Prove that $\lceil \frac{\sqrt{n^2+1+\sqrt{n^2}}}{\sqrt{n^2+3+\sqrt{n^2+2}}-\sqrt{n^2+2+\sqrt{n^2+1}}}\rceil = 2n^2+n+3$ First, the question: Prove that
$$\Bigg\lceil \frac{\sqrt{n^2+1+\sqrt{n^2}}}{\sqrt{n^2+3+\sqrt{n^2+2}}-\sqrt{n^2+2+\sqrt{n^2+1}}}\Bigg\rceil = 2n^2+n+3$$
The motive to this question is the following. I once saw the following "near-identity" in a facebook page
$$7(\sqrt{7+\sqrt{7}}-\sqrt{5+\sqrt{5}}) \approx \sqrt{6+\sqrt{6}}$$
So I thought of generating similar near-identities by directly calculating
$$E(k) = \frac{\sqrt{k+\sqrt{k}}}{\sqrt{k+1+\sqrt{k+1}}-\sqrt{k-1+\sqrt{k-1}}}$$
and then look for those entries which the expression is almost an integer (for example when k=6 we have $E(6) \approx 6.991418$)
Then I got the idea that we can also find near-identities of the form
$$N(\sqrt{k+3+\sqrt{k+2}} - \sqrt{k+2+\sqrt{k+1}}) \approx \sqrt{k+1+\sqrt{k}}$$
where $N, k$ is are positive integers, by calculating
$$F(k) = \frac{\sqrt{k+1+\sqrt{k}}}{\sqrt{k+3+\sqrt{k+2}}-\sqrt{k+2+\sqrt{k+1}}}$$
and then again look for almost-integer entries. Concretely in this case, my definition of "almost-integer" here is that the number is at most 0.001 away from the integer closest to it. Numerical evidence shows that the $k$'s for which $F(k)$ is an almost-integer are $25, 36, 49, 64, 81, \ldots$ which suggest something is at play. After some pattern-guessing, I ended up with the claim in the beginning.
Direct graph plotting seems to indicate that
$$f(x) = 2x+\sqrt{x}+3 - \Bigg(\frac{\sqrt{x+1+\sqrt{x}}}{\sqrt{x+3+\sqrt{x+2}}-\sqrt{x+2+\sqrt{x+1}}}\Bigg)$$
converges to 0 "from above" the x-axis, but that's as far as I can go.
Also, calculating $E(k)$ also yields the following two claims (which is also beyond my reach at this point)
$$\Bigg\lfloor \frac{\sqrt{2n(2n+1)+1+\sqrt{2n(2n+1)}}}{\sqrt{2n(2n+1)+1+\sqrt{2n(2n+1)+1}}-\sqrt{2n(2n+1)-1+\sqrt{2n(2n+1)-1}}}\Bigg\rfloor = 2n(2n+1)+n$$
and
$$\Bigg\lfloor \frac{\sqrt{2n(2n-1)+\sqrt{2n(2n-1)}}}{\sqrt{2n(2n-1)+1+\sqrt{2n(2n-1)+1}}-\sqrt{2n(2n-1)-1+\sqrt{2n(2n-1)-1}}}\Bigg\rfloor_{0.5} = 2n(2n-1)+n-0.5$$
where $\lfloor x \rfloor_{0.5}$ is the greatest integer or half-integer less than $x$, for example, $\lfloor 1.501 \rfloor_{0.5} = 1.5$
Any help would be highly appreciated.
(PS. this is not my homework, just a curiosity)

I do not know how much this could help you : for large values of $n$, the expansion of $$F(n)= \frac{\sqrt{n^2+1+\sqrt{n^2}}}{\sqrt{n^2+3+\sqrt{n^2+2}}-\sqrt{n^2+2+\sqrt{n^2+1}}}$$ is given by $$2 n^2+n+3-\frac{1}{4 n^2}+\frac{1}{16 n^3}+\frac{9}{16 n^4}-\frac{45}{64
   n^5}+O\left(\left(\frac{1}{n}\right)^6\right)$$ and the difference between $F(n)$ and $G(n)=2 n^2+n+3$ is always negative and smaller than $1$. The largest difference is obtained for $n=-2$ for which $F(-2)=8.48432$ and $G(-2)=9$.
