if $f(0)$ is not 0 then you can easily show that $f(x) = 1$ and $f(X) = -1$ are answers. otherwise: $f(0)=0$ let $y=-x$ so $f(-x)=-f(x)$ if $f(u) = 0$ then $f(x+u) = f(x)$ $f(u/2) = -f(-u/2) = -f(u/2) \rightarrow f(u/2) = 0$ we also have $f(2u)=0$ (and also $f(ku)=0$ by induction) so $f(\frac{n}{2^k}u)=0$ for every $n, k \in N$ so $f(x) = 0$ for every $x \in R$ if $f(u)=0$ only for $u=0$ now suppose $f(a) > 1$ so there is $x_0$ for which we have $f(x_0)=1$ now let $x = y = x_0$ so $f(x_0/2) = 1$ and because of continuity $f(0) = 1$ contradiction shows that $|f(x)|<1$ for every $x$ now let $f(x) = \tanh{(g(x))}$ so $g(x+y) = g(x) + g(y)$ so $g(x) = cx$ so $f(x) = tanh(cx)$

mohammadp will you suggest me any book on the techniques of solving functional equation?? because i really do not know the basic themes??

noman wrote: mohammadp will you suggest me any book on the techniques of solving functional equation?? because i really do not know the basic themes?? of course! here are some good pdfs

MohammadP wrote: if $f(0)$ is not 0 then you can easily show that $f(x) = 1$ and $f(X) = -1$ are answers. What we easily show is f(x)²=1. Passing from f(x)²=1 to f(x)=1 for all $x\in \mathbb{R}$ or f(x)=-1 for all $x\in \mathbb{R}$ is not direct. A function that verify f(x)² = 1 could coincide with f(x)=1 on some points, and with f(x)=-1 on others. We should prove that it's not the case, and we may use the hypothesis of continuity.

$f(x)=\frac{a^x+1}{a^x-1},1 or 0$ it is similar to china second round 2006,problem 3.

actually,suppose f non-constant $f(x+y)+1 =\frac{(f(x)+1)(f(y)+1)}{1+f(x)f(y)},f(x+y)-1 =\frac{(f(x)-1)(f(y)-1)}{1+f(x)f(y)}$ so let $\frac{f(x)+1}{f(x)-1}=e^{g(x)}$ then $g(x+y)=g(x)+g(y)$ hence g satisfies Cauchy's function since g is continuous hence $g(x)=cx$.

@littletush sorry but should we show that function $e^{g(x)}$ is exist? if we should , show me please . Thanks!

Just check $g(x=cx$ works.

anyone have an official solution ? This problem is so nice and classic !

There has just to be some changes to the solution of Mahamad to be correct: If $f(0)$ is not 0, then $P(0,0)$ gives $1+f(0)^2=2.$ $f(0)=1,-1$ $P(0,x)$ gives $f(x)= \pm 1$ each time and so by continuity we get $f(x)=1$ or $f(x)=-1.$ If $f(0)=0$ $P(x,-x)$ gives $f(-x)=-f(x)$ if $f(u) = 0$ with $u\ not= 0$ then $f(x+u) = f(x)$ $f(u/2) = -f(-u/2) = -f(u/2) \rightarrow f(u/2) = 0$ we also have $f(2u)=0$ (and also $f(nu)=0$ by induction) so $f(\frac{n}{2^k}u)=0$ for every $n, k \in N$ so $f(x) = 0$ for every $x \in R.$ if $f(u)=0$ only for $u=0$ now suppose there exist an $a$ such that $|f(a) \ge 1$ so there is $x_0$ for which we have $f(x_0)=1$ now let $x = y = 0.5 x_0$ so $f(x_0/2) = 1$ by $[f(0.5x_0)-1]^2=0$ and because of continuity $f(0) = 1$ or $f(0)=-1$ by the same argument. So $|f(x)|<1$ for every $x$ now let $f(x) = \tanh{(g(x))}$ (this may be done, by the domain of tanh) so $g(x+y) = g(x) + g(y)$ so $g(x) = cx$ so $f(x) = tanh(cx)$ (see the way of Litletush to finish this and so the compilation gave the solution)

SCP wrote: so $f(\frac{n}{2^k}u)=0$ for every $n, k \in N$ so $f(x) = 0$ for every $x \in R.$ How does the second line follow from then first? Could someone please offer a rigorous proof of this part? Thanks.

so $f(\frac{n}{2^k}u)=0$ can reach each value quitenear and so by continuity it follows

SCP wrote: so $f(\frac{n}{2^k}u)=0$ can reach each value quitenear and so by continuity it follows That's the intuition I get, but I would like a rigorous proof of why it follows.

$ f(x)=0 $ can be found (if $ x=u\not= 0 $) by plugging in $ u=\frac{x2^{k}}{n} $ into $ f(\frac{n}{2^{k}}u)=0 $.