Let $n$ and $a$ denote positive integers, and let $f_n(a)$ be a monotonically increasing function, with integer coefficients, of degree $n-1$.
If there exists some pair $(n,k)$ such that $$\frac{(4k+1)f_n(16k+3)}{(8k+1)f_n(16k+2)}=1, \tag{$\star$}$$ is there anything that can be said about $f$ [without knowing its coefficients, etc.]?
Evidently $\gcd(4k+1,8k+1)=1$, so we must have $(8k+1) \mid f_n(16k+3)$ and $(4k+1) \mid f_n(16k+2)$; in fact, there must exist a positive integer $m$ such that $$ f_n(16k+2) = (4k+1)m \qquad\text{and}\qquad f_n(16k+3)=(8k+1)m. $$ I'm just not sure how to proceed from there.
To be honest, I’ve been trying to find any function which satisfies ($\star$) [and the “monotonically increasing integer function” requirements], but haven’t found one yet. I’m close to calling the non-existence of such a function a conjecture, but I’m not quite there yet…
EDIT: @garondal found a counterexample to the [potential] “conjecture“ (see first answer).
Let $f(x)=67x^2-1111x$ and $n-1=2$ and $k=1$. This function is monotonically increasing for $x \geq 1$.
$f(19)=3078$ and $f(18)=1710$. Then you have that $\frac{5f(19)}{9f(18)}=1$.
Unfortunately I found this with brute force so I cannot offer any insights.
Edit for more general case: If you define $$f_n(x)=a_n x^n - a_{n+1} x^{n-1}$$ with $a_1=4$ and $$a_n=a_{n-1}\cdot \dfrac{5\cdot 19^{n-1}-9\cdot 18^{n-1}}{5\cdot 19^{n-2}-9\cdot 18^{n-2}}$$ you get a function with $\frac{5f_n(19)}{9f_n(18)}=1$. The sequence $(a_n)$ is $4, 67, 1111, 18\,193, 29\,3179, \ldots$. This seems to work but the functions are not monotonically increasing for $x\geq 1$.