The statement of the problem : Find all monotone functions $f : \mathbb R \rightarrow \mathbb R $ that verify the following $f(x) + f(y) = f ( xy - 2024x - 2024y + 2024*2025 ) ,\forall x,y \in (2024,\infty)$
My ideas (some of them are wrong ) : First of all, by making $x = y = 2025$ we easily get that f(2025) = 0. Thinking about the fact that it is monotonic, I thought that maybe I can prove that it is a constant(proving that it has different monotonies on certain intervals), which would result in $f \equiv 0$, which verifies so it might be the right path. I thought how we can exploit the fact that $f(2025) = 0$ so I thought about the solutions of the equation $$xy - 2024x - 2024y + 2024*2025=2025$$ now the bad part is that $x,y\in (2024,\infty)$ , so there aren't any solutions .The next idea was to think of an additive function, but it was unsuccessful.
Does it seem like a difficult problem? I want to see what ideas and solutions you have, and what helped you to intuit the answer, because maybe this will help me to develop a more analytical way of thinking, or at least make me understand the thought process behind the solution. Thank you very much for all the ideas and solutions !
We can simplify the RHS as: $$ f(xy-2024x-2024y+2024\cdot2025)=f((x-2024)(y-2024)+2024). $$ i.e, $$ f(x)+f(y)=f((x-2024)(y-2024)+2024) $$ Now, let $f(x)=g(x-2024)$ for some $g(x)$. This can also be written as $f(x+2024)=g(x)$. Therefore, $$ f(x)+f(y)=f((x-2024)(y-2024)+2024) \\ \implies g(x-2024)+g(y-2024)=g((x-2024)(y-2024)). $$ Substitute $u=x-2024$ and $v=y-2024$. Thus, $$ g(u)+g(v)=g(u\cdot v). $$ From this, we obtain that $g(x)=\log_a(x), a>0$. Thus, our original function, $f(x)=g(x-2024)=\log_a(x-2024)$ is the solution.
EDIT: Consider the function $h(x)=g(e^x)$, this function is additive, i.e, $h(x)+h(y)=h(x+y)$. Since both $g(x)$ and $e^x$ are monotonic, $h(x)$ is also monotonic. Additive monotonic functions are linear. Thus $h(x)=ax,a\in\mathbb{R}$. $g(e^x)=ax$ or $g(x)=a\ln(x)$.