YadisBeles wrote: Determine all functions $f:\mathbb{R}^+\rightarrow \mathbb{R}^+$ satisfying $f(x+y+f(y))=4030x-f(x)+f(2016y), \forall x,y \in \mathbb{R}^+$. Let $P(x,y)$ be the assertion $f(x+y+f(y))=4030x-f(x)+f(2016y)$ $P(2016x,x)$ $\implies$ $f(2017x+f(x))=4030\times 2016x$ and so $f(x)$ is surjective. Setting then in $P(x,y)$ $y$ such that $f(2016y)$ is as small positive as we want (possible since surjective), we get $f(x)\le 4030x$ $\forall x>0$ Let $U=\{x>0$ such that $f(x)=4030x\}$ 1) If $U=\emptyset$, then $f(x)=2015x$ $\forall x>0$ Let $x>0$ If $f(x)<2015x$, then $P(2015x-f(x),x)$ $\implies$ $f(2015x-f(x))=4030(2015x-f(x))$, impossible So $f(x)\ge 2015x$ $\forall x>0$ If $f(x)>2015x$, then $P(2016x,x)$ $\implies$ $f(2017x+f(x))=4030\times 2016x$ $<2015(2017x+f(x))$, impossible And so $\boxed{f(x)=2015x\text{ }\forall x>0}$ which indeed is a solution Q.E.D. 2) If $U\ne\emptyset$, then no solution 2.1) $U$ is not upperbounded Let $u\in U$ : $P(2016u,u)$ $\implies$ $f(6047u)=4030\times 2016u$ $P(2015\times 6047u-4030\times 2016u,6047u)$ $\implies$ $f(40602254u)=4030(40602254u)$ And so $40602254u\in U$Q.E.D. 2.2) $\forall u\in U$, $\exists$ a sequence $x_n$ such that $\lim_{n\to+\infty}x_n=u$ and $\lim_{n\to+\infty}f(x_n)=0$ $f(x)\in(0,4030x]$ implies that $\lim_{x\to 0^+}f(x)=0$ Choose then $x_n=u+\frac 1n+f(\frac 1n)$. we have $\lim_{n\to+\infty}x_n=u$ $P(u,\frac 1n)$ $\implies$ $f(x_n)=f(\frac{2016}n)$ and so $\lim_{n\to+\infty}f(x_n)=0$ Q.E.D. 2.3) No such $f(x)$ Let $x>0$ And $u\in U$ such that $u>\frac x{2016}+f\left(\frac x{2016}\right)$ Let $x_n>\frac x{2016}+f\left(\frac x{2016}\right)$ a sequence such that $\lim_{n\to+\infty}x_n=u$ and $\lim_{n\to+\infty}f(x_n)=0$ Let $y_n=x_n-\frac x{2016}-f\left(\frac x{2016}\right)$ $P(y_n,\frac x{2016})$ $\implies$ $f(x_n)=(4030y_n-f(y_n))+f(x)\ge f(x)$ Setting there $n\to+\infty$, we get $f(x)\le 0$, impossible. Q.E.D.

