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For-each-positive-integer-n-define-a-n-20-n-2-and-d-n-gcd-a-n-a-n-1-Find-the-set-of-all-values-that-are-taken-by-d-n-and-show-by-examples-that-each-of-these-values-are-attained-




Question Number 22379 by Tinkutara last updated on 16/Oct/17
For each positive integer n, define a_n  =  20 + n^2 , and d_n  = gcd(a_n , a_(n+1) ). Find  the set of all values that are taken by  d_n  and show by examples that each of  these values are attained.
$$\mathrm{For}\:\mathrm{each}\:\mathrm{positive}\:\mathrm{integer}\:{n},\:\mathrm{define}\:{a}_{{n}} \:= \\ $$$$\mathrm{20}\:+\:{n}^{\mathrm{2}} ,\:\mathrm{and}\:{d}_{{n}} \:=\:{gcd}\left({a}_{{n}} ,\:{a}_{{n}+\mathrm{1}} \right).\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{all}\:\mathrm{values}\:\mathrm{that}\:\mathrm{are}\:\mathrm{taken}\:\mathrm{by} \\ $$$${d}_{{n}} \:\mathrm{and}\:\mathrm{show}\:\mathrm{by}\:\mathrm{examples}\:\mathrm{that}\:\mathrm{each}\:\mathrm{of} \\ $$$$\mathrm{these}\:\mathrm{values}\:\mathrm{are}\:\mathrm{attained}. \\ $$
Answered by Tinkutara last updated on 21/Oct/17
Since d_n =gcd(a_n ,a_(n+1) )  So d_n  divides 20+n^2 .  Similarly d_n  divides 20+(n+1)^2 .  So d_n  divides (n+1)^2 −n^2 =2n+1  ⇒ d_n  divides 4(20+n^2 )=  (2n+1)(2n−1)+81  Since d_n  already divides 2n+1, d_n   must divide 81.  Hence d_n  is set of all divisors of 81.  ∴ Required set={1,3,9,27,81}
$${Since}\:{d}_{{n}} ={gcd}\left({a}_{{n}} ,{a}_{{n}+\mathrm{1}} \right) \\ $$$${So}\:{d}_{{n}} \:{divides}\:\mathrm{20}+{n}^{\mathrm{2}} . \\ $$$${Similarly}\:{d}_{{n}} \:{divides}\:\mathrm{20}+\left({n}+\mathrm{1}\right)^{\mathrm{2}} . \\ $$$${So}\:{d}_{{n}} \:{divides}\:\left({n}+\mathrm{1}\right)^{\mathrm{2}} −{n}^{\mathrm{2}} =\mathrm{2}{n}+\mathrm{1} \\ $$$$\Rightarrow\:{d}_{{n}} \:{divides}\:\mathrm{4}\left(\mathrm{20}+{n}^{\mathrm{2}} \right)= \\ $$$$\left(\mathrm{2}{n}+\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)+\mathrm{81} \\ $$$${Since}\:{d}_{{n}} \:{already}\:{divides}\:\mathrm{2}{n}+\mathrm{1},\:{d}_{{n}} \\ $$$${must}\:{divide}\:\mathrm{81}. \\ $$$${Hence}\:{d}_{{n}} \:{is}\:{set}\:{of}\:{all}\:{divisors}\:{of}\:\mathrm{81}. \\ $$$$\therefore\:{Required}\:{set}=\left\{\mathrm{1},\mathrm{3},\mathrm{9},\mathrm{27},\mathrm{81}\right\} \\ $$
Commented by Rasheed.Sindhi last updated on 22/Oct/17
Nice!
$$\mathrm{Nice}! \\ $$

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