<s>[INST] Motivation confusion in floor and ceiling function algebra From HMMT https://hmmt-archive.s3.amazonaws.com/tournaments/2019/nov/team/solutions.pdf:

Compute the sum of all positive real numbers $x \le 5$ satisfying $$x=\frac{\left\lceil{x^2}\right\rceil+\left\lceil x\right\rceil\cdot\left\lfloor x\right\rfloor}{\left\lceil x\right\rceil+\left\lfloor x\right\rfloor}.$$

The given solution says the following.

Note that all integer $x$ work. If $x$ is not an integer then suppose $n < x < n+ 1$. Then $x = n+\frac{k}{2n+1}$, where $n$ is an integer and $1 \le k \le 2n$ is also an integer, since the denominator of the fraction on the right hand side is $2n + 1$. We now show that all x of this form work.
Note that $$x^2=n^2+\frac{2nk}{2n+1}+\left (\frac{k}{2n+1}\right)^2=n^2+k-\frac{k}{2n+1}+\left (\frac{k}{2n+1}\right )^2.$$ For $\frac{k}{2n+1}$ between $0$ and $1$, $-\frac{k}{2n+1}+(\frac{k}{2n+1})^2$ is between $-\frac{1}{4}$ and $0$, so we have $n^2+k-1<x^2\le n^2+k$, and $\lceil x^2\rceil=n^2+k$.
Then, $$\frac{\left\lceil{x^2}\right\rceil+\left\lceil x\right\rceil\cdot\left\lfloor x\right\rfloor}{\left\lceil x\right\rceil+\left\lfloor x\right\rfloor}=\frac{n^2+k+n\cdot (n+1)}{2n+1}=n+\frac{k}{2n+1}=x,$$ so all $x$ of this form work...

To complete the solution there is more, but those parts I believe would be irrelevant to my question, although please comment to correct me if there is more context required, or click on the link, as there's only a paragraph more.
I don't understand, in the first paragraph, why $x=n+\frac{k}{2n+1}$ follows from $n < x < n+1$. More specifically, what is the motivation behind writing this? It seems that the paragraph after this claim is a proof that all $x$ of this form work, but in that proof, this fact is used. So then why do we have to "show that all $x$ of this form work", if it is used in the proof (and thus claimed without proof)?
 [/INST] Since all floors and ceilings are integral, it is clear that the right hand side is a rational number. Because $n < x < n+1$, we have $\left\lfloor{x}\right\rfloor = n$ and $\left\lceil{x}\right\rceil = n+1$. Therefore the denominator of the right hand side is $(n+1) + n = 2n+1$.  By hypothesis, this equals the left hand side which is $x$. Therefore, $x$ is a rational number whose denominator is $2n+1$ (which is not necessarily in reduced form). Since $n < x < n+1$, it follows that $x = n + \frac{k}{2n+1}$ where $1 \le k \le 2n$.
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