Solving the functional equation: $f\bigl(f(x)-x\bigr)=2x$ PROBLEM STATEMENT:

Find all functions $f :\mathbb{R}^{+}_{0} \to \mathbb{R}^{+}_{0} $ satisfying the functional equation
$$f\bigl(f(x)-x\bigr)=2x.$$

MY PROGRESS:
This question seemed pretty tough to me. All I could figure out was:

*

*$f(x) = 2x$ and  $f(x)=-x$ (considering the domain and co-domain to be all reals) satisfy the given equation.

*$f\bigl(f(0)\bigr)=0$.

*By substituting certain values for $x$, I found out that
$$f\bigl(n f(0)\bigr)= (-n+1) f(0)$$
for $n=1,3,-5,11, \dots$.

But these didn't help me much. What should be my approach? Clearly the general way of substitution and doing some algebraic manipulations didn't help you to proceed with the question. Any hints or ideas?
Also the solution to this question used some sort of sequences and recurrence relation that I couldn't understand at all. Any ideas in this respect?
Thanks.
Feel free to add tags and a better title as I am not good at that.
PS:
In general $$f\bigl(f(x)-nx\bigr)=(n+1)x,$$ for all integers $n$, is satisfied by $f(x)=(n+1)x$. Can this be proved explicitly?
 A: As Marktmeister has mentioned in a comment, for the functional equation
$$ f \bigl ( f ( x ) - x \bigr ) = 2 x \tag 0 \label 0 $$
to make sense for a given $ f : \mathbb R ^ + _ 0 \to \mathbb R ^ + _ 0 $ and all $ x \in \mathbb R ^ + _ 0 $, one must have $ f ( x ) \ge x $ for all $ x \in \mathbb R ^ + _ 0 $. In particular, $ f \bigl ( f ( x ) - x \bigr ) \ge f ( x ) - x $ for all $ x \in \mathbb R ^ + _ 0 $, which by \eqref{0} gives $ f ( x ) \le 3 x $ for all $ x \in \mathbb R ^ + _ 0 $. Let's iterate this method: define the sequences $ ( a _ n ) _ { n = 0 } ^ \infty $ and $ ( b _ n ) _ { n = 0 } ^ \infty $ of positive real numbers recursively by $ a _ 0 = 1 $ and for any nonnegative integer $ n $, $ b _ n = \frac 2 { a _ n } + 1 $ and $ a _ { n + 1 } = \frac 2 { b _ n } + 1 $. Note that similar to what we did above, by \eqref{0} and mathematical induction, one gets $ a _ n x \le f ( x ) \le b _ n x $ for all $ x \in \mathbb R ^ + _ 0 $ and all positive integers $ n $. Therefore, if we prove that as $ n $ tends to infinity, $ a _ n $ increasingly goes to $ 2 $ and $ b _ n $ decreasingly goes to $ 2 $, then we end up with $ f ( x ) = 2 x $ for all $ x \in \mathbb R ^ + _ 0 $, which indeed gives a solution to \eqref{0}. For this purpose, first use mathematical induction to prove that for any nonnegative integer $ n $, $ 1 \le a _ n < 2 $ and $ 2 < b _ n \le 3 $. Then, note that for any nonnegative integer $ n $,
$$ 0 < 2 - a _ { n + 1 } = 2 - \left ( \frac 2 { \frac 2 { a _ n } + 1 } + 1 \right ) = \frac { 2 - a _ n } { 2 + a _ n } \le \frac { 2 - a _ n } { 2 + 1 } = \frac { 2 - a _ n } 3 $$
and
$$ 0 < b _ { n + 1 } - 2 = \left ( \frac 2 { \frac 2 { b _ n } + 1 } + 1 \right ) - 2 = \frac { b _ n - 2 } { b _ n + 2 } < \frac { b _ n - 2 } { 2 + 2 } = \frac { b _ n - 2 } 4 \text . $$
Hence, by mathematical induction, we get
$$ 0 < 2 - a _ n \le \frac { 2 - a _ 0 } { 3 ^ n } = \frac 1 { 3 ^ n } $$
and
$$ 0 < b _ n - 2 \le \frac { b _ 0 - 2 } { 4 ^ n } = \frac 1 { 4 ^ n } $$
for all nonnegative integers $ n $, which proves what was desired.

Concerning the part you added later, if $ f : \mathbb R ^ + _ 0 \to \mathbb R ^ + _ 0 $ satisfies
$$ f \bigl ( f ( x ) - \alpha x \bigr ) = \beta x \tag 1 \label 1 $$
for some constants $ \alpha , \beta \in \mathbb R ^ + $ and all $ x \in \mathbb R ^ + _ 0 $, the same method as above works. Note that for all $ x \in \mathbb R ^ + _ 0 $ and all $ \gamma \in \mathbb R ^ + $, $ f \bigl ( f ( x ) - \alpha x \bigr ) \ge \gamma \bigl ( f ( x ) - \alpha x \bigr ) $ implies $ f ( x ) \le \left ( \frac \beta \gamma + \alpha \right ) x $ (by \eqref{1}), and $ f \bigl ( f ( x ) - \alpha x \bigr ) \le \gamma \bigl ( f ( x ) - \alpha x \bigr ) $ implies $ f ( x ) \ge \left ( \frac \beta \gamma + \alpha \right ) x $. Therefore, similar to before, we can define $ ( a _ m ) _ { m = 0 } ^ \infty $ and $ ( b _ m ) _ { m = 0 } ^ \infty $ by $ a _ 0 = \alpha $, $ b _ m = \frac \beta { a _ m } + \alpha $ and $ a _ { m + 1 } = \frac \beta { b _ m } + \alpha $, and inductively show that $ a _ m x \le f ( x ) \le b _ m x $ for all $ x \in \mathbb R ^ + $ and all nonnegative integers $ m $. Taking limits, you can then find out that $ f ( x ) = c x $ gives the unique solution, where $ c $ is the positive root of $ c ^ 2 - \alpha c - \beta $, i.e. $ c = \frac { \alpha + \sqrt { \alpha ^ 2 + 4 \beta } } 2 $. I leave verifying the details to you. In the special case where $ \alpha = n $ and $ \beta = n + 1 $, we have $ c = n + 1 $.
