the.sceth wrote: $ f(x + f(y)) > f(x + y),f(y)$ so there cannot exist a solution $ f(t)$ that is constant for sufficiently large $ t$. [Statement 1] If $ f(m) = f(n)$ then $ f(s + f(m)) = f(s + f(n)) \forall s \in \mathbb{R}^{ + }$ $ f(s + m) + f(m) = f(s + n) + f(n)$ $ f(s + m) = f (s + n)$. If $ m \neq n$ then $ f(t)$ is constant for sufficiently large $ t$. By Statement 1, $ f$ is injective. [Statement 2] Lets say $ m=1$, $ n=2$, then $ f(s+n)=f(s+1)$ where $ n$ is an integer...that does not make $ f(t)$ constant.

Altheman wrote: the.sceth wrote: $ f(x + f(y)) > f(x + y),f(y)$ so there cannot exist a solution $ f(t)$ that is constant for sufficiently large $ t$. [Statement 1] If $ f(m) = f(n)$ then $ f(s + f(m)) = f(s + f(n)) \forall s \in \mathbb{R}^{ + }$ $ f(s + m) + f(m) = f(s + n) + f(n)$ $ f(s + m) = f (s + n)$. If $ m \neq n$ then $ f(t)$ is constant for sufficiently large $ t$. By Statement 1, $ f$ is injective. [Statement 2] Lets say $ m = 1$, $ n = 2$, then $ f(s + n) = f(s + 1)$ where $ n$ is an integer...that does not make $ f(t)$ constant. I see the inadequacy in intuiting something not really trivial. This should cover it up - it can fit in the penultimate line of the quoted text. WLOG let $ m\geq n$, $ m - n: = S$, $ f(S): = T$, a constant. The direct implication is that $ f(s) = f(s + hS)$ where $ h$ is any positive integer. For any positive integer $ k$ and any positive real $ a$, $ f(a + S + f (kS)) = f(a + (k + 1)S) + f(kS)$ $ f(a + T) = f(a) + T$... $ [*]$ $ f(S + T) = 2T$ $ f(T) = 2T$ $ f(T + f(kS)) = f(T + kS) + f(kS)$ $ f(2T) = 3T$ Using $ [*]$ as an inductive step, $ f(AT) = (A + 1)T$ for the positive integer $ A$. If $ A$ and $ B$ are positive integers then $ f(AT + f(BT)) = f((A + B)T) + f(BT)$ $ (A + B + 2)T = (A + 2B + 2)T \forall B \in \mathbb{N}$ $ T = 0$ not in range of $ f$ $ S$ not in domain of $ f$ $ S = 0$ $ f$ is injective.

I think that the substitution g(x)=f(x)-x solves the problem !

1. $ f(x) > x$ for every positive $ x$. Indeed, if there exists $ x_0$ such that $ x_0 > f(x_0)$, then we know $ f(x_0 + f(x_0) - f(x_0)) = f(x_0) = f(2x_0 - f(x_0)) + f(x_0)$ - contradiction. 2. For every $ x > y$ we have $ f(x + f(y) - y) = f(x) + f(y)$. Let $ g(x) = f(x) - x$ as msecco said. $ ( g: \mathbb{R}^{ + }\to\mathbb{R}^{ + } )$ Then for every $ x > y > 0$ we have $ g(x + g(y)) = g(x) + y$ and if we fix $ x = x_0 > y$ and we assume $ g(y_1) = g(y_2)$ for two positive numbers less than $ x_0$, we will obtain that $ g(x_0 + g(y_1)) = g(x_0 + g(y_2))\Leftrightarrow g(x_0) + y_1 = g(x_0) + y_2$ - impossible unless $ y_1 = y_2$. Therefore, $ g(x)$ is injective and from this follows that $ f(x)$ is injective function, too. 3. For every $ x > y > 0$ we have $ g(x + g(y)) = g(x) + y = g(x + g(\alpha)) + (y - \alpha) = g(x + g(\alpha) + g(y - \alpha))$, $ \Rightarrow g(y) = g(y - \alpha) + g(\alpha)$. Since $ g: \mathbb{R}^{ + }\to\mathbb{R}^{ + }$, then $ g(x) = cx$ and after substituting, we find $ c = 1$.

(1) You got that function $ f(x) - x$ is injective, which doesn't at all mean that $ f$ is also injective. (2) You got that $ g$ satisfies the Cauchy equation and it's defined for the set of positive real numbers and gets only positive values. This doesn't mean that $ g(x)=cx$.

Your complaint in (1) is legitimate, but it was never used in the solution. You're wrong about (2). A function from $ \mathbb{R^+}$ to $ \mathbb{R^+}$ that satisfies Cauchy's equation must be linear. You can prove this by getting $ f(x) = cx$ on rationals and showing $ f$ is strictly increasing: if $ x > y$, then $ f(x) > f(y)$ because $ f(x) = f(y) + f(x-y)$ and $ f(x-y) > 0$. Note that any noncontinuous solution to Cauchy's equations (for functions on $ \mathbb{R}$) is dense in the plane, so if you restrict everything to one quadrant you must have continuity.

