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

Find all positive integers [MATH]n[/MATH] for which [MATH]\sqrt{n+\sqrt{1996}}[/MATH] exceeds [MATH]\sqrt{n-1}[/MATH] by an integer.

My solution:

Let [MATH]\sqrt{n+\sqrt{1996}}-\sqrt{n-1}=k[/MATH], where [MATH]k[/MATH] is a positive integer.

[MATH]\sqrt{n+\sqrt{1996}}=k+\sqrt{n-1}[/MATH]

Squaring both sides we get:

[MATH]n+\sqrt{1996}=k^2+n-1+2k\sqrt{n-1}[/MATH]

[MATH]\cancel{n}+\sqrt{1996}=k^2+\cancel{n}-1+\sqrt{4k^2(n-1)}[/MATH]

[MATH]\begin{align*}\sqrt{1996}+1&=k^2+\sqrt{4k^2(n-1)}\\&\ge k^2+\sqrt{4k^2}=k^2+2k\end{align*}[/MATH]

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 [MATH]n=500[/MATH].