# Solution(s) to $f(x+f(y))+f(y+f(x))=2f(f(x))f(f(y))$

I would like to know the solutions of the functional equation:

$$f(x+f(y))+f(y+f(x))=2f(f(x))f(f(y)), \forall x,y\in\mathbb{R}$$

where $f:\mathbb{R}\rightarrow\mathbb{R}$. I have already determined that $f\equiv 0$ is a solution, and I would not be surprised if it was the only solution, but I have been unsuccessful in both proving this and finding a counterexample. Letting $a=f(0)$, I know that any other solution must satisfy

• $f(a)=1$
• $f(f(x))=f(x+a)$ (and more generally $f^{n+1}(x)=f(x+na)$)
• $f(x+f(x))=f(f(x))^2=f(x+a)^2$ (and hence $f(x+f(x))>0$)
• $f(2a)=f(1)$ (and more generally $f((n+2)a)=f(na+1)=f^{n+1}(1)$)
• $f(3a)=f(a+1)=f(1)^2$
• $f(2f(x))=f(f(x+a)^2)$

However, I cannot see how to derive a reasonable description of another solution/a proof that such a solution is impossible from this information. I'd appreciate some help with this. Thanks!

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$f(a) = 0$ is another possibility. – Andrew Salmon Feb 27 '13 at 7:11
If $f(a)=0$, then $f\equiv 0$, which is why I did not mention the possibility in my description of a non-zero solution. If $f(a)=0$, then $f(f(x))=-f(x+a)$. We must also have $f(a)=a^2=0$, so $a=0$. Then $f(f(x))=-f(x)$, so $f(f(f(x)))=-f(f(x))=f(x)=f(-f(x))$. For every $p$ in the image of $f$ or the negative of the image, we must have $f(p)=-p$. However, when we let both $x$ and $y$ in the original equation equal $p$, we get $p=0$, so the range of $f$ is just $\{0\}$ and $f\equiv 0$. – rayradjr Feb 27 '13 at 7:51
$f(x)=1 \, \forall x \in \mathbb{R}$ works – Ivan Loh Feb 27 '13 at 11:13

Partial answer when the function have a root. Assume that there is $x_0\in \mathbb{R}$ such that $f(x_0)=0$. Set $y=x:=x_0$ in functional equation; implies that $f(0)^2=0$ i.e., $f(0)=0$. Set $y:=0$, we get that $$f(f(x))=-f(x)$$ Set $x:=f(x)$ and $y:=f(y)$ and also with recent relation we get that $$f(f(x)-f(y))+f(f(y)-f(x))=2f(x)f(y)$$ now in last relation, set $y:=x$ and conclude that $$f(x)=0.$$