Find all positive integers $\displaystyle n$ for which $\displaystyle \sqrt{n+\sqrt{1996}}$ exceeds $\displaystyle \sqrt{n-1}$ by an integer. (First Solution)

Find all positive integers $\displaystyle n$ for which $\displaystyle \sqrt{n+\sqrt{1996}}$ exceeds $\displaystyle \sqrt{n-1}$ by an integer.

My solution:

Let $\displaystyle \sqrt{n+\sqrt{1996}}-\sqrt{n-1}=k$, where $\displaystyle k$ is a positive integer.

$\displaystyle \sqrt{n+\sqrt{1996}}=k+\sqrt{n-1}$

Squaring both sides we get:

$\displaystyle n+\sqrt{1996}=k^2+n-1+2k\sqrt{n-1}$

$\displaystyle \cancel{n}+\sqrt{1996}=k^2+\cancel{n}-1+\sqrt{4k^2(n-1)}$

$\displaystyle \begin{align*}\sqrt{1996}+1&=k^2+\sqrt{4k^2(n-1)}\\&\ge k^2+\sqrt{4k^2}=k^2+2k\end{align*}$

since $n\ge 1$

Pressing on the calculator the square root of $1996$ gives us $\approx 44.6766$, so, we know $45^2\gt 1996$, thus we can have the following reasoning:

Since $45^2=2025\gt 1996$, we then have:

$\sqrt{45^2}\gt \sqrt{1996}$

$45+1\gt \sqrt{1996}+1\ge k^2+2k$

$46\gt  k^2+2k$

$k^2+2k-46 \lt 0$

Solving it for $k$, we have $-7\le k \le 5$ which implies $0\le k \le 5$.

Checking it for each case, there is only a solution to the problem, and that is when

$k=1$, which leads to $\displaystyle n=500$.