Let $P(x,y)$ be the assertion $f(x+y+f(y))=4030x-f(x)+f(2016y)...(1)$ $P(x,1):f(x+A)=4030x-f(x)+B$,where $A=1+f(1),B=f(2016)$ $P(x+A,y):f(x+y+f(y)+A)=4030x+4030A-f(x+A)+f(2016y)...(2)$ Plus $(1)$ and $(2)$,we have $4030(x+y+f(y))+B=8060x+4030A-4030x-B+2f(2016y)$ $\Rightarrow f(2016y)=2015(y+f(y))+B-2015A...(3)$ $P(x+f(x),y):f(x+f(x)+y+f(y))=4030(x+f(x))-f(x+f(x))+f(2016y)$ $\Rightarrow 4030(x+f(x))-f(x+f(x))-f(2016x)=2015(x+f(x))-f(x+f(x))+2015A-B=const$ So,$f(x+f(x))=2015(x+f(x))+C$,where $C$ is a constant From $P(2016y,y)$ we have $f(x)$ is surjective $\Rightarrow$ from $(3)$ we have $\forall a\ge N$ ($N$ is a constant),there exists a $x$ such that $x+f(x)=a$ So $\forall x\ge N$ we have $f(x)=2015x+C$ $P(x,N):2015(x+2016N+C)+C=f(x+2016N+C)=4030x-f(x)+2015(2016N)+C$ $\Rightarrow f(x)=2015x-2015C,\forall x\Rightarrow C=0\Rightarrow f(x)=2015x$,which is indeed a solution

Solution from CANBANKAN's stream: The answer is $f(x)\equiv 2015x$ which clearly works. Claim 1: $f$ is surjective Proof: $P(2016y,y)$ gives $f(2017y+f(y))=2016\cdot 4030y$. Since the range of RHS is $\mathbb{R}^+$, so is the range of LHS, which is a subset of the range of $f$. This is a very strong condition and turns the problem into a bounding problem. Claim 2: $f(x)\le 4030x$. Proof: Suppose $f(x)-4030x=C$ for some $C>0$. Rewriting our FE as $f(x)-4030x=f(2016y)-f(x+y+f(y)) \le f(2016y)$ and by surjectivity we can choose $y$ such that $f(2016y)<C$. Claim 3: $f(x)\ge 2015x$ Proof: The idea is to find a "linear" approximation/form of $f$. Note $P(x,y): f(x+y+f(y))+f(x)= 4030x+f(2016y)$ (1) $P(x+y+f(y), y): f(x+2y+2f(y))+f(x+y+f(y))=4030(x+y+f(y))+f(2016y)$ (2) (2)-(1): $f(x+2y+2f(y))-f(x)=4030(y+f(y))$ Suppose $f(x)<2015x$. Let $C=2015x-f(x)$. By Claim 2, we find $y<\frac{C}{10^{10}}$, and $n$ such that $x-2n(y+f(y)) > 0 \ge x-(2n+2)(y+f(y))$. Then note $f(x-2n(y+f(y))) = f(x)-4030n(y+f(y)) =2015x-C-4030n(y+f(y)) =2015(x-2n(y+f(y)) - C < 4030(y+f(y))-C<0$, contradiction. Claim 4: $f(x)\equiv 2015x$. Proof: Suppose $C=f(x)-2015x>0$, then taking $y<\frac{C}{10^{10}}$, $f(x+y+f(y))+f(x)=4030x+f(2016y)$ $2015(x+y+f(y))+2015x+C\le 4030x+f(2016y)$ $C\le f(2016y)\le 2016*4030y \le \frac{2016\cdot 4030}{10^{10}}C$ contradiction.

The answer is $f(x) = 2015x \ \ \forall x\in \mathbb{R}^+$. It is easy to see that it satisfies the condition. Now we prove that it is the only one. Plug in $(x+f(x), y)$ we have $$f(x+f(x)+y+f(y)) = 4030(x+f(x))-f(x+f(x))+f(2016y)$$Note that the left hand side is symmetric with respect to $x$ and $y$, hence $$4030(x+f(x)) -f(x+f(x)) -f(2016x) = c_1$$, where $c_1$ is a constant. Plug in $(x+f(x)+z, y)$ we have \[ \begin{aligned} f(x+f(x)+y+f(y)+z) &= 4030(x+f(x)+z)-f(x+f(x)+z)+f(2016y) \\ &= 4030(x+f(x)+z)-4030z +f(z) -f(2016x) + f(2016y) \end{aligned} \]Note that the left hand side is symmetric with respect to $x$ and $y$. Hence $$2015(x+f(x))-f(2016x) = c_2$$, where $c_2$ is a constant. Combine two results we conclude $$f(2016x)-2015(x+f(x)) = c_3$$, where $c_3$ is a constant. And we have $$f(x+y+f(y)) = 4030x-f(x) +f(2016y) = 4030x -f(x) +2015(y+f(y)) +c_3$$Now we prove that $f$ is injective. If $f(a) = f(b)$, compare $(a, b)$ and $(b, a)$ we have $$ f(a+b+f(a)) = 4030a -f(a) +2015(b+f(b))+c_3 = 4030b-f(b)+2015(a+f(a))+c_3 $$Which means $a = b$, so $f$ is injective. Plug in $(2016y, y)$ we have \begin{align} f(2017y+f(y)) = 2\cdot 2015 \cdot 2016 y \end{align}If there exists $z$ such that $f(z) < 2015 z$, then plug in $(2015z-f(z), z)$ we have $$ f(2015z-f(z)) = 4030(2015z-f(z))$$Hence if we plug in $y(2015z-f(z))/2016$ to $(1)$ we have $$f(2017y+f(y) = 2\cdot 2015 \cdot 2016 y = f(2016y)$$so $f(y) + y = 0$ since $f$ is injective, which is absurd. Now we have $f(z) \ge 2015 z$ for all $z \in \mathbb{R}^+$, combine $(1)$ be have $$ 2\cdot 2015 \cdot 2016 y = f(2017y+f(y)) \ge 2015\cdot (2017y+f(y)) \ge 2015(4032y) $$Hence the equality holds, which conclude $f(y) = 2015y \ \forall y \in \mathbb{R}^+$, as desired.

