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]

Another method to prove $7 ≥ \sqrt 2+\sqrt 5 + \sqrt {11}$

Show with proof which of these two values is smaller:

$7$, or $\sqrt 2+\sqrt 5 + \sqrt {11}$

In my (few) previous blog post (Which is greater), I mentioned of how I proved for $\sqrt 2+\sqrt 5 + \sqrt {11}$ is smaller than $7$.

But that doesn't mean that solution is the only way out to prove for that kind of problem.