Mehrshad wrote: Find all functions $f:\mathbb{R}^+\to\mathbb{R}^+$ such that for all $x,y,z\in\mathbb{R}^+$ $$f(x+f(y)+f(f(z)))=z+f(y+f(x))$$ Let $P(x,y)$ be the assertion $f(x+f(y)+f(f(z)))=z+f(y+f(x))$ Let $c=f(1)$ Looking at $f(z)$, we immediately get that $f(x)$ is injective Comparing $P(x,f(y),z)$ and $P(y,f(x),z)$ and using injectivity, we get $x+f(f(y))=y+f(f(x))$ And so $f(f(x))=x+a$ for some $a\ge 0$ So $P(1,1,x)$ becomes $f(x+a+c+1)=x+f(c+1)$ And so $f(x)=x+u$ $\forall x>m$ for some $m=a+c+1$ and $u=f(c+1)-a-c-1$ Setting $x$ great enough to have $x>m$ and $x+u>m$, $P(x,y,z)$ becomes $x+f(y)+f(f(z))+u=z+y+x+2u$ And so $f(y)=y+u-a$ $\forall y>0$ Plugging this in original equation, we get $\boxed{f(x)=x\quad\forall x>0}$, which indeed fits

Very Nice problem ! It is very easy to see $f$ is injective Let $P(x,y,z) : f(x+f(y)+f(f(z)))=z+f(y+f(x))$ $P(f(f(x)),f(y),z) : f(f(x)+f(f(y))+f(f(z)))=z+f(f(y)+f(f(x)))$ ( ) $P(f(f(y)),f(x),z) : f(f(x)+f(f(y))+f(f(z)))=z+f(f(x)+f(f(y)))$ ( ) By ( ) and ( ) we undrestand : $f(f(y)+f(f(x)))=f(f(x)+f(f(y)))$ $\Leftrightarrow$ $f(y)+f(f(x))+f(x)+f(f(y))$ $P(f(x),y,z) : f(f(x)+f(y)+f(f(z)))=z+f(y+f(f(x))$ ( ) $P(f(y),x,z) : f(f(y)+f(x)+f(f(z)))=z+f(x+f(f(y))$ ( ) By ( ) and ( ) we undrestand $f(f(x))+y=f(f(y))+x$ Now we know $f(f(x))+y=f(f(y))+x$ and $f(y)+f(f(x))+f(x)+f(f(y))$ . If we subtract these two terms, the result obtained is equal to the following term $f(y)-y=f(x)-x$ . then we put $y=1$ and we can see $f(x)=x+c$ is solution ! . From the insertion in the main equation, we understand that the function of the problem is the function $f(x)=x$

Clearly $f$ is bijective. Setting $y=f(y)$ implies, $$f(x+f(f(y))+f(f(z)))=z+f(f(y)+f(x)) \qquad(\dagger).$$Using the symmetry of RHS and injectivity we get $f(f(y))=y+k.$ Similarly, setting $x=f(f(x))$ in $(\dagger)$ implies $$f(f(f(x))+f(f(y))+f(f(z)))=z+f(f(y)+f(f(x)))$$where we have a symmetric LHS so that $m+f(y)=f(f(y))=y+k.$ Checking we get $m=k$; the identity works.

Are you sure? Setting $x\to f(f(x))$ gives $f(f(y)+f(f(f(x))))$ in the RHS

$ P(1,1,z) \rightarrow f $ is injective ang surjective $ P(x,f(y),z) , P(y,f(x),z) $ we find that $ f(f(y))+x=f(f(x))+y $ $ \rightarrow f(f(x))= x+a ...(1) $ from $ (1) $ we find that $ f(f(f(x)))=f(x)+a=f(x+a) $ $ P(x,f(y),z) \rightarrow f(x+y+z+2a) = z+f(f(y)+f(x)) $ $ \rightarrow f(x+y+z)+2a = z +f(f(x)+f(y)) $ $ z=a \rightarrow f(x+y)+3a=a+f(f(x)+f(y)) $ $ P(f(x),f(y)) \rightarrow f(f(x)+f(y))+2a = f(x+y+2a)=f(x+y)+2a $so we have $ f(x+y)=f(f(x)+f(y)) $ ,$ a=0$ , $ff(x)=x$ then $ x+y = f(x)+f(y) \rightarrow f(x)+y =x+f(y) $ then $ f(x)=x+c $ and from $ff(x)=x $ we find $f(x)=x$ #

