TITLE: Show that 13 is the largest prime which divides two consecutive terms of $n^2 + 3$.
QUESTION [5 upvotes]: Show that 13 is the largest prime which divides two consecutive terms of $n^2 + 3$.

The integers are $39$ and $52$. First of all, I set the variable for the number as $k$. So, $k|n^2 +3$ and $k|n^2 + 2n+ 4$ which imply that $k|2n+1$. $n=6$ over here. And the fact that 13 is the largest 'prime' makes me feel it is hard to prove. That's all I have managed to get. I need a few hints to set me in the right direction. Thanks.

REPLY [11 votes]: If $k\mid n^2+3$ and $k\mid n^2+2n+4$, then as you noted, $k\mid 2n+1$.
But then also from $k\mid n^2+3$ we have $k\mid 2n^2+6$, and from $k\mid 2n+1$ we have $k\mid 2n^2+n$.
Hence $k\mid (2n^2+n)-(2n^2+6))=n-6$. And so $k\mid 2n-12$.
From $k\mid 2n-12$ and $k\mid 2n+1$, we obtain $k\mid 13$.