<s>[INST] $\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$ is independent of $n$ 
If $n$ is a positive integer, prove that
$$\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$$
is independent of $n$.

Taking
$$f(n)=\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$$
I could prove that
$$f(n+25)=f(n)$$
But, I have no idea how to show
$$f(n+k)=f(n)\\\forall k\in \{1,2,\dots 24\}$$
Of course, we can check these finite number of cases by force. But is there any other elegant method to handle this?
 [/INST] Here is a variation which is also related to Bill Dubuque's hint. We want to show that
\begin{align*}
f(n)&=\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor\\
&=\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor+4
\end{align*}
is independent from $n$. Since
\begin{align*}
f(0)&=\left\lfloor\frac{13}{25}\right\rfloor-\left\lfloor\frac{-\left\lfloor\frac{-17}{25}\right\rfloor}{3}\right\rfloor+4\\
&=0-0+4=4
\end{align*}
we claim the following is valid for all non-negative integers $n$:
\begin{align*}
\color{blue}{\left\lfloor\frac{8n+13}{25}\right\rfloor=\left\lfloor\frac{n-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor}\tag{1}
\end{align*}

We start with the right-hand side of (1) and obtain
\begin{align*}
\color{blue}{\left\lfloor\frac{n-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor}
&=\left\lfloor\frac{-\left\lfloor\frac{-24n-17}{25}\right\rfloor}{3}\right\rfloor
=\left\lfloor\frac{\left\lceil\frac{24n+17}{25}\right\rceil}{3}\right\rfloor\tag{2.1}\\
&=\left\lfloor\frac{\left\lfloor\frac{24n+41}{25}\right\rfloor}{3}\right\rfloor
=\left\lfloor\frac{24n+41}{75}\right\rfloor\tag{2.2}\\
&=\left\lfloor\frac{24n+39}{75}\right\rfloor\tag{2.3}\\
&\,\,\color{blue}{=\left\lfloor\frac{8n+13}{25}\right\rfloor}
\end{align*}
and the claim follows.

Comment:

*

*In (2.1) we bring $n$ into the inner floor symbol and use the rule
$$\lfloor -x\rfloor=-\lceil x\rceil$$


*In (2.2) we use the rule
\begin{align*}
\left\lceil \frac{n}{m} \right\rceil =\left \lfloor \frac{n+m-1}{m} \right\rfloor
\end{align*}
and also the nested division rule
\begin{align*}
\left\lfloor\frac{\left\lfloor x/m\right\rfloor}{n}\right\rfloor=\left\lfloor\frac{x}{mn}\right\rfloor
\end{align*}


*In (2.3) we can check for $0\leq n<25$ the validity of $\left\lfloor\frac{24n+41}{75}\right\rfloor=\left\lfloor\frac{24n+39}{75}\right\rfloor$. We can alternatively note the difference between these two expressions inside the floor symbols is $\frac{2}{75}$. So, different natural numbers would occur, iff there is a natural number $m$ with
\begin{align*}
\frac{24n+39}{75}=m-\frac{1}{75}\qquad\quad\text{and}\qquad\quad\frac{24n+41}{75}=m+\frac{1}{75}
\end{align*}
This is equivalent with the existence of a solution of the linear congruence relation
\begin{align*}
24n+41&\equiv 1\pmod{75}\\
24n&\equiv 35\pmod{75}\\
\end{align*}
But this has no solution, since
\begin{align*}
35&\not\equiv 0\pmod{\gcd(24,75)}\\
35&\not\equiv 0\pmod{3}\\
\end{align*}
</s>