I was solving a problem when this random thought popped into my head. Suppose you have a function, say $f(x)=x^2-1=(x-1)(x+1)$. The roots of this function are $-1$ and $-1$. We can write these roots as $f(a)$ for some $a$, because solving $f(x)=-1$ gives $x=0$ and solving $f(x)=1$ gives $x=\pm \sqrt{2}$. Therefore, we can write the function as either $f(x)=(x-f(\sqrt{2}))(x-f(0))$ or $f(x)=(x-f(-\sqrt{2}))(x-f(0))$.
My questions: 1. Has any work been done on such representations of f(x)? 2. If not, does this seem like a nice idea or is it just a trivial thing that won't lead anywhere? 3. Can you think of any way to reverse the process? For example, given that $f(x)=(x-f(\sqrt{2}))(x-f(0))$, is it possible to find $f(x)$'s that "work"?
Thanks!
EDIT: Example: For $f(x)=x$ the only root is $x=0$. So we can write it (as we did above) as $f(x)=x-f(0)$.
For the reverse problem, we want to solve $f(x)=x-f(0)$ for $f(x)$. Plugging $x=0$ yields $f(0)=-f(0)$ or $2f(0)=0$ or $f(0)=0$. Therefore $f(x)=x-f(0)=x$. So the only function that "works" is indeed $f(x)=x$.
EDIT (2): Harder example: Solve $f(x)=-(x-f(0))$.
HINT
$$\begin{align} f(x) &= f(0) - x \\ f(1) &= f(0) - 1 \\ f(2) &= f(0) - 2 \\ &= f(1) - 1 \\ &\vdots \\ \implies f(n) &= f(n-1) -1 \end{align}$$
Therefore, $f(x) = -x$.