<s>[INST] The only positive integers that divide successive numbers of the form $n^2+3$ are $1$ and $13$ I stuck with this problem, I don't know how to start with. 
Prove that the only positive integers that can divide successive numbers of the form $n^2+3$ are 1 or 13.
 [/INST] We apply the euclidean algorithm on the polynomials:
$n^2+3,(n+1)^2+3$
We get:
$(n+1)^2+3=(n^2+3)+(2n+1)$
$2(n^2+3)=n(2n+1)+(-n+6)$
$2n+1=-2(-n+6)+13$
From here:
If $d$ divides $(n+1)^2+3$ and $(n^2+3)$ then $d$ divides $2n+1$.
If $d$ divides $n^2+3$ and $(2n+1)$ then $d$ divides $(-n+6)$.
If $d$ divides $2n+1$ and $-n+6$ then $d$ divides $13$.
Putting it all together we have that if $d$ divides $n^2+3$ and $(n+1)^2+3$ then $d$ divides $13$.
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