The answer is $f(x) = 2015 x$, which clearly works. Now we show it's the only solution. Claim: $f$ is surjective. Proof: Setting $y = \frac{x}{2016}$ gives that $4030x$ is in the image of $f$, as desired. Let $g(x) = x + f(x)$. The equation becomes \[ g(x + g(y)) = 4032x - g(x) + g(y) + g(2016 y) - 2016 y \] Let $P(x,y)$ denote this assertion and $z$ be any positive real number. $P(x + g(z), y): g(x + g(y) + g(z)) = 4032(x + g(z)) - (4032x - g(x) + g(z) + g(2016z) - 2016z) + g(y) + g(2016y) - 2016 y$, which implies \[ g(x + g(y) + g(z)) = 4032 g(z)+ g(x) - (g(z) + g(2016 z) - 2016 z) + g(y) + g(2016 y) - 2016 y\] Now swapping $y,z$ and adding the two equations gives that $ 2g(x + g(y) + g(z)) = 4032(g(y) + g(z)) + 2g(x)$, so \[ g(x + g(y) + g(z)) = g(x) + 2016(g(y) + g(z)) \] Setting $x \to g(x)$ gives \[g(g(x) + g(y) + g(z)) = g(g(x)) + 2016(g(y) + g(z))\] If we let $Q(x,y,z)$ be this assertion, then $Q(x,y,z) - Q(y,x,z)$ gives that $g(g(x)) - g(g(y)) = 2016g(x) - 2016g(y)$, implying that $g(g(x)) - 2016 g(x)$ is constant for all positive reals $x$. So, this means there exists a constant $c$ so that $g(x + f(x)) = 2016(x + f(x)) + c$, so $f(x + f(x)) = 2015(x + f(x)) + c$. $P(x + f(x), y): f(x + f(x) + y + f(y)) = 4030(x + f(x)) - (2015(x + f(x)) + c) + f(2016 y) $, so \[ f(g(x) + g(y)) = 2015(x + f(x)) + f(2016 y) - c\] Now swapping $x,y$ and comparing gives that $2015(x + f(x)) + f(2016 y) = 2015(y + f(y)) + f(2016 x)$, meaning that $f(2016 x) - 2015(x + f(x))$ is constant. Let $d = f(2016 x) - 2015(x + f(x))$. Setting $x$ with $f(2016x)$ arbitrarily small gives that $d \le 0$ and $g$ can take any positive real number at least $-d$. Hence $g(x) = 2016 x + c \forall x \ge -d$. Now, we have that $g(x + g(y) + g(z)) = g(x) + 2016(g(y) + g(z))$, so fixing $y,z$ such that $g(y) + g(z) \ge -d$ but $x < -d$ gives that $g( x+ g(y) + g(z)) = 2016x + 2016(g(y) + g(z)) + c$, so $2016 x + 2016(g(y) + g(z)) + c = g(x) + 2016(g(y) + g(z))$, so $g(x) = 2016 x + c$ for all positive reals $x$, so $f(x) = 2015 x + c$. Since $f$ is surjective and always positive, we require $c = 0$, as desired.