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.


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Question Number 22379 by Tinkutara last updated on 16/Oct/17

$For each positive integer n, define a_{n} =$ $20 + n^{2}, and d_{n} = g c d (a_{n}, a_{n + 1}) . Find$ $the set of all values that are taken by$ $d_{n} and show by examples that each of$ $these values are attained .$
	Answered by Tinkutara last updated on 21/Oct/17
	$Since d_n =gcd(a_n ,a_(n+1) ) So d_n divides 20+n^2 . Similarly d_n divides 20+(n+1)^2 . So d_n divides (n+1)^2 −n^2 =2n+1 ⇒ d_n divides 4(20+n^2 )= (2n+1)(2n−1)+81 Since d_n already divides 2n+1, d_n must divide 81. Hence d_n is set of all divisors of 81. ∴ Required set={1,3,9,27,81}$
	$S i n c e d_{n} = g c d (a_{n}, a_{n + 1})$ $S o d_{n} d i v i d e s 20 + n^{2} .$ $S i m i l a r l y d_{n} d i v i d e s 20 + {(n + 1)}^{2} .$ $S o d_{n} d i v i d e s {(n + 1)}^{2} - n^{2} = 2 n + 1$ $\Rightarrow d_{n} d i v i d e s 4 (20 + n^{2}) =$ $(2 n + 1) (2 n - 1) + 81$ $S i n c e d_{n} a l r e a d y d i v i d e s 2 n + 1, d_{n}$ $m u s t d i v i d e 81 .$ $H e n c e d_{n} i s s e t o f a l l d i v i s o r s o f 81 .$ $∴ R e q u i r e d s e t = {1, 3, 9, 27, 81}$
	Commented by Rasheed.Sindhi last updated on 22/Oct/17

	$Nice!$
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