<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's an approach that involves checking considerably fewer than $24$ cases, with it also having the advantage that it helps to provide a somewhat intuitive explanation of why the relation holds.
First, note that $\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor = \left\lfloor\frac{n-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor - 4$. Since $-4$ doesn't affect whether or not your $f(n)$ is always a constant, I'll ignore it. Next, set
$$g(n) = \left\lfloor\frac{8n+13}{25}\right\rfloor, \; \; h(n) = \left\lfloor\frac{n-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor, \; \; f_1(n) = g(n) - h(n) \tag{1}\label{eq1A}$$
As you've indicated, $f(n)$ (and, thus, my $f_1(n)$) has a period of $25$. Thus, we just need to prove that $f_1(n)$ doesn't change for any consecutive set of $25$ integers, so induction can then be used for any larger values, & also smaller values, to show $f_1(n)$ is constant for all integers, not only just the positive ones.
Note the numerators inside the floor functions in $g(n)$ and $h(n)$ are both increasing functions, both $g(n)$ and $h(n)$ are integral valued step functions, and the increase of each step is just one. I'll show that $f_1(n)$ is a constant for $0 \le n \le 24$ by checking the end-points and each value of $n$ where $h(n)$ increases. At those points, I'll show that
$$8n + 13 = 25(h(n)) + r, \; \; 0 \le r \le 12 \tag{2}\label{eq2A}$$
to confirm $g(n) = h(n)$ and $g(n-1) = g(n) - 1$.
For $0 \le n \le 16$, we have $h(n) = \left\lfloor\frac{n+1}{3}\right\rfloor$, and for $17 \le n \le 24$, we have $h(n) = \left\lfloor\frac{n}{3}\right\rfloor$. Next, $g(0) = h(0) = 0$ and
$$h(2) = 1, \; 8(2) + 13 = 25(1) + 4 \tag{3}\label{eq3A}$$
For $n$ up to $16$, the value of $h(n)$ increases by $1$ for each increase in $n$ by $3$, with $8n + 13$ increasing by $8(3) = 25 - 1$ each time. This means $g(n)$ also increases by $1$, but with the value of $r$ decreasing by $1$ each time, starting from $4$ as indicated in \eqref{eq2A}. Thus, we get a match up to $h(14) = 5, \; 8(14) + 13 = 25(5) + 0$.
The next $h(n)$ increase is at $n = 18$, with $h(18) = 6, \; 8(18) + 13 = 25(6) + 7$. Thus, as discussed above, we'll continue to have matching increases with $g(21) = h(21)$ and $g(24) = h(24)$. This confirms $g(n) = h(n)$ for $0 \le n \le 24$.

Note: Since the remainder $r$ is $0$ at $n = 14$, the next increase in $h(n)$ must occur when $n$ increases by $4$ instead of $3$, with this accomplished by the increase in value of $\left\lfloor\frac{n-m}{25}\right\rfloor$ at $n = m$ for any $15 \le m \le 17$, not just $m = 17$ as used in the question.
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