obviously, f is injective $P(x,f(y),z),P(y,f(x),z)\Rightarrow f(f(x))=x+c $ $$\Rightarrow f(x+f(y)+z+c)=z+f(y+f(x)):Q(x,y)$$$$\begin{cases}Q(x,t+r+c,z):f(x+f(t+r+c)+z+c)=z+f(t+r+c+f(x))\\Q(r,x,t):z+f(t+r+c+f(x))=z+t+f(x+f(r))\\Q(r,x,z+t):z+t+f(x+f(r))=f(z+t+r+c+f(x))\end{cases}$$and because of injectivity we have $x+f(t+r+c)=t+r+f(x)$ and if we fix $t+r$ we get $f(x)=x+\alpha$ and if we plug this, we have $f(x)=x(\forall x)$

pco wrote: Mehrshad wrote: Find all functions $f:\mathbb{R}^+\to\mathbb{R}^+$ such that for all $x,y,z\in\mathbb{R}^+$ $$f(x+f(y)+f(f(z)))=z+f(y+f(x))$$ Let $P(x,y)$ be the assertion $f(x+f(y)+f(f(z)))=z+f(y+f(x))$ Let $c=f(1)$ Looking at $f(z)$, we immediately get that $f(x)$ is injective Comparing $P(x,f(y),z)$ and $P(y,f(x),z)$ and using injectivity, we get $x+f(f(y))=y+f(f(x))$ And so $f(f(x))=x+a$ for some $a\ge 0$ So $P(1,1,x)$ becomes $f(x+a+c+1)=x+f(c+1)$ And so $f(x)=x+u$ $\forall x>m$ for some $m=a+c+1$ and $u=f(c+1)-a-c-1$ Setting $x$ great enough to have $x>m$ and $x+u>m$, $P(x,y,z)$ becomes $x+f(y)+f(f(z))+u=z+y+x+2u$ And so $f(y)=y+u-a$ $\forall y>0$ Plugging this in original equation, we get $\boxed{f(x)=x\quad\forall x>0}$, which indeed fits I don't understand why $f$ is injective and how you find it by looking at $f(z)$ I am a noob and beginner, kindly describe it to me...

lian_the_noob12 wrote: I don't understand why $f$ is injective and how you find it by looking at $f(z)$ I am a noob and beginner, kindly describe it to me... Fix $x$, $y$ to $x_0$, $y_0$. So \begin{align*} z_1+f(y_0+f(x_0))&=f(x_0+f(y_0)+f(f(z_1))) \\ &=f(x_0+f(y_0)+f(f(z_2))) \\ &=z_2+f(y_0+f(x_0)) \\ \implies z_1&=z_2. \end{align*}So $f$ is injective.

Mehrshad wrote: Find all functions $f:\mathbb{R}^+\to\mathbb{R}^+$ such that for all $x,y,z\in\mathbb{R}^+$ $$f(x+f(y)+f(f(z)))=z+f(y+f(x))$$ I have solution but i'm not sure of this solution because it's weird Firstly, we begin by claiming $f$ is injective, which is easy. Then, we claim the following : Claim : $f(x+f(y)) = f(y+f(x))$ for all $x,y \in \mathbb{R}^+$. Proof : Assume there exist some $x,y\in \mathbb{R}^+$ where $f(x+f(y))$ and $f(y+f(x))$ are distinct. WLOG $f(x+f(y)) > f(y+f(x))$. Now, we assert $z = f(x+f(y)) - f(y+f(x))$ to get \begin{align*} f(x+f(y)+f(f(z))) &= z + f(y+f(x))\\ f(x+f(y)+f(f(z))) &= f(x+f(y))\\ x+f(y)+f(f(z)) &= x+f(y)\\ f(f(z)) &= 0 \end{align*}Which is a contradiction, because $f(f(z))>0$. Claim proven Hence $f(f(x)+y) = f(f(y)+x)\implies f(x)+y = f(y)+x \implies f(x)-x = f(y)-y$ for all positive reals $x,y$ So $f(x) - x$ is constant, let $f(x)-x=k$ for all pos real $x$. Substituting $f(x)=x+k$ into the equation gives $k=0$. So, the solution is $f(x)=x \; \forall \; x \in \mathbb{R}^+$

wait I thought this problem was not that hard imma write a sol for this EDIT: ohmygodiwassowrong