Yes you're right. Sorry for my dull post

My Solution: 0) $ f(x + f(y)) = f(x + y) + f(y)\ \forall \ x,y\in \mathbb R^ +$. 1) $ f(x) > x\ \forall \ x\in \mathbb R^ +$. 2) Putting $ x: = x - y$ in (0) $ f(x + [f(y) - y]) = f(x) + f(y)\ \forall \ x > y\in \mathbb R^ +$. 3) In (0) Let $ x: = f(z) - z$ where $ 0 < z < y$. We have $ f(f(y) + [f(z) - z]) = f(y + [f(z) - z]) + f(y)$. Using (2) : $ [f(f(y)) + f(z)] = [f(y) + f(z)] + f(y)\implies f(f(y)) = 2f(y)\ \forall \ y\in \mathbb R^ +$. 4) Putting $ y: = f(y)$ in (2) and using (3) we have $ f(x + f(y)) = f(x) + 2f(y)\ \forall \ x > f(y)$, $ x,y\in \mathbb R^ +$. 5) Equating (4) and (0) we have $ f(x + y) = f(x) + f(y)\ \forall \ x > f(y)$, $ x,y\in \mathbb R^ +$. 6) If $ f(a) = f(b)$ for some $ a\neq b\in \mathbb R^ +$ then putting $ y: = a$ in (0) and then putting $ y: = b$ in (0) we get $ f(x + a) = f(x + b)\ \forall \ x\in \mathbb R^ +$. That is, $ f$ becomes eventually periodic. But (5) implies that for each $ y$, all large enough $ x$s satisfy $ f(x) > f(y)$. Contradiction. So $ f$ is injective. 7) Applying (5) in RHS of (0) we get $ f(x + f(y)) = f(x + 2y)\ \forall \ x > f(y) - y$ but (6) gives $ f(y) = 2y\ \forall \ y\in \mathbb R^ +$.

The following solution is due to Andreas Dwi Maryanto Gunawan, from Indonesia. For any positive real numbers $ z$, we have that \[ f(x+f(y))+z=f(x+y)+f(y)+z\] \[ f(f(x+f(y))+z)=f(f(x+y)+f(y)+z)\] \[ f(x+f(y)+z)+f(x+f(y))=f(x+y+f(y)+z)+f(x+y)\] \[ f(x+y+z)+f(y)+f(x+y)+f(y)=f(x+2y+z)+f(y)+f(x+y)\] \[ f(x+y+z)+f(y)=f(x+2y+z)\] sehingga \[ f(a)+f(b)=f(a+b)\] and by Cauchy in positive reals, then $ f(x)=\alpha x$ for all $ x \in (0, \infty)$. Now it's easy to see that $ \alpha=2$, then $ f(x)=2x$ for all positive real numbers $ x$. Is this OK? Please check this solution...

I think it's OK..

This problem was proposed by Paisan Nakmahachalasint, leader of Thailand IMO team in 2008 and 2009.

@Raja Oktovin. I doubt you will see this, but this was one of the neatest solutions in my opinion. i really loved it.

can we extend this result if the function is defined from all reals to all reals? (R to R instead of R+ to R+)

Raja Oktovin wrote: The following solution is due to Andreas Dwi Maryanto Gunawan, from Indonesia. For any positive real numbers $ z$, we have that \[ f(x+f(y))+z=f(x+y)+f(y)+z\] \[ f(f(x+f(y))+z)=f(f(x+y)+f(y)+z)\] \[ f(x+f(y)+z)+f(x+f(y))=f(x+y+f(y)+z)+f(x+y)\] \[ f(x+y+z)+f(y)+f(x+y)+f(y)=f(x+2y+z)+f(y)+f(x+y)\] \[ f(x+y+z)+f(y)=f(x+2y+z)\] sehingga \[ f(a)+f(b)=f(a+b)\] and by Cauchy in positive reals, then $ f(x)=\alpha x$ for all $ x \in (0, \infty)$. Now it's easy to see that $ \alpha=2$, then $ f(x)=2x$ for all positive real numbers $ x$. Is this OK? Please check this solution... i think it is wrong !!! Because we have : $f(x)+f(y)=f(x+y)\: \: \forall x,y\in \mathbb{R}^{+}\: ,x> y$

@alibez; i don't understand! Of course, It's true. $f$ - additive and $R^+ \rightarrow R^+$, so it's monotone.

we have : $f(x+y+z)+f(y)=f(x+2y+z)\Rightarrow f(a+b)=f(a)+f(b)\: \:\: \: \forall a> b$ but we know if : $f(a+b)=f(a)+f(b)\: \: \: \forall a,b\in \mathbb{R}^{+}\Rightarrow f(x)=cx$

in this problem we can prove $f$ is additive but it is very hard .

