<s>[INST] Find all possible values of $\gcd(n^2+3, (n+1)^2+3)$. Let $n \in \mathbb{N}$. Find all possible values of $\gcd(n^2+3, (n+1)^2+3)$. 
I began this problem giving some values for $n$ and I found that $\gcd(n^2+3, (n+1)^2+3)=1$ for most of $n$ I tried, but  if $n=6$, then $\gcd=13$.
Then I tried to prove that only for $n=6$, $\gcd \neq 1$ but I couldn't. 
Can someone help me with this problem?  
 [/INST] I think that I just solve it:
Let $d=gcd(n^2+3, (n+1)^2+3).$ By the property $\gcd(a,b)=\gcd(a,b-a)$, we obtain:
$$d=\gcd(n^2+3,2n+1).$$
Since $d \mid n^2+3,$ we have $n^2+3=ds$ for some integer $s$. 
Similarly, $2n+1=dr$ for some integer $r$, and by this relation we obtain: $4n^2=d^2 r^2 -2dr +1 \iff 4ds=4(n^2+3)=d^2 r^2 -2dr +13$.
Hence, it must be clear that $13=dq$ for some integer $q$. Then $d \mid 13$.
So $d=13$ or $d=1$ since $13$ is prime. 
It is correct?
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