<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] $$\gcd(n^2+3,(n+1)^2+3)=\gcd(n^2+3,2n+1)=\gcd(4n^2+12,2n+1)=$$
$$=\gcd(4n^2+4n+1-4n+11,2n+1)=\gcd(4n-11,2n+1)=$$
$$\gcd(4n+2-13,2n+1)=\gcd(13,2n+1).$$
Can you end it now?
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