If $n,k\in\mathbb{Z}^+$ and $n+k|n^2$ then $k>\sqrt{n}$ This is what I tried, can someone please see if this is adequate and if not, provide how to do it?  
Suppose that $n,k\in\mathbb{Z}^+$ with $n+k|n^2$.
Suppose for contradiction that $k\leq\sqrt{n}$.
By definition, $n^2 = m(n+k)$ for some $m\in\mathbb{Z}.$
Then $k = \frac{n^2-mn}{m}$.
By assumption that $k\leq \sqrt{n}$, we have  
$$\frac{n^2-mn}{m}\leq\sqrt{n}\\ \frac{n^2(n-m)^2}{m^2}\leq n \\ n^3 - 2mn^2 + m^2n - m^2\leq 0$$
which is a contradiction since the LHS goes to $\infty$ as $n\rightarrow \infty$.
Hence, $k>\sqrt{n}$.
 A: Let $n^2=(n-k+t)(n+k)$ for some positive integer $t$.  Then, we have $(n+k)t=k^2$.  As $k\geq 1$ and $t\geq 1$, we have
$$k^2 \geq (n+1)\cdot 1>n\,.$$
P.S.:  The motivation for the definition of $t$ is as follows: I thought of the identity $$n^2-k^2=(n-k)(n+k)\,.$$  If $n^2$ (which is greater than $n^2-k^2$) is divisible by $n+k$, then the quotient $\frac{n^2}{n+k}$ must be an integer greater than $n-k$.

In fact, we have stronger bounds: $$\frac{1}{2}+\sqrt{n+\frac{1}{4}}\leq k \leq n^2-n\,.$$ The left inequality becomes an equality if and only if $n=k^2-k$ and $t=1$, whereas the right inequality becomes an equality if and only if $k=n^2-n$ and $t=(n-1)^2$.

A: Let $n^2=(n-m)(n+k)$, then $(k-m)n=km$. We then have $m\le k-1$, and therefore,
$$
\frac1{k-1}\le\frac1m=\frac1k+\frac1n\tag1
$$
Multiply $(1)$ by $nk(k-1)$
$$
kn\le(k+n)(k-1)\iff k^2-k-n\ge0\tag2
$$
Since $k\ge0$, we get
$$
k\ge\frac12+\sqrt{n+\frac14}\tag3
$$

I just noticed that directly from $(2)$, we get
$$
k^2\ge n+k\gt n\implies k\gt\sqrt{n}\tag4
$$
without having to solve the quadratic equation.
A: I hope this helps. From your hipotesis $km = n(n-m)$. Due to $k,n\in \mathbb{Z}^{+}$ then $n-m \geq 1$. Now $n^2 \geq n$ and $(n-m)^{2} \geq m^2$ therefore
$k^2m^2 = n^2(n-m)^2 \geq nm^2$ i.e. $km \geq \sqrt{n}m$. This implies $km-\sqrt{n}m\geq 0$ or $m(k-\sqrt{n})\geq 0$ i.e. $k\geq \sqrt{n}$
