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? 

Since $p \mid 2n+1$ we have $p\mid(2n+1)^2=4n^2+4n+1$.
We also have $p\mid2(n^2+3)+2((n+1)^2+3)=4n^2+4n+14$. Thus we have $p\mid13$ and $p=13$.
Also, any $n=13k+6, k\in \mathbb{Z}$ satisfies the condition.
