Taiwan IMO Selection Camp

It's trivial that if $f$ maps $x$ to $2015x$, then the functional equation holds. And we'll show that it is the only function that satisfies our requirement. As usual, there are some parts in our proof. For simplicity, let $k = 2015$, then the functional equation becomes
$$f(x + y + f(y)) = 2kx - f(x) + f((k+1)y) \quad \forall x, y \in \mathbb{R}^+ \quad (1)$$

Step 1:
$$f((k+1)a) - f((k+1)b) = k(f(b) + b - f(c) - c) \quad \forall b, c \in \mathbb{R}^+.$$
Set $x = a + b + f(b)$, $y = c$ in (3), then we get
$$\begin{aligned} d &= 2k(a+b+f(b)) - f(a+b+f(b)) + f((k+1)c) \\ &= 2k(a+b+f(b)) - (2ka - f(a) + f((k+1)b)) + f((k+1)c) \end{aligned}$$
where $d = f(a+b+f(b)+c+f(c))$. Similarly, when $x = a+c+f(c), y=b$, then
$$d = 2k(a + c + f(c)) - (2ka - f(a) + f((k+1)c)) + f((k+1)b)$$
Consequently, we must have
$$f((k+1)b) - f((k+1)c) = k(f(b) + b - f(c) - c).$$
Since there isn't any restriction to $b, c$, the conclusion follows.

Step 2: $\forall b, c \in \mathbb{R}^+$, we have
$$\begin{aligned} & 2kb - f(b) + f((k+1)[(k+2)c + f(c)]) \\ & 2k(b + (k+1)(2k+1)c) - f(b + (k+1)(2k+1)c) + f((k+1)c). \end{aligned}$$
Take $x = (k+1)b, y = b$ in original equation
$$f((k+2)b + f(b)) = 2k(k+1)b, \quad \forall b \in \mathbb{R}^+ \quad (2)$$
Apply the same trick, we use double counting to obtain the result, again. First, set $x = b, (k+2)c + f(c)$ in original equation, then
$$e = 2kb - f(b) + f((k+1)[(k+2)c + f(c)])$$
where $e = f(b + (k + 2)c + f(c) + 2k(k + 1)c)$.
Second, set $x = b + (k + 1)(2k + 1)c$, $y = c$, we have
$$e = 2k(b + (k + 1)(2k + 1)c) - f(b + (k + 1)(2k + 1)c) + f((k + 1)c).$$
Those equalities implies our statement.

Step 3: $\forall b, c \in \mathbb{R}^+$, $f(b+c) = f(b) + kc$ holds.
Rewrite the identity in Step 2, we know
$$\begin{aligned} & f(b + (k + 1)(2k + 1)c) + f((k + 1)[(k + 2)c + f(c)]) - f((k + 1)c) \\ &= f(b) + 2k(k + 1)(2k + 1)c \end{aligned}$$
Furthermore, by using Eq. (1) we find
$$\begin{aligned} & f((k + 1)[(k + 2)c + f(c)]) - f((k + 1)c) \\ &= k(f((k + 2)c + f(c)) + (k + 2)c + f(c) - f(c) - c) \\ &= k(k + 1)(2k + 1)c. \end{aligned}$$
Thus, the previous equation becomes
$$f(b + (k + 1)(2k + 1)c) = f(b) + k(k + 1)(2k + 1)c,$$
which means $f(b+c) = f(b) + kc$ holds.

Step 4: $f(a) = ka \forall a \in \mathbb{R}^+$.
Note that $f(b)+kc = f(b+c) = f(c)+kb$ for all $b, c \in \mathbb{R}^+$.
As the result, $f(a) - ka$ is a constant. Suppose $f(x) = kx + l$ for some $l \in \mathbb{R}$, then
$$\text{LHS} = f(x + (k + 1)y + l) = kx + k(k + 1)y + (k + 1)l$$
$$\text{RHS} = kx + k(k + 1)y.$$
So $l = 0$, and the problem is solved.