Conjecture about $A f(x) = f(g(x)) + f(h(x))$

Question

Let a given real $A$ satisfy $0 < A < 2$.

Conjecture :

For any real entire nonconstant $f(x)$ there exist real entire $g(x)$ and $h(x)$ such that

$A f(x) = f(g(x)) + f(h(x))$

or

$ A f(x) = f(g(x)) - f(h(x))$

is satisfied.

Is this true ?

If its true , then how to compute $g(x)$ and $h(x)$ efficiently ?

edit

I changed the question due to HansEnglers comment. Basically I assumed the equations to hold on a real interval but because of analytic continuation the equation must hold everywhere.

edit

edit 2

I added the condition that $f(x)$ needs to be a nonconstant function. ( I am tempted to wonder if I need to say nonpolynomial )

This edit is thanks to the comment of user43208.

edit 2

Answer 1

The conjecture is false. Consider $ A = 1 $ and $ f ( x ) = \frac 1 2 \left ( 1 + \frac 1 { 1 + x ^ 2 } \right ) $, both satisfying the requirements of the problem. for any $ x \in \mathbb R $ we have $ \frac 1 2 < f ( x ) \le 1 $. Therefore for any $ y , z \in \mathbb R $ we have $ f ( y ) + f ( z ) > 1 $ and $ f ( y ) - f ( z ) < \frac 1 2 $. This means that $ A f ( x ) = f ( y ) + f ( z ) $ and $ A f ( x ) = f ( y ) - f ( z ) $ never happen for any $ x , y , z \in \mathbb R $.