Find all functions$ f : R_+  \to R_+$ such that $f(f(x)+y)=x+f(y)$ , for all $x, y \in R_+$

(Folklore)

[hide=PS]Using search terms [color=#f00]+ ''f(x+f(y))'' + ''f(x)+y[/color]''
I found the same problem [url=https://artofproblemsolving.com/community/c6h1122140p5167983]in Q[/url], [url=https://artofproblemsolving.com/community/c6h1597644p9926878]continuous in R[/url], [url=https://artofproblemsolving.com/community/c6h1065586p4628238]strictly monotone in R[/url] , [url=https://artofproblemsolving.com/community/c6h583742p3451211 ]without extra conditions in R[/url]

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1. Given the functional equation \( f(f(x) + y) = x + f(y) \) for all \( x, y \in \mathbb{R}_+ \).

2. Let's denote this equation as \((\alpha)\):
   \[
   f(f(x) + y) = x + f(y)
   \]

3. Substitute \( x \) with \( f(x) + y \) in \((\alpha)\):
   \[
   f(f(f(x) + y) + y) = (f(x) + y) + f(y)
   \]

4. By \((\alpha)\), we have:
   \[
   f(f(f(x) + y) + y) = x + f(y) + f(y)
   \]

5. Therefore:
   \[
   f(x + y + f(y)) = x + 2f(y)
   \]

6. Now, substitute \( y \) with \( x + y \) in \((\alpha)\):
   \[
   f(f(x) + (x + y)) = x + f(x + y)
   \]

7. By \((\alpha)\), we have:
   \[
   f(x + y + f(x)) = x + f(x + y)
   \]

8. From the above, we can infer:
   \[
   f(x + y + f(y)) = x + 2f(y)
   \]
   and
   \[
   f(x + y + f(x)) = x + f(x + y)
   \]

9. By comparing these two equations, we get:
   \[
   x + 2f(y) = x + f(x + y)
   \]

10. Therefore:
    \[
    f(x + y) = f(x) + f(y)
    \]

11. The functional equation \( f(x + y) = f(x) + f(y) \) is the Cauchy functional equation. Since \( f : \mathbb{R}_+ \to \mathbb{R}_+ \) and \( f \) is strictly increasing, the only solution is:
    \[
    f(x) = cx
    \]
    for some constant \( c \).

12. Substitute \( f(x) = cx \) back into the original equation \((\alpha)\):
    \[
    f(f(x) + y) = f(cx + y) = c(cx + y) = c^2x + cy
    \]
    and
    \[
    x + f(y) = x + cy
    \]

13. Equating both sides, we get:
    \[
    c^2x + cy = x + cy
    \]

14. Therefore:
    \[
    c^2x = x
    \]

15. Since \( x \neq 0 \), we have:
    \[
    c^2 = 1
    \]

16. Since \( f : \mathbb{R}_+ \to \mathbb{R}_+ \), \( c \) must be positive. Thus, \( c = 1 \).

17. Therefore, the function is:
    \[
    f(x) = x
    \]

The final answer is \( \boxed{f(x) = x} \).