Question: 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.

Answer: 
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 $13$.
Let $p$ be a prime such that $p$ divides the GCD of $n^2+3$ and $(n+1)^2+3$.

It follows $$2n+1=Np\iff n=\frac{Np-1}{2}$$ Hence $$\left(\frac{Np-1}{2}\right)^2+3=Rp\iff N^2p^2-2Np+1+12=4Rp$$ Consequently $p=13$, only possibility for primes $p$, the other factor being the trivial one $1$.

We must, of course, see at the other condition $(n+1)^2+3$ in order to verify compatibility.It gives immediately, $p|13$ also. 

An easy calculation gives the numbers in question: $$\color{red}{n=13N+6}$$