MHB What is the maximum value of $d_n$ and for which value of $n$ does it occur?

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The maximum value of \( d_n \) is determined to be 401, which occurs when \( n \) is of the form \( 401k + 200 \) for integer \( k \). The relationship stems from the fact that \( d \) divides both \( a_n \) and \( a_{n+1} \), leading to the conclusion that \( d \) must also divide \( 401 \). The calculations show that \( d \) cannot exceed 401, and it achieves this maximum when \( 2n + 1 \) is an odd multiple of 401. Thus, the maximum value of \( d_n \) is confirmed to be 401 for specific values of \( n \).
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$ a_n=100+n^2.\,\, n=1,2,3,----$

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

find :max($d_n)$
 
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Albert said:
$ a_n=100+n^2.\,\, n=1,2,3,----$

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

find :max($d_n)$
[sp]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 $\max d_n = 401.$[/sp]
 
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
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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