alibez wrote: in this problem we can prove $f$ is additive but it is very hard . Here is a solution to a more general f.e. http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3158470#p3158470

Let $P(x,y)$ be the assertion that $f(x+f(y)) = f(x+y) +f(y)$. Clearly $f(y) > y$ for all $y$, because otherwise $P(y-f(y), y)$ implies $f(2y-f(y)) = 0$, a contradiction. Now let $g : \mathbb{R}^{+}\to\mathbb{R}^{+}$ be a function with $g(x) = f(x) - x$ for all $x$. $P(y,x)$ gives that $g(g(x)+x+y) = g(x+y)+x$ for all $x,y > 0$ after the substitution $f(x) = g(x)+x$. Let $Q(x,y)$ be the assertion that $g(g(x)+x+y) = g(x+y)+x$. First, summing the assertions $Q(x,y), Q(x, g(x)+y), \cdots Q(x, (k-1)g(x)+y)$ for an integer $k$ gives that $g(kg(x)+x+y) = g(x+y)+kx$ for all $x,y > 0$ and positive integers $k$. Let $R(x,y,k)$ be the assertion $g(kg(x)+x+y) = g(x+y)+kx$ for positive integers $k$. Claim 0: If there exists $d$ with $g(d) = m$ for some $m$, then for any $q > m$, we can find $e$ with $g(e) = q$. Proof: Consider an integer $k$ with $k > \frac{q-m}{d}$. Then $R(\frac{q-m}{k},d-\frac{q-m}{k}, k)$ gives that $g(kg(\frac{q-m}{k})+d) = g(d)+k\frac{q-m}{k} = q$, so we take $e = kg(\frac{q-m}{k})+d$, so $g(e) = q$. We can let $r$ be the greatest lower bound of the range of $g$. By Claim 0, the range of $g$ contains all numbers more than $r$, and doesn't contain any number less than $r$. (We do not need to look at whether $r$ is in the range of $g$.) Claim 1: For any integers $p,q$ and any $x > 0$, $qg(px) = pg(qx)$. Proof: Assume for contradiction that this is not true. WLOG $qg(px) < pg(qx)$. Let $k$ be an integer such that $k > \frac{r}{pg(qx)-qg(px)}$. Then $k(pg(qx)-qg(px)) > r$, so there exists a positive $b$ with $g(b) = k(pg(qx)-qg(px))$. Now let $y$ be a positive real such that $k(qg(px)+pqx)+y > b$. $R(px, (pq-p)x+y, kq)$ gives that $g(kqg(px)+pqx+y) = g(pqx+y)+kpqx$. $R(qx, (pq-q)x+y, kp)$ gives that $g(kpg(qx)+pqx+y) = g(pqx+y)+kpqx$. Combining these two, we get $g(kqg(px)+pqx+y)=g(kpg(qx)+pqx+y)$. $Q(b, k(qg(px)+pqx+y)+y-b)$ gives that \[g(g(b)+k(qg(px)+pqx+y)) = g(k(qg(px)+pqx+y))+b.\] However, since $g(b) = k(pg(qx)-qg(px))$, $g(g(b)+k(qg(px)+pqx+y)) =g(kpg(qx)+pqx+y)$. Thus $0 = b$, a contradiction. By Claim 1, we have arbitrarily small numbers in the range of $g$, so by Claim 0, $g$ is surjective. Claim 2: $g$ is increasing. Proof: Consider $a > b > 0$. Since $g$ is surjective, consider $z$ with $g(z) = a-b$. Take a positive integer $k$ with $k > \frac{z}{b}$. Then by Claim 1, $g(\frac{z}{k}) = \frac{a-b}{k}$. $R(\frac{z}{k}, b-\frac{z}{k},k)$ implies that $g(kg(\frac{z}{k}) + b) = g(b)+k\frac{z}{k}$, or $g(a) = g(b) +z$. Thus $g(a) > g(b)$. proving our claim. Let $c = f(1)$. By Claim 1, we have $g(x) = cx$ for all rationals $x$. By Claim 2, and the fact that the rationals are dense in the reals, we have $g(x) = cx$ for all reals $x$. If we plug this in, we get that $c^2x+cx+cy=cx+cy+x$, so $c = 1$. Thus $g(x) = x$, so $f(x) = 2x$. Since $f(x) = 2x$ is a valid solution to the original equation, the only solution for $f(x)$ is $f(x) = 2x$.

