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
$Given A=n^2 −2n+2 , B=n^2 +2n+2 n ∈ N^∗ −{1}. Show that ∀ divisor of A which divise n can also divise 2. Show that all common divisor of A and B can divise 4n.$
$G i v e n A = n^{2} - 2 n + 2, B = n^{2} + 2 n + 2$ $n \in N^{*} - {1} .$ $S h o w t h a t \forall d i v i s o r o f A w h i c h d i v i s e$ $n c a n a l s o d i v i s e 2 .$ $S h o w t h a t a l l c o m m o n d i v i s o r o f$ $A a n d B c a n d i v i s e 4 n .$
	Answered by 1549442205PVT last updated on 16/Oct/20
	$i)Suppose d∣A and d∣n.Then { (( A=n^2 −2n+2 ⋮d)),((n⋮d)) :}⇒ { (( A=(n^2 −2n+2 )⋮d)),((C=(n^2 −2n)⋮d)) :} ⇒2=(A−C)⋮d⇒d∣2 (q.e.d) ii)Now suppose d is a common divisor of A=n^2 −2n+2 and B= A=n^2 +2n+2 ⇒ { (( A=n^2 −2n+2 ⋮d)),(( B=n^2 +2n+2⋮d )) :} ⇒4n=(A−B)⋮d⇒d∣4n (q.e.d)$
	$i) Suppose d ∣ A and d ∣ n . Then$ ${\begin{cases} A = n^{2} - 2 n + 2 ⋮ d \\ n ⋮ d \end{cases} \Rightarrow {\begin{cases} A = (n^{2} - 2 n + 2) ⋮ d \\ C = (n^{2} - 2 n) ⋮ d \end{cases}$ $\Rightarrow 2 = (A - C) ⋮ d \Rightarrow d ∣ 2 (q . e . d)$ $ii) Now suppose d is a common divisor$ $of A = n^{2} - 2 n + 2 and B = A = n^{2} + 2 n + 2$ $\Rightarrow {\begin{cases} A = n^{2} - 2 n + 2 ⋮ d \\ B = n^{2} + 2 n + 2 ⋮ d \end{cases}$ $\Rightarrow 4 n = (A - B) ⋮ d \Rightarrow d ∣ 4 n (q . e . d)$
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