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Given-A-n-2-2n-2-B-n-2-2n-2-n-N-1-Show-that-divisor-of-A-which-divise-n-can-also-divise-2-Show-that-all-common-divisor-of-A-and-B-can-divise-4n-

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Question Number 118193 by mathocean1 last updated on 15/Oct/20
Given A=n^2 −2n+2 , B=n^2 +2n+2 n ∈ N^∗ −{1}. Show that ∀ divisor of A which divise n can also divise 2. Show that all common divisor of A and B can divise 4n.
GivenA=n2−2n+2,B=n2+2n+2n∈N∗−{1}.Showthat∀divisorofAwhichdivisencanalsodivise2.ShowthatallcommondivisorofAandBcandivise4n.
Answered by 1549442205PVT last updated on 16/Oct/20
i)Suppose d∣A and d∣n.Then { (( A=n^2 −2n+2 ⋮d)),((n⋮d)) :}⇒ { (( A=(n^2 −2n+2 )⋮d)),((C=(n^2 −2n)⋮d)) :} ⇒2=(A−C)⋮d⇒d∣2 (q.e.d) ii)Now suppose d is a common divisor of A=n^2 −2n+2 and B= A=n^2 +2n+2 ⇒ { (( A=n^2 −2n+2 ⋮d)),(( B=n^2 +2n+2⋮d )) :} ⇒4n=(A−B)⋮d⇒d∣4n (q.e.d)
i)Supposed∣Aandd∣n.Then{A=n2−2n+2⋮dn⋮d⇒{A=(n2−2n+2)⋮dC=(n2−2n)⋮d⇒2=(A−C)⋮d⇒d∣2(\boldsymbolq.\boldsymbole.\boldsymbold)ii)NowsupposedisacommondivisorofA=n2−2n+2andB=A=n2+2n+2⇒{A=n2−2n+2⋮dB=n2+2n+2⋮d⇒4n=(A−B)⋮d⇒d∣4n(\boldsymbolq.\boldsymbole.\boldsymbold)

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