Somehow got a weird infimum solution This is equivalent to \[ f(x + (f(y) - y)) = f(x) + f(y) \]holding for $x > y$. Note that $f(x) \ne x$ due to there being no roots. Furthermore, $f(x) < x$ can not hold, or else $f(x - k(f(y) - y))$ would be negative for sufficiently large positive integer $k$ by the expression. Claim: Fix $y$ and let $c$ be the infimum of $f(x)$ on the interval $[y, y + f(y) - y)$. Then, $c + kf(y)$ is the infimum on the interval $[y + k(f(y) - y), y + (k+1)(f(y) - y))$. Proof. Follows since shifting by a constant preserves the infimum. $\blacksquare$ Note that for each $y$, these intervals are covering and the infimums lie on a line with slope $\frac{f(y)}{f(y) - y}$. Then, $\frac{f(y)}{f(y) - y}$ must be constant over all $y$ or else that leads to contradiction. As such, this implies that $\frac{y}{f(y)}$ is constant, and thus $f(x) = cx$ for some $c$. Checking shows that only $f(x) = 2x$ works.

$P(0,y)$ gives $f(f(y)) = 2y$, then $P(x,f(y))$ gives $f(x + 2f(y)) = f(x + f(y)) + 2f(y) = f(x + y) + 3f(y)$. This then implies that $f(x - f(y) + f(y)) = f(x - f(y) + y) ) + f(y), f(x - f(y) + 2f(y)) = f(x - f(y) + y) + 3f(y)$, so $f(x + f(y)) - f(x) = 2f(y)$, or $f(x + y) + f(y) - f(x) = 2f(y)$, giving $f(x + y) = f(x) + f(y)$. Now we have $f(x + f(y)) = f(x + y) + f(y) = f(x + 2y)$. Now we just need to show $f$ injective to finish. Consider two different values $a,b$ of $y$ giving the same $f(y)$. Then we have $f(x + a) = f(x+b)$, thus we can find numbers $c,d$ with arbitrarily large difference such that $f(c) = f(d)$. Then, we can set $y = c$ and $x + f(y) = d$, which forces $f(x + y) = 0$, which is impossible. Since $f$ injective, we can easily get $f(y) = 2y$.

Wasn't thinking of posting this one since I took some hints... But someone check this please... Also the above solution is totally flawed as of now(at the time I'm posting) since we're talking $x,y > 0$. Solution: The answer is $f(x) = 2x$ for all $x \in \mathbb{R}^+$. This clearly works and we now show that this is the only possible solution. Let $P(x,y)$ denote the assertion to the functional equation. Claim: $f(x) \ge x$ and $f$ is an injective function. Proof: Assume for the sake of contradiction there is some positive real $k$ such that $f(k) < k$. Then $P(k-f(k), x)$ gives $f(2x - f(x)) = 0$ which is impossible for the given co-domain. Therefore $f(x) \ge x$ for all $x$. Once again, assume on the contrary that there exists some positive reals $a>b>0$ such that $f(a) = f(b)$. This time $P(x,a)$ and $P(x,b)$ yield us that $f$ is eventually periodic with period $T \coloneqq a-b$. Consider some huge enough $\alpha \in \mathbb{R}^+$ for periodicity. Choose a positive integer $m$ which is large enough such that $f(\alpha)< \alpha + mT$. Combining everything \[\alpha \le f(\alpha) < \alpha + mT \le f(\alpha + mT) = f(\alpha)\]which is absurd. Hence the claim holds true. $\square$ The latter part of the solution is mainly grinding substitutions until you get a nice one. Our primary goal here is to show that $f(2x + y) = f(2x) + f(y)$ for all $x > y$ which is almost a Cauchy like condition. $P(x-y, y)$ and $P(x,x-y+f(y))$ for $x > y$ in sequence would yield us that \begin{align} f(x + f(x) + f(y)) = f(2x) + 2f(y) + f(x) \tag{1.1} \label{1.1} \end{align} $P(x+f(x), y)$ for $x > y$ we will get \begin{align} f(x + f(x) + f(y)) = f(x + f(x) + y) + f(y) \tag{1.2} \label{1.2} \end{align} Combine $(1.1)$ and $(1.2)$ to get \begin{align} f(x + f(x) + y) = f(2x) + f(y) + f(x) \,\, \text{for $x>y$.}\tag{1.3}\label{1.3} \end{align} Consider $P(y + x, x)$ for $x > y$ to get \begin{align*} f(x + y + f(x)) = f(2x + y) + f(x) \end{align*}and finally combine this with $(1.3)$ to get \begin{align} f(2x + y) = f(2x) + f(y) \tag{1.4} \label{1.4} \end{align}for $x>y$ which is the partial Cauchy condition advertised above. Replace $x$ by $2y$ in $(1.4)$. This would fetch us $f(y) + f(4y) = f(5y)$. Now consider $P(3x,x)$ to get \[f(3x + f(x)) = f(4x) + f(x) = f(5x) \iff f(x) = 2x\]due to injectivity. This completes the solution. $\blacksquare$

