commutative wrote:

Find all possible functions

which satisfy the following condition:

This reminds me of the formula for tanh(x+y). In fact, let

and

, and let

Then the functional equation (*) says that

Therefore

and so

It is clear from (*) (with x=y=0) that f(0)=0, and hence g(0)=0. Now

*suppose that f* (and hence g)

*is differentiable at 0*, with f'(0)=α (and hence by the chain rule g'(0)=α too). Therefore

.

But then

, from which it follows that g is everywhere differentiable and

That differential equation, with the initial condition g(0)=0, has the solution

. Conclusion:

So, provided that f is differentiable at 0, there is indeed just a one-parameter family of solutions. I haven't thought about how to get from

to

, but that ought to be "just algebra".

**Edit.** I could have done that much more simply. The map τ:x→tanh(x) is an isomorphism from the additive group (

**R**,+) of reals to the group (G,*) (in

**commutative**'s Remark 2). So if f is an endomorphism of (G,*) then

is an endomorphism of (

**R**,+). But it is well known that if such an endomorphism is continuous (or differentiable at 0) then it is of the form x→αx for some real number α. If the map is not continuous then you have to think of it as a real-valued linear functional on

**R** regarded as a vector space over the rationals. In that case, it can be defined arbitrarily on each element of a Hamel basis of the

**Q**-vector space

**R** (provided you're happy to use the Axiom of Choice).