\[a_n=an^2+bn+c\]
.We can find the greatest common divisor of successive terms by repeated subtraction or addition of multiples of successive terms.
\[ a_n=an^2+bn+c\]
\[a_{n+1}=a(n+1)^2+b(n+1)+c=an^2+(2a+b)n+a+b+c \]
\[\begin{equation} \begin{aligned} & gcd(a_n, a_{n+1} ) \\ &= gcd(an^2+bn+c,an^2+(2a+b)n+a+b+c) \\ &= gcd(an^2+bn+c,(an^2+(2a+b)n+a+b+c)\\ &-(an^2+bn+c) = 2an+a+b) \\ &= gcd(2(an^2+bn+c)-n(2an+a+b) \\ &= (a-b)n-2c, 2an+a+b)) \\ &= gcd(2a((a-b)n-2c)-(a-b)(2an+a+b) \\ &= b^2-a^2-4ac, 2an+a+b ) \end{aligned} \end{equation}\]
Hence
\[gcd(a_n, a_{n+1} ) =b^2-a^2-4ac\]
or some divisor thereof.Example here.