We claim the only solution is $f(x)=2x$, which obviously works. Conversely, let $f$ be a function that satisfies the given equation. Claim : For all $y>0$ we have $f(y)>y$. Proof : Suppose for the sake of contradiction that $f(y)\le y$ for some $y>0$. Clearly we have $f(y)<y$, because any fixed point $y$ is such that $f(y)=0$ by simply looking at the statement, which is not possible. We can thus take $x=y-f(y)>0$ and get $f(x+y)=0$, which is again a contradiction. $\square$ This motivates us to define a function $g:\mathbb R_+^{*}\rightarrow \mathbb R_+^{*}$ such that $g(x)=f(x)-x$ for all $x>0$. The equation that we have for $g$ is \[\forall x,y\in \mathbb R_+,\quad g(x+y+g(y))=g(x+y)+y\quad (*),\]and we need to prove that $g$ is the indentity function. Claim : $g$ is injective. Proof : Suppose $g(x)=g(y)$ for some reals $x,y>0$. We then have \[g(x+y)+y=g(x+y+g(y))=g(x+y+g(x))=g(x+y)+x,\]i.e. $y=x$. $\square$ Claim : For every positive integer $n$ and for every real number $x>0$, we have $g(nx)=n\cdot g(x)$. Proof : Note that we can interprate $(*)$ as follows : For all real numbers $y>0$ and $A>y$, we have \[g(A+g(y))=g(A)+y\quad (1)\]In particular, note that iterating gives \[g(A+2g(y))=g(A+g(y)+g(y))=g(A+g(y))+g(y)\]We thus get by induction that $g(A+kg(y))=g(A)+ky$ for every positive integer $k$ and real numbers $A>y>0$. In particular, we have that for all $k\in \mathbb N_0$ and $y>0$, \[g(B+kg(y))=g(B)+ky=g(B+g(ky)),\]where $B>ky$ is some constant. By injectivity this implies that $g(ky)=k\cdot g(y)$, as desired. $\square$ Claim : $g$ is surjective. Proof : Let $m$ be the infimum of $g$. Note that the $RHS$ of $(1)$ takes all real values greater than $m$, whereas the $LHS$ takes values of $g$ where the input is greater than $m$. It follows that $m$ must be zero by injectivity. In particular, $g$ takes arbitrarily small values. Given any $t>0$, we can thus take some $A>0$ such that $g(A)<t$, and choose a positive integer $k$ large enough so that \[y=\frac{t-g(A)}{k}<A.\]We then have $g(A+kg(y))=g(A)+ky=t$, which proves that $g$ is indeed surjective. $\square$ We're now ready to finish off the problem. Take some $y>0$, and let $y_0>0$ be such that $g(y_0)=0$. Then, we choose a positive integer $k$ large enough, so that $A=k\cdot y_0>y$. We then have \begin{eqnarray*} g(ky+g(g(y_0)))=g(A+g(y))=g(A)+y&=&g(ky_0)+y\\ &=& kg(y_0)+y\\ &=& (k+1)y\\ &=& g((k+1)y_0) \end{eqnarray*}By injectivity, we must have \[ky+g(g(y_0))=(k+1)y_0\iff k(y-y_0)=T,\]where we let $T=y_0-g(g(y_0))$. Note that this holds for all $k\in \mathbb N$ large enough, thus it must be the case that $y=y_0$, which proves that $g$ is the identity. $\blacksquare$

The solution is $f\equiv 2x$. $P(x,y+f(z)) \implies f(x+f(y+f(z))) = f(x+y+f(z)) + f(y+f(z)) = f(x+y+z) + f(z) + f( y+ z) + f(z)$. Also, \begin{align*} f(x+f(y+f(z))) &= f(x+ f(y+z) + f(z)) \\ &= f(x+f(y+z) +z) + f(z) \\ &= f(x + z + f(y+z)) + f(z) \\ &= f(x+z+y+x) + f(y+z) + f(z) .\end{align*} Comparing, we get that $f(x+y+2z) = f(x+y+z) + f(z)$. Now for any $b>a>0$, set $z = a$ and $x = y = \dfrac{b-a}{2}$. So $f(b+a) = f(b) + f(a)$. Also, $f(2x) = f(x+ \varepsilon) + f(x - \varepsilon) = f(x) + f(\varepsilon) + f(x-\varepsilon) = f(x) + f(x) = 2f(x)$. Thus $f$ is Cauchy and linear $\implies f\equiv 2x$.

solved with ThapaKazi pog

We are given \[f(x+f(y))=f(x+y)+f(y)\]We claim that $f(x)=2x$ for all $x$, which clearly works. Call the given assertion as $P(x,y)$. Then, $P(f(y),x)$ gives \[f(x+f(y))+f(x)=f(f(x)+f(y))=f(x)+f(y)+f(x+y)\]This means that if $f(x_1)+f(y_1)=f(x_2)+f(y_2)$, then $x_1+y_1=x_2+y_2$. Note that $P(f(x),y)$ gives $f(f(x)+f(y))=f(y)+f(f(x)+y)$. So, we have \[f(f(x)+y)+f(y)=f(f(y)+x)+f(x)\Longrightarrow f(x)+2y=f(y)+2x\Longrightarrow f(x)-2x=f(y)-2y\]So, $f(x)-2x$ is a constant. So, $f(x)=2x+c$. Substituting back in the original FE, we get \[2x+4y+3c=f(x+f(y))=f(x+y)+f(y)=2x+4y+2c\Longrightarrow c=0,\]and $f(x)=2x$ works. So, the only solution is $f(x)=2x$.

