Only a Mathematician Can Solve This Equation

[Perfectly Innocent]

This post is a sincere request for help. I assume that the following problem can be represented by a differential equation and that only a mathematician can solve it.

Let A={real numbers x such that |x|<1}
Let S={real numbers Z such that 1 < Z < infinity}

Define ^ on A by the rule x^y = (x+y)/(1+xy).
It follows that (A, ^) is an Abelian group:

x^y=y^x
x^0=x
x^(-x)=0
x^(y^z)=(x^y)^z

I'm looking for a function from AxS->S (also written ^) such that, for any x, y, in A and any Z in S:

x^(y^Z) = (x^y)^Z

0^Z=Z
x^Z > Z if x>0
x^Z < Z if x<0

Thanks

[AKG]

This post is a sincere request for help. I assume that the following problem can be represented by a differential equation and that only a mathematician can solve it.

Let A={real numbers x such that |x|<1}
Let S={real numbers Z such that 1 < Z < infinity}

Define ^ on A by the rule x^y = (x+y)/(1+xy).
It follows that (A, ^) is an Abelian group:

x^y=y^x
x^0=x
x^(-x)=0
x^(y^z)=(x^y)^z

I'm looking for a function from AxS->S (also written ^) such that, for any x, y, in A and any Z in S:

x^(y^Z) = (x^y)^Z

0^Z=Z
x^Z > Z if x>0
x^Z < Z if x<0

ThanksI don't have an answer, but I think I have a start:

x^Z = Z + x(Z - 1)

if x = 0, then:
0^Z = Z + 0(Z - 1) = Z, as required

if x < 0, then:
x^Z = Z + x(Z - 1)

since Z > 1, Z-1 > 0, and thus x(Z-1) < 0, therefore x^Z < Z, as required. Also, x must be greater than -1, so Z + x(Z - 1) must be greater than 1, therefore the value of the operation is greater than one, making it an element of S, as required.

It can be shown easily that it works in the case where x > 0.

x^(y^Z)
= x^[Z + y(Z - 1)]
= [Z + y(Z - 1)] + x(Z + y(Z - 1) - 1]
= Z + yZ - y + xZ + xyZ - xy - x

(x^y)^Z
= [(x + y)/(1 + xy)]^Z
= Z + ... (it's not going to work)

At first, I thought it wouldn't work for this condition. Then, after doing it I thought it did work, because for some reason I thought you wanted x^(y^Z) = y^(x^Z). But I figured you wanted associativity, and realized I wasn't showing that, and realized what I was showing wasn't what you wanted. After looking at what you said, I also realized what the first bit was for (defining x^y). If this helps any, then there you go, but it might not. I might think about this some more and see if I can come up with something.