2

I have given the sequence $A_i = K+i^2$ for a given integer $K$ and it looks like: $K+1,K+4,K+9,\dots$

Now I want to consider $gcd(A_i,A_{i+1})$, where gcd(A,B) denotes the greatest common divisor between A and B. So is there something special of gcd of consecutive perfect squares shifted by $K$?

Can I simplify this expression? Any help would be greatly appreciated!

Bill Dubuque
  • 272,048
kryomaxim
  • 2,882

1 Answers1

1

$A_n=K+n^2$

$A_{n+1}=K+(n+1)^2$

$gcd(y,x)|(x\pm y)$

$I:gcd(A_n,A_{n+1})|(2n+1)$

$II:gcd(A_n, A_{n+1})|(2K+2n^2+2n+1)$

$2,5,10,17,26,37,50,65- x,1,5,1,1,1,1,5$

$3, 6,11, 18, 27, 38, 51, 66, 83,102- x,3,1,1,9,1,1,3,1,1 $

$4, 7,12,19, 28, 39,52,67,84,103- x,1,1,1,1,1,13,1,1,1 $

$g|(4K+4n^2+4n+2)\land g^2|(4n^2+4n+1)\implies g|(4K+1)$

$gcd(A_n, A_{n+1})| (4K+1) \land gcd(A_n, A_{n+1})|(2n+1) $

TurlocTheRed
  • 5,683
  • When you update yuur answer to include deductions already posted in comments or dupes then you should acknowledge such. – Bill Dubuque Aug 03 '23 at 19:19