<s>[INST] Show that the prime must be $13$ if $p\mid n^2+3$ and $p\mid (n+1)^2+3$ Let $p$ be a prime for which it exists a $n\in \mathbb{Z}$ with $p\mid n^2+3$ and $p\mid (n+1)^2+3$. I want to show that it must hold that $p=13$ and that there are infinitely many integers $n$ so that the above relations hold. 
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We have that $p\mid n^2+3$ and $p\mid (n+1)^2+3 \Rightarrow p\mid n^2+2n+1+3$, so we get $p\mid (n^2+2n+1+3)-(n^2+3) \Rightarrow p\mid 2n+1$, right?  
How could we continue? 
 [/INST] We have $p\mid n^2+3$ and $p\mid (n+1)^2+3$. So in particular $p$ divides the difference $2n{+}1$ (and $p$ is odd), as you say.
This gives us $2n\equiv p{-}1$ and thus $n \equiv 2^{-1}(p-1) \bmod p$.
Then working all $ \bmod p$
$\begin{align}
n^2 &\equiv (2^{-1}(p-1))^2 \\
&\equiv 4^{-1}(p^2-2p+1) \\
&\equiv 4^{-1} \\
\text{ and since }\quad n^2+3&\equiv 0 \\
4^{-1}&\equiv -3 \\
1&\equiv -12 \\
13&\equiv 0 \\
\end{align}$
This gives $p \mid 13 $ and thus $p=13$ is the only answer.
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