5
$\begingroup$

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.

$\endgroup$

marked as duplicate by user91500, Alex M., user223391, colormegone, Pragabhava Mar 24 '16 at 19:28

This question was marked as an exact duplicate of an existing question.

10
$\begingroup$

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 $k\mid 13$.

$\endgroup$
4
$\begingroup$

Let $a_n=n^2+3$. If $p|a_n$ then $n^2=-3\pmod p$.

A quick calculation shows that $a_{n+1}-a_n=2n+1$ so if $p$ also divides $a_{n+1}$ we must have $2n=-1\pmod p$. Square this to see that it implies $4n^2=1\implies -12=1 \pmod p$. Thus, $p|13$ and we are done.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.