A new (hopefully correct) solution: Clearly, $f(x)>x, \forall x>0$. (if $f(k) = k$, than $P(x, k)$ implies $f(k) = 0$, and if $f(k)<k$, than $P(k-f(k), k)$ implies $f(2k-f(k))=0$, both of which result in contradictions.) Also, note that $f$ is injective. Suppose for the sake of contradiction that there exists $a, b$ such that $a>b$ and $f(a) = f(b)$. Than $P(x, a) - P(x, b)$ gives that $f(x+a) = f(x+b), \forall x>0$. So $f(x) = f(x+a-b), \forall x>b$. Using this, we get $f(x) = f(x+n(a-b)) > x+n(a-b), \forall n \in \mathbb{N}$ which is clearly impossible. Therefore $f$ is injective. For $x>y$, $P(x-y, y)$ implies $f(x+f(y)-y) = f(x)+f(y), \forall x>y$. (1) For $x<y$, $P(f(x)-x, y)$ implies $f(f(x)-x+f(y)) = f(y+f(x)-x)+f(y) = f(x)+2f(y)$ (by (1)). (2) (1) is the same as $f(y+f(x)-x) = f(x) + f(y) \forall y>x$. (3) Using $P(f(y), x)$ in (3) and substracting (2) we get $f(f(y)) = 2f(y), \forall y>0$. $P(f(x), y)$ in the original relation: $f(f(x) + f(y)) = f(f(x) + y) + f(y) = f(x) + f(f(y) + x)$ (by symmetry). (4) $P(x, f(x))$ in (4): $f(2f(x)) +f(x) = f(2f(x) + x)$. (using that $f(f(x)) = 2f(x)$) (5) The original relation is $f(x+f(y)) = f(x+y) + f(y)$. $P(f(y)+y, y)$ in this relation: $f(f(y)+2y) + f(y) = f(2f(y) + y) = f(y) + f(2f(y))$ (using (5)) $\implies f(2f(y)) = f(f(y) + 2y)$. By injectivity, this is just $2f(y) = f(y) + 2y$, so $f(y) = 2y, \forall y>0$. This function satisfies the conditions.

The answer is $f(x)=2x$, which clearly works. Claim: $f(y)>y$. If $f(y)=y$ for any $y$, then $f(y)=0$, contradiction. Thus, $f(y)\neq y$. Suppose FTSOC that $f(y)<y$. Then, $x=y-f(y)>0$ gives $f(x+y)=0$, contradiction. Now, let $f(x)=x+g(x)$ for some function $g$. We know that $g$ is also over the positive reals due to the above claim. The equation becomes $$g(x+y+g(y))=g(x+y)+y.$$If $x+y=a$ and $y=b$, then this substitution is valid for any $a>b$, so $$g(a+g(b))=g(a)+b$$for positive reals $a>b$. Suppose FTSOC that there exists $x$ for which $g(x)\neq x$. Claim: There exists $y$ such that $g(y)>y$. Of course, if $g(x)>x$ it is true, so assume $g(x)<x$. Then, take $t>>x$. The condition says that starting from $g(t)$, if we increase the input by $g(x)$, the output increases by $x$. As $g(x)<x$, doing this sufficiently many times brings us to some $y$ such that $g(y)>y$. Starting from $r>>y$ ($y$ such that $g(y)>y$) and increasing the input by $g(y)$ and the output by $y$ sufficently many times allows us to find arbitrarily large $s$ such that $g(s)<s$. Let $x$ be such that $g(x)<x$, and let $y$ be a sufficiently large number for which $g(y)<y$. Note that, starting from $y$, we can decrease the input by $g(x)$ to decrease the output by $x$, as long as the new input is greater than $x$ (if we go from $t+g(x)$ to $t$, simply plug $b=x$ and this is valid as long as $t>x$). Furthermore, as $g(x)<x$, the "slope" of this drop is greater than $1$. Thus, if we pick sufficiently large $y$ for which $g(y)<y$, then we will reach negative values, as if $y$ is large enough the fact that the input can't go under $x$ is negligible, contradiction.

