2016/006) show that 13 is the largest number that divides 2 numbers of the form $(n^2+3),(n+1)^2+3$

 let us find the GCD

$GCD(n^2+3,(n+1)^3+3) = GCD(n^2+3, n^2 + 2n + 4)$
$=GCD(n^2+3, n ^2+2n+4-(n^2+3))= GCD(n^2+3,2n+1)$
$=GCD(2*(n^2+3), 2n+ 1)$ we can multiply 1st one by 2 as 2nd one is odd
$=GCD(2n^2+6,2n++1)$
$=GCD(2n^2+6-n(2n+1),2n+1)$
$=GCD(6-n, 2n+ 1)$
$= GCD(12-2n,2n+1)$ we can multiply 1st one by 2 as 2nd one is odd
$= GCD(13,2n+1)$
So 13 is the largest number that divides both and when 2n+1 = 13 mod 13 or n= 6 mod 13