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
 A: Your proof is interest but seems perfectly valid.  As (n-1)(n+1) = $n^2 -1 < n^2$ and $n^2 < n(n+1)$ (for n $\ge 2$; and we can test n = 1) so $n - 1 < \frac {n^2}{n+1} < n$ wouldn't need the $\frac {2n+1}{n+1}$ step (although it was clever.)
=====
Integer - Integer = Integer by induction.
So (Integer + Non-Integer) - First Integer = Non-Integer.  So if (Integer + Non-Integer were an integer than Non-Integer would be an integer too, which it isn't. So we can safely assume Integer + non-integer is not an integer.
A: 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
$\frac{n^2}{n+1}$.
A: For if it were some integer $k$, then $\dfrac{n^2}{n+1} = k \Rightarrow n^2 = kn + k \Rightarrow n^2 -kn - k = 0$ and this implies this quadratic equation has integer solution $n$, thus $\triangle = b^2-4ac = (-k)^2 - 4(1)(-k)= k^2 + 4k$ must be a perfect square. But $(k+1)^2 < \triangle < (k+2)^2$, thus it cannot be a perfect square, proving the assertion....
A: Suppose that $\frac{n^2}{n+1}\in\mathbb{Z}$. Then $n^2=k(n+1)$ for some $k\in \mathbb{Z}$. Now \begin{array} ((n-1)(n+1)&=n^2-1\\&<n^2\\&<n^2+n\\&=n(n+1),\end{array} so $(n-1)<k<n$, which contradicts that $k\in\mathbb{Z}$, whence $\frac{n^2}{n+1}\in(\mathbb{R}-\mathbb{Z})$ 
