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For-each-positive-integer-n-define-a-n-20-n-2-and-d-n-gcd-a-n-a-n-2-Find-the-set-of-all-values-that-are-taken-by-d-n-

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Question Number 22635 by Rasheed.Sindhi last updated on 21/Oct/17
For each positive integer n, define a_n =20+n^2 ,and d_n =gcd(a_n ,a_(n+2) ). Find the set of all values that are taken by d_n .
$$\mathrm{For}\:\mathrm{each}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{n},\:\mathrm{define} \\ $$$$\mathrm{a}_{\mathrm{n}} =\mathrm{20}+\mathrm{n}^{\mathrm{2}} ,\mathrm{and}\:\mathrm{d}_{\mathrm{n}} =\mathrm{gcd}\left(\mathrm{a}_{\mathrm{n}} ,\mathrm{a}_{\mathrm{n}+\mathrm{2}} \right). \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{all}\:\mathrm{values}\:\mathrm{that}\:\mathrm{are} \\ $$$$\mathrm{taken}\:\mathrm{by}\:\mathrm{d}_{\mathrm{n}} . \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 21/Oct/17
a_n =20+n^2 , d_n =gcd(a_n ,a_(n+2) ) −−−−−−−−−−−−− ∵ d_n =gcd(a_n ,a_(n+2) ) ∴ d_n ∣ 20+n^2 ∧ d_n ∣ 20+(n+2)^2 ∴ d_n ∣ {20+(n+2)^2 }−(20+n^2 ) d_n ∣ 4(n+1)⇒d_n ∣ (n+1)........(i) Now, 20+n^2 =(n+1)(n−1)+21 So d_n ∣ 20+n^2 ⇒d_n ∣ (n+1)(n−1)+21....(ii) (i)&(ii): d_n ∣ 21 ∴The required set is{1,3,7,21} (Method of proof used by Mr Tinkutara in answer of Q# 22379)
$$\mathrm{a}_{\mathrm{n}} =\mathrm{20}+\mathrm{n}^{\mathrm{2}} ,\:\mathrm{d}_{\mathrm{n}} =\mathrm{gcd}\left(\mathrm{a}_{\mathrm{n}} ,\mathrm{a}_{\mathrm{n}+\mathrm{2}} \right) \\ $$$$−−−−−−−−−−−−− \\ $$$$\because\:\mathrm{d}_{\mathrm{n}} =\mathrm{gcd}\left(\mathrm{a}_{\mathrm{n}} ,\mathrm{a}_{\mathrm{n}+\mathrm{2}} \right) \\ $$$$\therefore\:\mathrm{d}_{\mathrm{n}} \:\mid\:\mathrm{20}+\mathrm{n}^{\mathrm{2}} \:\wedge\:\mathrm{d}_{\mathrm{n}} \:\mid\:\mathrm{20}+\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\therefore\:\mathrm{d}_{\mathrm{n}} \:\mid\:\left\{\mathrm{20}+\left(\mathrm{n}+\mathrm{2}\right)^{\mathrm{2}} \right\}−\left(\mathrm{20}+\mathrm{n}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\mathrm{d}_{\mathrm{n}} \:\mid\:\mathrm{4}\left(\mathrm{n}+\mathrm{1}\right)\Rightarrow\mathrm{d}_{\mathrm{n}} \mid\:\left(\mathrm{n}+\mathrm{1}\right)……..\left(\mathrm{i}\right) \\ $$$$\mathrm{Now},\:\mathrm{20}+\mathrm{n}^{\mathrm{2}} =\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}−\mathrm{1}\right)+\mathrm{21} \\ $$$$\mathrm{So}\:\mathrm{d}_{\mathrm{n}} \:\mid\:\mathrm{20}+\mathrm{n}^{\mathrm{2}} \Rightarrow\mathrm{d}_{\mathrm{n}} \mid\:\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}−\mathrm{1}\right)+\mathrm{21}….\left(\mathrm{ii}\right) \\ $$$$\:\left(\mathrm{i}\right)\&\left(\mathrm{ii}\right):\:\mathrm{d}_{\mathrm{n}} \mid\:\mathrm{21}\:\:\:\:\:\: \\ $$$$\therefore\mathrm{The}\:\mathrm{required}\:\mathrm{set}\:\mathrm{is}\left\{\mathrm{1},\mathrm{3},\mathrm{7},\mathrm{21}\right\} \\ $$$$\left(\mathrm{Method}\:\mathrm{of}\:\mathrm{proof}\:\mathrm{used}\:\mathrm{by}\:\mathrm{Mr}\:\mathrm{Tinkutara}\:\mathrm{in}\right. \\ $$$$\left.\mathrm{answer}\:\mathrm{of}\:\mathrm{Q}#\:\mathrm{22379}\right) \\ $$$$ \\ $$

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