<s>[INST] $\forall n\in\mathbb{Z}$, find the $\gcd(n^2+1, (n+1)^2+1)$? 
$\forall n\in\mathbb{Z}$, find the $\gcd(n^2+1, (n+1)^2+1)$. 

I think this is a simple exercise, but I get this: 
$(n+1)^2+1=n^2+2n+2$. 
$n^2+2n+2 = (n^2+1)+(2n+1)$ 
then $\gcd(n^2+1, (n+1)^2+1)=\gcd(n^2+1, 2n+1)$ 
and $\displaystyle n^2+1 = \frac{n(2n+1)}{2}+\left(-\frac{n}{2}+1\right)$ 
then $\gcd(n^2+1, 2n+1)=\gcd(2n+1, \frac{n}{2}-1)$. 
But gcd is on integer numbers and $\dfrac{n}{2}-1$ is not always an integer number, so, I need some help?
 [/INST] Your step that says $\displaystyle n^2+1 = \frac{n(2n+1)}{2}+\left(-\frac{n}{2}+1\right)$ is correct but not useful.  We are working in the naturals and this threatens (if $n$ is not even) to takes us out of the naturals.  
If we just make a spreadsheet and try the small numbers we find that the GCD is $1$ unless $n\equiv 2 \pmod 5$, in which case it is $5$.  We can see that if $n \equiv 2 \pmod 5$, $n^2+1 \equiv (n+1)^2+1 \equiv 0 \pmod 5$.  Now we need to ask if there are any other numbers $k$ where $n^2\equiv (n+1)^2 \equiv -1 \pmod k$.  This requires that $2n+1 \equiv 0 \pmod k$.  We can factor $k$ demand that $2n+1 \equiv 0 \pmod p$ for some prime $p$ dividing $k$.  If we square that we get $4n^2+4n+1\equiv 4n-3\equiv 0, n \equiv \frac 34 \pmod p$ where we take advantage of the fact that $\Bbb Z_p$ is a field to divide.  Now $n^2 \equiv -1 \pmod p$ says $9 \equiv -16 \pmod p$ and the only prime is $5$.  So the $\gcd$ is $1$ unless $n \equiv 2 \pmod 5$ in which case it is $5$.
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