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I have to find all surjective functions from $R^+$ to $R^+$, which satisfy $$f(f(a)+b)+f(f(b)+a)=f(2a+f(2b))$$

Does this inspire anyone?

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  • $\begingroup$ One is $f(a)=a$ & $f(b)=b$ $\endgroup$ – Atul Mishra Jul 1 '17 at 15:10
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Let's look for injective solutions. Then, since LHS is invariant under interchanging the role of $a$ and $b$, so it has to be RHS. By injectivity (and replacing a and b with half of them) you get $a+f(b)=b+f(a)$ for every $a$ and $b$. If $a \neq b$ you get $\frac{f(a)-f(b)}{a-b}=1$, which means that $f$ is differentiable and $f(x)=x+c,$ $c$ constant. Surjectivity forces $c=0$.

Other solutions would be not injective. I'm still thinking about it. I think it would be comfortable if you also add 0 to domain and codomain. In this case, $f^2(0)=0$, where $f^2(x)=f(f(x))$. I suppose further properties can be shown much easily for $f^2$ than for $f,$ initially. Then you can get properties of $f$ with arguments like $f(x)=f(f(y))$ for some $y$. Hope it will be helpful.

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If you are interested and if it could be helpful, I can write the proof that $f$ solves the equation above for every $x,y>0$ iff $F(x,y)=f(x+f(y))$ solves $$F(x,y)=F(y,x), \, F(2x,2y)=2F(x,y) \, \forall x,y>0$$ and $f$ is surjective iff $F$ is. Note that in that case $\lim F(\frac{x}{2^n},\frac{y}{2^n}) = 0 \, \forall x,y>0,$ but F may have no limit exist at the origin.

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