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

Commented by RasheedSoomro last updated on 09/Jul/17

G ⊚^(⌢) ⊚^(⌢) _(⌣) D approach!

$$\:\mathrm{G}\:\underset{\smile} {\overset{\frown} {\circledcirc}}\:\:\overset{\frown} {\circledcirc}\:\mathrm{D}\: \\ $$$$\mathrm{approach}! \\ $$

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