Edit: This solution is wrong, the substitution $P(2y-f(y),y)$ does not give $f(3y-f(y))=0$ but $f(3y-f(y))+f(y)=f(2y)$. Most solutions use the substitution $g(x)=f(x)-x$, but I do not think any have chosen $g(x)=f(x)-2x$. We shall prove $f(x)=2x$ is the only solution, which clearly works. Let $P(x,y)$ denote the assertion. First of all, suppose FTSOC that $2y>f(y)$, then take $x=2y-f(y)>0$ and $P(x,y)$ gives $f(3y-f(y))=0$, which is impossible. Hence $ f(y)\geq 2y$ for all $y$. Now we may define $g: \mathbb{R}_{+} \rightarrow \mathbb{R}_{\geq 0}$ by $g(x)=f(x)-2x$. Then rewriting the FE in terms of $g$ (and swapping $x$ and $y$) gives $$g(x+y)=g(x)+g(y+2x +g(x)).$$We wish to prove $g\equiv 0$. Note that $g$ is nonnegative, so $g(x+y)\geq g(x)$ for all $x$ and $y$. In particular, this implies that $g(y+2x+g(x))\geq g(x)$. Hence $$g(x+y)\geq 2g(x)$$for all $x,y$. Now suppose FTSOC that $g(c)=d\neq0$ for some $c$. This gives a contradiction because then $g$ “blows up”, namely take any $x>c$. Then for any positive integer $k$, we have $g(x)\geq 2^k d >0$ by $k$ applications of $g(x+y)\geq 2g(x)$. This implies $g(x)$ in unbounded, which is impossible, Hence $g\equiv 0$, or $f(x)=2x$ for all $x$.

The answer is $\boxed{2x}$, which works. Denote the given assertion as $P(x,y)$. We begin with a claim: Claim 1: We have $f(x) > x$ for all $x \in \mathbb{R}^+$ Proof: If $f(x) = x$, we have $f(y)=0$, a contradiction. So, suppose that $f(x)<x$. Then, $P(x-f(x),x)$ yields \[f(2x-f(x)) = 0,\] another contradiction. Hence, the only option is for $f(x)>x$ and this applies for all $x$ in the domain of $f$. $\square$ Let $g(x) = f(x)-x$, which is well-defined due to Claim 1. The given equation becomes \[g(x+y+g(y)) = g(x+y)+y,\] which becomes \[g(a+g(b)) = g(a)+b,\] when we substitute $(a,b) = (x+y,y)$. Denote the above assertion as $Q(a,b)$. Claim 2: $g$ is injective Proof: Suppose that there exist $k_1$ and $k_2$ such that $g(k_1)=g(k_2)$. Comparing $Q(a,k_1)$ and $Q(a,k_2)$ yields \[g(a)+k_1 = g(a+g(k_1)) = g(a+g(k_2)) = g(a)+k_2,\] which implies injectivity. $\square$ Now, plugging in $Q(a,a+b)$ gives \[g(a+g(a+b)) = g(a)+(a+b) = g(a+g(b))+a = g(a+g(a)+g(b)),\] which means $g(a+b) = g(a)+g(b)$ due to Claim 2. Since $g$ is additive and bounded over $(0, \infty)$, we see that $g$ is linear. Plugging this in yields $g(x) = x, -x$, but the latter is absurd.

Note that \begin{align*} f(x+f(y+f(z))) & = f(x+y+f(z))+ f(y+f(z)) \\ \ f(x+f(y+z)+f(z)) & =f(x+y+z)+f(z)+f(y+z)+f(z) \\ \ f(x+z+f(y+z))+f(z) & =f(x+y+z)+f(y+z)+2f(z) \\ \ f(x+z+y+z)+f(y+z) & =f(x+y+z)+f(y+z)+f(z) \\ f(x+y+z+z) & =f(x+y+z)+f(z).\end{align*}Thus it follows that $f(t+z)=f(t)+f(z)$ for all $t\neq z$. Now note that $f(2f(y))=2f(y)+f(2y)$ which we get by putting $x=f(y)$ in the equation. Now pick $x>\max(f(y),y)$. Then $$f(x+f(y))=f(x+y)+f(y)\implies f(x)+f(f(y))=f(x)+f(y)+f(y) \implies f(f(y))=2f(y).$$So we get $$2f(y)+f(2y)=f(2f(y))=f(f(f(y)))=2f(f(y))=4f(y)$$which implies that $f(2y)=2f(y)$. Thus, $f(x+y)=f(x)+f(y)$ holds for all $x,y>0$. And it is well-known that in this case we must have $f(x)=cx$. Checking this in the equation we see that only $f(x)=2x$ for all $x$ or $f(x)=0$ for all $x$ works.

