Prove that for $n$ an integer larger than $0$, $\frac{n^2}{n+1}$ is never an integer. I have a couple questions.
$1)$ Is my proof is considered valid $2)$ If it is good may you tell me how to improve it or if you have a better proof $3)$ how may I prove the obvious assumption I made that an integer and a non integer may never add to an integer.
Sorry for all the questions, here is my proof:
Playing around with the equation:
$$\frac{n^2+2n+1}{n+1}=n+1$$
I get:
$$\frac{n^2}{n+1}+\frac{2n+1}{n+1}=n+1$$
Let $I$ represent an integer, $N$ a non integer. I recall that:
$N+I=I$ is not possible
So either $\frac{n^2}{n+1}$ and $\frac{2n+1}{n+1}$ are both integers or they are both non
integers. 
$\frac{2n+1}{n+1}=f(n)$ is increasing for $n>0$ and is bounded by horizontal asymptote  $y=2$. Thus if $f(n)$ was an integer larger than $0$ it could only be $1$. This would imply $n=0$ which does not satisfy our bound on $n$. Thus for $n$ an integer larger than $0$:
$\frac{2n+1}{n+1}$ is not an integer.
Recall:
$I+N=I$ is not possible 
Thus it is not possible (with $n$ larger than zero) for:
$\frac{n^2}{n+1}$ to be an integer

Note that
$\frac{2n+1}{n+1}-2
=\frac{2n+1-2(n+1)}{n+1}
=\frac{-1}{n+1}
$.
Since
$\frac{-1}{n+1}$
is not an integer,
$\frac{2n+1}{n+1}$
is not an integer.
To handle
$\frac{n^2}{n+1}$,
use
$(n+1)(n-1)
=n^2-1
$.
$\frac{n^2}{n+1}-(n-1)
=\frac{n^2-(n-1)(n+1)}{n+1}
=\frac{n^2-(n^2-1)}{n+1}
=\frac{1}{n+1}
$.
Again,
this is not an integer,
so neither is
$\frac{n^2}{n+1}$.
