parmenides51 wrote:
Find all functions f:(0,+∞)→R satisfying f(x)−f(x+y)=f(xy)f(x+y) for all x,y>0.
(Austria)
Let P(x,y) be the assertion f(x)−f(x+y)=f(xy)f(x+y)
If f(u)=−1 for some u>0, then P(u,1) implies contradiction
So f(x)≠−1 ∀x>0
So P(x,y) may be written as
New assertion Q(x,y) : f(x+y)=f(x)f(xy)+1
Subtracting Q(x,y) from Q(y,x), we get f(x)(f(yx)+1)=f(y)(f(xy)+1) (1)
Q(x+y,z) ⟹ f(x+y+z)=f(x+y)f(xz+yz)+1 =f(x)f(xy)+1f(xz)f(xy)+1+1
And so f(x+y+z)=f(x)f(xz)+f(xy)+1
Swapping there x,y and subtracting, we get
f(x)f(yz)+f(x)(f(yx)+1)=f(y)f(xz)+f(y)(f(xy)+1)
Using (1) above, we conclude f(x)f(yz)=f(y)f(xz)
Replacing there (x,y,z) by (x,xy,x), we get f(x)f(y)=f(xy)f(1)
If f(1)=0, this implies S1 : f(x)=0∀x>0 which indeed is a solution
If f(1)=c≠0, we get f(xy)=1cf(x)f(y) and f(x)≠0 ∀x>0
Q(x,y) becomes then f(x+y)=f(x)f(y)cf(x)+f(y)
Swapping x,y, we get c=1 and so f(xy)=f(x)f(y) and f(x+y)=f(x)f(y)f(x)+f(y)
This means that 1f(x) is both additive and multiplicative and so is identity (since allzero is here impossible)
And so S2 : f(x)=1x∀x>0 which indeed is a solution.