Solved with OronSH. We claim that the only solution is $f(x) = 2x$. From trying to set the LHS argument equal to each of the two RHS arguments, we get that $f(x) > x$ for all $x$. Let $g : \mathbb R_{> 0} \to \mathbb R_{> 0}$ be defined as $g(x) = f(x) - x$, so that our FE becomes \[g(z+g(y)) = g(z)+y, \qquad ( \diamondsuit )\]where $z=x+y > y$. By varying $y$, we have injectivity of $g$. By iterating the above equation, we get \[g(z + ng(y)) = g(z) + ny. \]So, we can vary the variables while fixing $ny$ to get, due to injectivity, that \[z+ng(y)=z+n'g(y') \implies ng(y) = n'g(y')\]whenever $ny = n'y'$. So, $g(x)$ is linear over $k \mathbb Q$ for all positive reals $k$, giving us "quasi-additivity". Claim: We have $g(g(x)) = x$ for all $x$. Proof: Taking $z = ng(y)$ for sufficiently large $n$ (we just need $z > y$, as always) and using our quasi-additivity gives us \[g(g(y)) = g((n+1)g(y)) - g(ng(y)) = y.\] Claim: We have $g(y) = y$ for all $y$. Proof: From wrapping $(\diamondsuit)$ in $g$'s we have \begin{align*}z+g(y) &= g(g(z)+y) \\ g(z)+g(y) &= g(z+y) \end{align*}(with $z > y$, as always). Combining this with an analogous manipulation to $(\diamondsuit )$ gives us that \[g(z+y) = g(z)+g(y)\]for all $z \neq g(y)$. Going back to $(\diamondsuit)$ and setting $z=x+y$, we get \[g(x+y+g(y))=g(x+y)+y.\]Now setting $y=g(y)$ we get \[g(x+g(y))+g(y)=g(x+g(y)+y)=g(x+y+g(y))=g(x+y)+y.\] Now setting $x=y$ gives \[g(y+g(y))=y+g(2y)-g(y).\]By quasi-additivity, $g(2y)=2g(y)$ so $g(y+g(y))=y+g(y)$, so $g$ is additive for $z = g(y)$ as well. Thus $g$ is additive and bounded, so $g(x)=cx$ by Cauchy. Checking, we see $c=1$, so $g(x)=x$ and $f(x)=2x$.

Claim : $f(x)>x$ Proof: Suppose that $f(x_0)<x_0$ for some $x_0$, proving this is not possible suffices as if $f(x)=x$ we get: \[ f(2x)=f(2x)+f(x) \]However as $f(x)>0$ this is not possible. Let $P(x, y)$ denote the assertion: $ f\left(x + f\left(y\right)\right) = f\left(x + y\right) + f\left(y\right)$ Let $x_0-f(x_0)=c$. $P(x_0-f(x_0), x_0)$ \[f(x_0)=f(2x_0-f(x_0))+f(x_0)\]This is a contradiction so $f(x)>x$. Claim : If we let $g(x)=f(x)-x$ we get $g(x)=x$. Proof: Clearly $ g: \mathbb{R}^{ + }\to\mathbb{R}^{ + } $. Let $Q(x, y)$ denote: \[g(x+g(y))=g(x)+y\]This is equivalent to $g(x+y+g(y))=g(x+y)+y$ as we can let $x=x-y$. First I will show $g(x)\geq x$. Suppose $g(y_0)<y_0$. $Q(y_0-g(y_0), y_0)$ \[g(y_0)=g(y_0-g(y_0))+y\]Thus we must have $g(y)\geq y$. Now suppose $g(x_0)=x_0+c$ and $g(y_0)=y_0+k$ for $c>k$. $Q(y_0, x_0)$ \[g(x_0+y_0+c)=x_0+y_0+k\]However as $k<c$ this is a contradiction, thus $k=c$ and as $g(x_0+y_0+c)=x_0+y_0+c$ we also must have $c=0$. Thus $g(x)=x$ for all $x$. Thus $f(x)=2x$ for all $x$ which clearly works.

Pick $x = y - f(y)$ to get $f(2y - f(y)) = 0,$ which is a impossible, implying $f(y) \geq y.$ Define $g\equiv f - x\colon\mathbb R_{>0}\to\mathbb R_{>0}.$ We get $g(x + y + g(y)) = g(x + y) + y.$ We also get $g(x + y) = g(g(x) + g(y)).$ Now, assume $g$ isn't injective. Let $g(a) = g(b) = c.$ Taking $(x, a)$ into our first equation, $g(x + a + g(a)) = g(x + a) + a,$ while we take $(x + a - b, b)$ to get $g(x + a + g(b)) = g(x + a) + b,$ which tells us that $a = b.$ Thus, we conclude $x + y = g(x) + g(y).$ We have \begin{align*} a + b & = g(a) + g(b) \\ a + c & = g(a) + g(c) \\ c + b & = g(c) + g(b) \end{align*}Adding the first two and then subtracting the last, we get $2a = 2g(a),$ or $g(a) = a,$ or $\boxed{f\equiv 2x}.$