Find all functions of $\displaystyle R\to R$ with the property
$\displaystyle f(xf(x) + f(y)) = (f(x))^2 + y$
I might be barking up the wrong tree here but can't you apply $\displaystyle f^{-1}$ to both sides and get $\displaystyle x f(x) + f(y) = x f(x) + f^{-1}(y)$ .....?
In which case the problem boils down to finding all functions such that $\displaystyle f(y) = f^{-1}(y)$, that is, functions which are their own inverse ......
Such functions are called involutions.
$\displaystyle
f(xf(x) + f(y)) = (f(x))^2 + y
$
$\displaystyle f(xf(x)+f(y))-(f(x))^2=y$
$\displaystyle (\sqrt{f(xf(x)+f(y)}+f(x))(\sqrt{f(xf(x)+f(y))}-f(x)=y$
Does this help in any way? I was thinking that somehow I could get the $\displaystyle f(xf(x)+f(y))$ to cancel out but I couldn't get it to work.
The RHS can be multiplied by$\displaystyle \frac{\sqrt{f(xf(x)+f(y))}-f(x)}{\sqrt{f(xf(x)+f(y)}-f(x)}$ to give:
$\displaystyle \sqrt{f(xf(x)+f(y)}+f(x)=\frac{y(\sqrt{f(xf(x)+f(y ))}-f(x))}{f(xf(x)+f(y))-(f(x))^2}$
but I don't think that's helping.
It's an interesting question. I hope someone here can solve it!
EDIT: Perhaps you could try case analysis. When x=0:
[MATh]f(f(y))=(f(0))^2+y[/tex]
f(0) has an unknown value so it might not help. However, we have successfully eliminated a term which is beneficial!
1) $\displaystyle f$ is onto. (it's actually one-to-one as well, but we don't need that!)
Proof: let $\displaystyle z \in \mathbb{R}.$ choose $\displaystyle y=z-(f(0))^2.$ then: $\displaystyle f(f(y))=(f(0))^2+y=z. \ \ \Box$
2) $\displaystyle f(0)=0.$
Proof: by 1) there exists $\displaystyle x \in \mathbb{R}: \ f(x)=0.$ thus: $\displaystyle x=x+(f(x))^2=f(xf(x)+f(x))=f(0)=f(f(x))=(f(0))^2+x .$ thus: $\displaystyle f(0)=0. \ \ \Box$
3) $\displaystyle f(x)=\pm x.$
Proof: we have: $\displaystyle f(f(x))=(f(0))^2+x=x.$ thus: $\displaystyle x^2=(f(f(x)))^2=f(f(x)f(f(x))+f(0))=f(xf(x))=(f(x) )^2.$ thus: $\displaystyle f(x)=\pm x. \ \ \Box$