Proving divisibility in elementary number theory problem Find all positive integers n such that $(n+1)\mid(n^2+1)$.
What I have done so far.
I noticed that $ n^2 + 1 = (n + 1 - 1)^2 + 1 = (n + 1)^2 -2(n + 1) + 2$.
Hence, for the relation to be true, we must have that $(n+1)\mid 2$, that is $n=1$.
How would I prove it?
 A: Your argument is wonderful.
Why must $n = 1$? Since the only positive integers that divide $2$ are $1$ and $2.$ So the only solution for positive $n$ such that $(n + 1)$ divides $2$ is when for $n = 1 \implies n + 1$. $\;(n = 0$ would given us $n + 1 = 1 \mid 2$, but $n = 0 \ngeq 1$.)
Done!
Of course, you are using the fact that $\,a\mid (am + an + x) \iff a\mid a(m + n) + x \implies a \mid x,\;$ but that is fairly easy discern, given your argument.
A: Your argument is very good. There is a very minor flaw. In principle, you should have written that $n+1$ divides $n^2+1$ if and only if $n+1$ divides $2$. Since you did not write that, and only wrote the equivalent of "any solution must divide $2$," you should have verified that $n=1$ actually works. 
I will describe two other approaches  that are (in this case) less nice than yours. 
$1.$ Divide the polynomial $x^2+1$ by $x+1$, in the usual way. We get
$$\frac{x^2}{x+1}= x-1+\frac{2}{x+1}.$$
It follows that for any integer $n\ne -1$, we have
$$\frac{n^2+1}{n+1} =n-1 +\frac{2}{n+1}.$$
Thus $n+1$ divides $n^2+1$ if and only if $n+1$ divides $2$. The only positive solution is $n=1$.
$2.$ We use congruence notation. If you have not met it yet, it is coming soon. Note that $n\equiv -1\pmod{n+1}$. It follows that $n^2\equiv 1\pmod{n+1}$. But from $n+1$ divides $n^2+1$, we conclude that $n^2\equiv -1\pmod{n+1}$. 
From $n^2\equiv -1\pmod{n+1}$ and $n^2\equiv -1\pmod{n+1}$, we conclude by subtraction that $2\equiv 0\pmod{n+1}$, that is, $n+1$ divides $2$. The only (positive) $n$ that works is given by $n=1$. Finally, if $n=1$, then indeed $n+1$ divides $n^2+1$. 
