$\gcd(n^2+3,(n^2+2n+1)+3)=\gcd(2n+1,n^2+3)=d$ $2n+1 \equiv 0 \mod d$ $n^2+3 \equiv 0 \mod d$ $13\equiv 0 \mod d$ $d=13$ $n=13k+6$ Otherwise $d=1$ so maximum $d=13$

$d\left| {{n}^{2}} \right.+3;d\left| {{n}^{2}}+2n+4\Rightarrow d\left| 2n-1 \right. \right.$ thus $d\left| 2{{n}^{2}}-n,d\left| 2{{n}^{2}}+6\Rightarrow d\left| n+6\Rightarrow d\left| 2n+12,d\left| 2n-1\Rightarrow d\left| 13 \right. \right. \right. \right. \right. \right.$