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find max(dn)

Albert

Well-known member
Jan 25, 2013
1,225
$ a_n=100+n^2.\,\, n=1,2,3,----$

$d_n=(a_n,a_{n+1})$

find :max($d_n)$
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
$ a_n=100+n^2.\,\, n=1,2,3,----$

$d_n=(a_n,a_{n+1})$

find :max($d_n)$
If $d$ divides $a_n$ and $a_{n+1}$ then $d$ divides $a_{n+1} - a_n = 2n+1$. So $d$ divides $2a_n - n(2n+1) = 200-n$. Therefore $d$ divides $(2n+1) + 2(200-n) = 401$. But $401$ divides $a_{200}$ ($= 40100$) and $a_{201}$ ($= 40501$). Thus $\max d_n = 401.$
 

kaliprasad

Well-known member
Mar 31, 2013
1,322
dn= (100 + n^2, 100 + (n+1)^2)
= (100+n^2 , 100 + n^2 + 2n + 1)
= (100+n^2, 2n + 1) subtracting 1st term from second
= ( 200+2n^2 , 2n + 1) doubling 1st as 2nd is odd
= (200+ 2n^2- n(2n+1), 2n+ 1)
= ( 200 - n , 2n + 1)
= (400- 2n ,2n + 1) doubling 1st as second is odd
= ( 400 -2 n + 2n+ 1,2n+1)
= (401,2n + 1)
it shows
1) cannot be > 401
2) is 401 when 2n +1 = odd multiple of 400
or 2n + 1 = 401(2k+ 1) = 802k + 401

or 2n = 802k + 400

or n = 401 k + 200
we are able to find n as well for which it is 401